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Modern Physics for Scientists and EngineersInstructors’ Solutions

Joseph N. Burchett

January 21, 2011

Introduction - Solutions

1. We are given a spherical distribution of charge and asked to calculate thepotential energy of a point charge due to this distribution. There are twosituations to consider which deliver quite different results. The problem doesnot specify whether the point charge of interest is inside or outside the chargedistribution so we must take both possibilities into consideration. Before weget started, let’s set up an appropriate description of our charge distributionso that we may proceed with each of the above situations. It is most usefulto express the charge distribution in terms of a charge density ρ:

ρ =Q

43πR3

This is the charge per unit volume as found by merely dividing the totalcharge Q by the total volume of the sphere. We now handle the “outside thesphere” configuration. All of the charge inside the sphere may be consideredas though it is concentrated at a single point. We may then use the Coulomb’sLaw result for the force acting on the point charge q:

F =1

4πε0

Qq

r2

We integrate to find the potential energy:

V = − Qq

4πε0

∫ r

∞

1

r′2dx =

Qq

4πε0

1

r

This result is also identical to the one obtained in the text in the case of thetwo-point charge configuration. We see that this should in fact be the casesince we started with the same quantity of force.

Now, to handle the “inside the sphere” situation, we shall use the abovestated charge density. We easily see that the distance r from the center of the

3

4

distribution of charge to the point charge is less than the radius R. Imaginenow a sphere of radius r′ that is inside the larger spherical charge distributionbut centered at the same point. Only the charge within this inner sphere willact on the point charge. We may now use our charge density to find thecharge contained in the smaller sphere:

Qin =4

3πr′3ρ

The force on a charge q due to the inner sphere is then given by Coulomb’sLaw:

F =q

4πε0

4πr′3ρ

3r′2=qρr′

3ε0Before we integrate to find the potential energy, we must make an importantdistinction from the “outside the sphere” procedure above. We were able tointegrate directly from infinity to the location of our point charge because theamount of source charge taken into account did not change (Q). However, aswe enter the sphere, the amount of source charge begins to decrease until wereach our point at distance r from the center. We split the integration intotwo parts corresponding to both regions to find the potential energy:

V = −[

∫ R

∞

Qq

4πε0

1

r′2dr′ +

∫ r

R

qρr′

3ε0dr′]

=Qq

4πε0

1

R− ρq

6ε0(r2 −R2)

Note that the first integral above was identical to the one for the outsideconfiguration, just evaluated at r = R. Now, we just substitute our chargedensity ρ and simplify:

V =Qq

4πε0

1

R− Qqr2

8πε0R3+

Qq

8πε0R

=3Qq

8πε0R− Qqr2

8πε0R3

2. First, convert to SI (1 eV = 1.6 × 10−19 J) then calculate speed fromthe definition of kinetic energy:

KE =1

2mv2

5

1.6× 10−19J =1

2(9.109× 10−31kg)v2

v = 5.92× 105 m

s

Multiply by the mass of the electron (from Appendix A) to find the momen-tum:

p = (9.109× 10−31kg) · (5.92× 105m

s= 5.39× 10−25 kg

m

s

3. Looking at the definition of kinetic energy:

KE =1

2mv2

Let’s solve for velocity:

v =

√2(KE)

m

Let KE ′ = 4KE.

v =

√2(KE ′)

m=

√2(4KE)

m= 2

√2(KE)

m

= 2v

So, the speed will increase by a factor of 2 if the kinetic energy is increasedby a factor of 4. By the definition of momentum:

p = mv ⇒ m(2v) = 2p

We should also expect the momentum of the particle above to increase bya factor of 2. In fact, as long as the mass does not change, the momentumshould increase or decrease by the same factor as the changing speed.

4. We see that the given function is identidcal in form to that of I.19,comparing the two:

ψ(x, t) = Ae× 10i(kx−ωt) = Ae× 10(αx+βt)

6

Therefore, we may relate α and β to their physical parameters k and ω by:k = α and ω = −β. FRom (I.8), k = 2π/λ and:

λ =2π

α

5. The frequency and wavelength of light are related by equation I.23:

f =c

λ

The constant c refers to the speed of light which we know to be 3.00×108 m/s.Taking care to convert our wavelength, which is given in nanometers, tometers:

f =3.00× 108m

s

500× 10−9m= 6× 1014 Hz

6. We may use (I.6) which relates the wavelength and frequency via thevelocity of the wave (the speed of light):

λf = v ⇒ λ(600× 1012s−1) = 3× 108 m/s

λ = 5× 10−7 m

7. Equation I.25 relates the photon energy to the wavelength:

E =hc

λ

Substituting the value hc = 1240 eV ·nm:

E =1240 eV ·nm

500nm= 2.48 eV

8. I.25 relates the wavelength and energy:

E =hc

λ⇒

7

λ =hc

E=

1240 eV · nm5.4 eV

= 229.6 nm

9. The energy of a quantum of light (the photon) must be equal to thedifference in energies of the two states:

Ephoton = E2 − E1

According to equation I-25, the energy also obeys the following:

Ephoton =hc

λ

Equating the two and solving for λ, we get:

λ =hc

E2 − E1

10. Once again, we use equation I.25 , however to ensure the correct unitsfor the result, we convert 1A = .1 nm.

E =1240 eV · nm

.12 nm= 10333 eV

1

The Wave-Particle Duality -Solutions

1. The energy of photons in terms of the wavelength of light is given by eq.1.5. Substituting the given wavelength into eq. 1.5:

Ephoton =hc

λ=

1240 eV · nm200 nm

= 6.2 eV

2. The 100 Watt beam of light will carry 100 J of energy every second.We then want to find the number of 200 nm photons required to deliver 100J of energy. The energy per photon in electron volts is given by (1.5):

E =1240 eV · nm

200 nm= 6.2 eV

Now, convert the energy to Joules:

E = 6.2 eV × 1.6× 10−19 J

1 eV= 9.92× 10−19 J

Then, find the number of photons:

100 J

9.92× 10−19J/photon= 1.0× 1020 photons

9

10 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS

3. Considering that we are given the power of the laser in units of milli-watts, the power maybe expressed as: 1 mW = 1 × 10−3J/s. This gives usa strong hint how to proceed. We are hoping to find the rate of emissionof photons, so that if we can find the energy of a single photon, we can usethe power as stated above to calculate the number of photons. The energyof a single photon can be calculated as in Problem 1 by substituting thewavelength of the light into equation 1.5:

Ephoton =hc

λ=

1240 eV · nm632.8 nm

= 1.960 eV

We now convert to SI units:

1.960 eV × 1 J

1.6× 10−19 eV= 3.14× 10−19 J

Now, using the given laser power:

Rate of emission =power

energy per photon=

1× 10−3 Js·photon

3.14× 10−19J= 3.18×1015 photons

s

4. The maximum kinetic energy of emitted photoelectrons may be calcu-lated via (1.6) and work functions of the metals may be found in table 1.1:

a)Na: (KE)max =hc

λ−W

=1240 eV · nm

200 nm− 2.28 eV

= 3.92 eV

b)Al: (KE)max =1240 eV · nm

200 nm− 4.08 eV

= 2.12 eV

c)Ag: (KE)max =1240 eV · nm

200 nm− 4.73 eV

= 1.47 eV

11

5. This problem concerns the photoelectric effect. Given the work functionof the material and the emitted electron kinetic energy, we wish to calculatethe wavelength of the light incident to the surface. Equation 1.6 providesthe following:

(KE)max =hc

λ−W where W is the work function of the material.

The hc/λ term describes the energy supplied by the incoming photons. Byviewing the work function as an energy threshold for producing photoelectriccurrent, we see that the amount by which the photon energy (hc/λ) exceedsthe work function is the resultant maximum kinetic energy of the emittedelectrons. We may thus write:

hc

λ= (KE)max +W

= 2.3 eV + 0.9 eV

= 3.2 eV

Using hc = 1240 eV · nm,

λ =1240 eV · nm

3.2 eV= 387.5 nm

6. Here, we are given the stopping potential of a photoelectric experimentand wish to determine the work function of the metal. Since 0.72 eV is thenecessary potential energy to cease the flow of electrons, the maximum ki-netic energy of the electrons being emitted must equal 0.72 eV. Our problemis then reduced to solving eq. 1.6 for the work function:

W =hc

λ− (KE)max =

1240 eV · nm460 nm

− 0.72 eV = 1.98 eV

7. Given that the work function from problem 6 is 1.98 eV, we find themaximum kinetic energy of the photoelectrons by (1.6):

(KE)max :1240 eV · nm

240 nm− 1.98 eV = 3.19 eV


Therefore a stopping potential of 3.19 eV must be applied as a stoppingpotential to prohibit the photoelectrons from reaching the anode. This cor-responds to the increase:

3.19 eV − 0.72 eV = 2.47 eV

8. Section 1.1.1 gives the equation for just this situation:

W =hc

λ0

= 3.44 eV

9. Given that the work function for sodium is 2.28 eV , we find the maximumkinetic energy of the emitted electrons by (1.6):

hf = 6.626× 10−34 J · s× 1200× 1012s−1

= 7.951× 10−19 J

= 4.97 eV

(KE)max = hf − 2.28 eV

= 4.97 eV − 2.28 eV

= 2.69 eV

10. Equation 1.7 is the expression for transitions where m = 2. Substi-tuting into (1.8),

1

λ= R

(1

4− 1

n2

)and thus:

1

λ= 1.0972× 105cm−1

(1

4− 1

n2

)Taking the reciprocal:

λ = 9.114× 10−6

(1

4− 1

n2

)−1

13

We may then rewrite,1

4− 1

n2=n2 − 4

4n2

And we then get (1.7):

λn = 36456n2

n2 − 4× 10−8 cm

The equations are equivalent.

11. For this atomic transition, the energy of the emitted photon must equalthe difference in energy of the two states of hydrogen (n = 2 and n = 5).Equation 1.22 gives us those energies, thus:

Ephoton = E5 − E2 = −13.6 eV

52− (−13.6 eV

22) = 2.86 eV

From eq. 1.12:

λ =hc

∆E=

1240 eV · nm2.86 eV

= 434.2 nm

12. We may use (1.7) to find the wavelength of the photons, then use (1.12)to find their corresponding energies.

λ = 3645.642

42 − 4× 10−8cm

= 4860.8× 10−8cm

= 486.08 nm

E =hc

λ= 2.55 eV

13. Using 1.7:

λ = 3645.632

32 − 4× 10−8 cm

= 6.562× 10−5 cm


14. From figure 1.6, we see that the energy of an n=2 electron is -3.4 eV.Therefore, 3.4 eV is the required energy to just ionize the atom. Now, use(1.12) to find the wavelength:

λ =hc

∆E

=1240 eV · nm

3.4 eV= 364.7 nm

15. We must first find the energy of the UV photons, which may be donevia eq. 1.5:

Ephoton =hc

λ=

1240 eV · nm45 nm

= 27.6 eV

If electrons are to be emitted, we need to overcome the energy that is keepingthem in a bound state. For hydrogen atoms, equation 1.22 gives us thatenergy. We need only supply the n quantum number for the state of theatom (for ground state, n = 1):

E = −13.6 eV

n2= −13.6 eV

12= −13.6 eV

The kinetic energy of these emitted electrons should then be equal to thedifference between the energy provided by the incident light and the groundstate electron energy: KE = 27.6 eV − 13.6 eV = 14.0 eV For the velocitycalculation, it will be useful to convert from eV to J:

14.0 eV · 1.6× 10−19J

1 eV= 2.24× 10−18J (1.1)

We can then find the velocity using the definition of kinetic energy andthe mass of the electron:

KE =1

2mV 2 ⇒ v =

√2(KE)

m=

√2.24× 10−18J

9.11× 10−31kg= 2.21× 106 m

s(1.2)

15

16. We may use (1.8) where each transition will have its own values ofm and n:

a.) m=1, n=2 :1

λ= 1.0972× 105 cm−1

(1

12− 1

22

)λ = 1.215× 10−5 cm

b.) m=2, n=5 :1

λ= 1.0972× 105 cm−1

(1

22− 1

52

)λ = 4.340× 10−5 cm

c.) m=3, n=5 :1

λ= 1.0972× 105 cm−1

(1

32− 1

52

)λ = 1.282× 10−4 cm

17. It should be first noted that the wavelength of the emitted light increasesas the photon energy decreases. Thus, the longest acceptable wavelengthwill correspond to a transition to the nearest excited state. Transitions tohigher-lying states will require more energy and thus correspond to a shorterwavelength. We then examine the energy of the transition from the n = 3state of the hydrogen atom. Equation 1.22 gives:

E3 = −13.6 eV

n2= −13.6 eV

32= −1.51 eV

The next available state is the state corresponding to n=4 whose energy maybe found by eq. 1.22 again:

E4 = −13.6 eV

n2= −13.6 eV

42= −0.85 eV

So, in order for the absorption to take place, the incident photons mustcontain at least as much energy as the difference in these energies (−0.85 eV −(−1.51 eV ) = .66 eV ). We may now solve eq. 1.5 for the wavelength:

λ =hc

∆E=

1240 eV · nm.66 eV

= 1878 nm

Thus, any light of wavelength greater than 1878 nm would not possess suffi-cient energy to be absorbed by the hydrogen atom.


18. Using (1.26),

p =h

λ=

6.626× 10−34 J · s.2× 10−9 m

= 3.313× 10−24 N · s

The kinetic energy may then be calculated:

KE =p2

2me

= 6.025× 10−18 J

Convert to eV:6.025× 10−18 J = 37.65 eV

19. Using a procedure such as that of Example 1.5, we can relate the kineticenergy to the de Broglie wavelength by the following equation:

λ =h√

2m(KE)

We need only convert the kinetic energy from eV to J and substitute theappropriate masses.

KE = 20 eV × 1.6× 10−19 J

1 eV= 3.2× 10−19J

Electron : λ =6.626× 10−34J ·s√

2(9.11× 10−31kg)(3.2× 10−18J)= 2.74× 10−10m

Proton : λ =6.626× 10−34J ·s√

2(1.67× 10−27kg)(3.2× 10−18J)= 6.41× 10−12m

An α-particle is the most commonly occurring isotope of the helium atomconsisting of two protons and two neutrons. We can thus approximate itsmass: m = 2(1.672× 10−27) + 2(1.674× 10−27) = 6.692× 10−27kg.

α− particle : λ =6.626× 10−34J ·s√

2(6.692× 10−27kg)(3.2× 10−18J)= 3.20× 10−12m

17

20. From (1.12), we may find the corresponding wavelength for 40 keVX-rays:

λ =hc

∆E=

1240 eV · nm40× 103eV

= .031 nm

We then find the momentum possessed by an electron with the same wave-length from (1.25):

p =h

λ=

6.626× 10−34 J · s.031× 10−9 m

= 2.137× 10−23 kgm

s

The kinetic energy of the electron may be calculated as follows:

E =p2

2me

=2.137× 10−23 kg m

s

2(9.109× 10−31kg

= 2.507× 10−16 J

To find the necessary potential for such electrons, we merely convert thisenergy quantity to eV:

2.507× 10−16 J × 1 eV

1.6× 10−19 J= 1566 eV

Therefore, 1566 V would be needed as an accelerating potential difference.

21. We may write the kinetic energy as KE = p2/2m, the momentumas p = ~k (1.27), and k = 2π/λ. Substituting k into (1.27):

p = ~ · 2π

λ=h

λ

Substituting our given wavelength:

p =6.626× 10−34 J · s.01× 10−9 m

= 6.626× 10−23 kgm

s


This is now simply a matter of using the mass of each particle in our firstequation above:

Electron : KE = 15000 eV

Proton : KE = 8.2 eV

Neutron : KE = 8.2 eV

22. Visible light has corresponding wavelength between about 400 nm and700 nm. Therefore, from 1.25:

p (400 nm) =6.626× 10−34 J · s

400× 10−9 m

= 1.657× 10−27 kgm

s

p (700 nm) =6.626× 10−34 J · s

700× 10−9 m

= 9.466× 10−28 kgm

s

We then divide each momentum by the electron mass to obtain the speeds:

λ = 400 nm : v = 1815m

s

λ = 700 nm : v = 1035m

s

23. We first convert to SI and then use E = p2/2m to find the momen-tum:

40, 000 eV = 6.4× 10−15J

p =√

2m(KE)

=√

2(9.109× 10−31 kg)(6.4× 10−15 J

= 1.081× 10−22 kgm

s

19

Since k = 2π/λ, the proton must have the same value of k if it is to havethe same wavelength. From (1.27), p = ~k where k = 2π/λ. Therefore, aproton having the same de Broglie wavelength should also have the samemomentum. We now use the mass of the proton to find its kinetic energy:

KE =(1.081× 10−22 kg m

s)2

2(1.673× 10−27 kg)

= 3.492× 10−18 J

= 21.8 eV

24. The Davisson-Germer experiment measured the scattering of electronsby a crystal during which the interference patterns characteristic of light wereshown to also occur with particles. We can solve Bragg’s Law (eq. 1.24) forthe inter-atom spacing of the crystal in terms of the de Broglie wavelengthand the scattering angle:

2d sin(θ) = nλ⇒ d =nλ

2 sin(θ)

We may use the relation derived in example 1.5 to find the de Broglie wave-length from kinetic energy after first converting our kinetic energy to SIunits:

KE = 54 eV × 1.6× 10−19

1 eV

λ =h√

2m(KE)=

6.626× 10−10m√2(9.11× 10−31kg)(8.64× 10−18 J)

= 1.66× 10−10m

We let n=1 and solve Bragg’s Law for d, the spacing in the crystal:

d =λ

2 sin(θ)=

1.66× 10−10m

2 sin(50)= 1.09× 10−10m

2

The Schrodinger WaveEquation - Solutions

1. Equation 2.17 gives the energies for the particles in an infinite potentialwell:

E =n2h2

2mL2

Part (a) asks for the size of the region, which we can find by solving eq. 2.17for L:

L =

√n2h2

8mE

We need only supply some information such as the energy (1.0 eV as given),the mass of the electron (9.11× 10−31kg from Appendix A), and the appro-priate value of n. The lowest energy will be found when the electron is atthe ground state (n=1). We must also convert 1.0 ev to 1.6× 10−19J . Nowwe are ready:

L =

√12(6.63× 10−34 J ·s)2

8(9.11× 10−31 kg)(1.6× 10−19 J)= 6.14× 10−10 m

While the derivation of eq. 2.17 is correct, let’s do a little dimensional anal-ysis to better “see through” our length calculation above. The quantumnumber n is dimensionless as is of course the constant 8, so let’s examine theoperations of units of the rest of the quantities:√

n2h2

8me⇒

√J2 · s2

kg · J=

√J · s2

kg

21

22 2. THE SCHRODINGER WAVE EQUATION - SOLUTIONS

Now substitute 1 J = 1 kgm2

s2· s2:√

J · s2

kg=

√kgm

2

s2· s2

kg=√m2 = m

It should be emphasized that the m here is the abbreviation for the unit oflength meters, not the quantity of mass.

Part (b) can be approached in a couple of different ways. The essentialquantity we must obtain now is the energy of the first level above groundstate (also called the first excited state). The energy required to excite theelectron must be:

∆E = En2 − En1 = E2 − E1

The quantum numbers n=2 and n=1 correspond to the first excited stateand ground state respectively. So, for E2 from equation 2.17:

E2 =22h2

8mL2=

(6.63× 10−34 J ·s)2

2(9.11× 10−31 kg)(6.14× 10−10 m)2

= 2.14× 10−18 J × 1 eV

1.6× 10−19 J

= 4.00 eV

Here we used the result of the size of the well from part (a). The othermethod would involve solving eq. 2.17 for everything except n and using ourgiven quantities as follows:

E =n2h2

2mL2⇒ h2

8mL2=E

n2

h2

8mL2= (1.0 eV )(12) = 1.0 eV

En = n2(1.0 eV )⇒ E2 = 4.0 eV

Now for our desired result:

∆E = 4.0eV − 1.0eV = 3.0eV

23

2. Equation (2.20 states that the wavefunction for odd n is:

Ψ(x) =

√2

Lcos(nπxL

)=

√2

10 nmcos

(3πx

10 nm

)where − L

2≤ x ≤ L

2

Now, evaluate at the given x-values:

Ψ(0) = .447

Ψ(2) = −.139

Ψ(4) = −.364

Ψ(8) = 0

Ψ(10) = 0

The last two are equal to zero because x > L/2 in each of these cases.

3. As with the atomic transitions we dealt with in chapter 1, the emit-ted photon energy will equal the difference of the energies of the two statescorresponding to n = 3 and n = 2. So our task becomes finding each of theseenergies via equation 2.17:

E3 =32h2

8mL2=

9(6.63× 10−34 J ·s)2

8(9.11× 10−31 kg)(10× 10−9 m)2

= 5.43× 10−21 J · 1 eV

1.6× 10−19 J= .034 eV

E2 =22h2

8mL2=

4(6.63× 10−34 J ·s)2

8(9.11× 10−31 kg)(10× 10−9 m)2

= 2.41× 10−21 J · 1 eV

1.6× 10−19 J= .015 eV

∆E = .034 eV − .015 eV = .019 eV

This value of ∆E will then be equal to the energy of the photon and we mayuse eq. 1.5 to calculate the wavelength of the light:

KE =hc

λ⇒ λ =

hc

KE


λ =1240 eV · nm.019 eV

= 65263 nm

4. Since E = p2/2m, the momentum squared may be written as:

p2 = 2mE

where the energy is given by 2.17:

p2 = 2mn2h2

8mL2

=n2h2

4L2

The average value of a function is found via the method prescribed on page30.

〈f(x)〉 =

∫ ∞−∞

f(x) |Ψ(x)|2 dx

Using (2.20) for Ψ(x),

⟨p2⟩

=n2h2

4L2

∫ L/2

−L/2

2

Lcos2

(nπxL

)dx for odd n

⟨p2⟩

=n2h2

4L2

∫ L/2

−L/2

2

Lsin2

(nπxL

)dx for even n

By a similar argument to that used in example 2.2, the integral over sin2 (nπx/L)is equal to the integral over cos2 (nπx/L) and the average value of the squareof momentum is the same for the even and odd n:

⟨p2⟩

=n2h2

4L2

∫ L/2

−L/2

1

2dx

=n2h2

8L

5. Note the typo, “Draw the wave function...” should be “Find the wavefunction...”

25

We are given the following information: finite well of depth 0.3 eV andthickness L = 10 nm, we are dealing with conduction electrons if GaASin their lowest state, and these electrons have an effective mass of 0.067times the electron mass. From section 2.3, we see the process of solvingthe finite square well which culminates in the numerical solution of tanθ =√

(θ20/θ

2)− 1, for the even solutions, and −cotθ =√

(θ20/θ

2)− 1 for the oddcase. Our variables θ and θ0 correspond to the following expressions:

θ =kL

2, θ0

2 =mV0L

2

2~2(from eq. 2.30)

Notice that the latter of the two is a squared quantity where the former isnot, but we’ll deal with that a little later on. We are interested in the groundstate of the electron, which will correspond to n=1, leading us to an evensolution which may be found by equation 2.29:

tanθ =

√θ0

2

θ2− 1

In the text near the end of section 2.3, the value of θ02 is given to be 13.2.

Our equation becomes:

tanθ =

√13.22

θ2− 1

By using a computer algebra system or calculator to graph the left-hand andright-hand sides, we find the first intersection at θ ≈ 1.46. Thus, we maysolve θ = kL/2 for k:

k =2θ

L=

2(1.46)

10× 10−9 m= 2.92× 108 m−1

For the equations outside the well, we must solve for κ as well. Let’s firstsolve equation 2.24 for E:

k =2mE

~2

2

⇒ E =k2~2

2m=

(2.92× 108 m−1)2(1.054× 10−34 J · s)2

2(6.103× 10−32 kg)

E = 7.760× 10−21 J × 1 eV

1.6× 10−19 J= .049 eV


Now for κ (from eq. 2.26):

κ =

√2(6.103× 10−32 kg)(.3 eV − .049 eV )(1.6× 10−19 J

eV)

(1.054× 10−34 J · s)2= 6.64×108 m−1

We know have our necessary information for the wave equations:

Ψ(x) =

B exp(−(6.64× 108 m−1)x) : x ≤ −5 nmA cos((2.44× 108 m−1)x) : −5 nm ≤ x ≤ 5 nmB exp(−(6.64× 108 m−1)x) : x ≥ 5 nm

6. We may use a similar method to that for problem 5 however we mustfirst calculate a new value for θ2

0 corresponding to the new well depth. By(2.30):

θ20 =

mV0L2

2~2

=.067(9.109× 10−31 kg)(.2 eV )(1.6× 10−19 J/eV )(10× 10−9 m)2

2(1.054× 10−34J · s)= 8.8

Eq (2.29) then becomes:

tanθ =

√8.82

θ2− 1

which yields a solution of θ ≈ 1.41. We then find k, E, and κ:

k =2θ

L= 2.82× 108 m−1

E =k2~2

2m= .045 eV

κ =

[2m(V0 − E)

~2

] 12

= 5.22× 108 m−1

The wave equations are then:

Ψ(x) =

B exp(−(5.22× 108 m−1)x) : x ≤ −5 nmA cos((2.82× 108 m−1)x) : −5 nm ≤ x ≤ 5 nmB exp(−(5.22× 108 m−1)x) : x ≥ 5 nm

27

8. The methodology of example 2.1 holds until we apply the boundaryconditions which are now:

Ψ(0) = 0

Ψ(L) = 0

First examining the even solutions, we apply the boundary conditions:

Ψ(0) = A cos 0 = 0

Ψ(L) = A cos kL = 0

Upon examining Ψ(0), we reject these solutions. Now turning to the oddsolutions, Ψ(x) = B sin(kx), we again apply our boundary conditions:

Ψ(0) = A sin 0 = 0

Ψ(L) = A sin kL = 0

We see that the first of these is automatically satisfied while second is satisfiedprovided that kL = nπ. Therefore,

Ψ(x) = B sin(nπxL

)where 0 ≤ x ≤ L

The energy levels are found from (2.12) where k = nπ/L

E =~2n2π2

2mL2

Comparing these with the results of example 2.1, we see that the wavefunc-tions are all odd while the energy levels remain the same.

9. Outside the finite potential well, V = V0, thus our time-independentSchrodinger equation is:

− ~2

2m

d2ψ

dx2+ V0ψ = Eψ

If we multiply both sides by −2m/~2:

d2ψ

dx2− 2mV0

~2ψ = −2mE

~2ψ


Moving 2mE/~2 to the LHS:

d2ψ

dx2+

2m(E − V0)

~2ψ = 0

We make the following substitution:

k =

(2m(E − V0)

~2

) 12

Since E is greater than V0, k is in fact a real number and our Schrodingerequation becomes:

d2ψ

dx2+ k2ψ = 0

And, we may readily confirm by substitution that the general form of thesolution of this equation is a linear combination of the functions, A cos(kx)and B sin(kx), and is oscillatory in nature.

10. First evaluating the second derivative,

dψ

dx= −Amωx

~e−mωx

2/2~

d2ψ

dx2= −Amω

~e−mωx

2/2~ + Am2ω2x2

~2e−mωx

2/2~

Substituting into the left hand side of (2.33):

−~2

2m

d2ψ

dx2+

1

2mω2x2ψ =

A~ω2

e−mωx2/2~

Setting this equal to the right-hand side,

~ω2Ae−mωx

2/2~ = Eψ

This is in fact satisfied given E = ~ω2

, which is consistent with (2.34) whenn = 0.

11. The normalized wave function must satisfy the normalization condition(equation 2.18): ∫ ∞

−∞|ψ(x)|2 dx = 1

29∫ ∞−∞

Ae−mωx2\2~Ae−mωx

2\2~dx =

∫ ∞−∞

A2e−mωx2\~

Therefore,1

A2=

∫ ∞−∞

e−mωx2\~dx = 2

∫ ∞0

e−mωx2\~dx

This looks very similar to the integral we are given if we let a = mω/~. Then:

1

A2=

√πmω~

=

√π~mω

A =(mωπ~

) 14

12. b.) The normalization condition is given by (2.18). For our givenwavefunction, we need only integrate the piece where x > 0:∫ ∞

0

A2x2e−2ax dx = 1

To solve the integral, use integration by parts twice and solve for A:

1

A2=

1

4a3

A = 2a23

c.) In finding the most probable position of the particle, we maximize theprobability as given by (2.9). This leads to:

|ψ(x)|2 = 4a3x2e−2ax

Setting the derivative equal to zero:

d

dx|ψ(x)|2 = 8a3xe−2ax − 8a4x2e−2ax = 0

Now, solve for x:

x =1

a


d.) By (2.21), the average value of the position of a particle is given by:

< x > =

∫ ∞0

x |ψ(x)|2 dx

=

∫ ∞0

4a3x3e−2ax

After three rounds of integrating by parts:

< x >=3

2a

13. a) From 0 ≤ x ≤ L, V = 0, and our Schrodinger equation becomes:

− ~2

2m

d2ψ

dx2= Eψ or

d2ψ

dx2+

2mE

~2ψ = 0

For x ≥ L, V = V0 and:

− ~2

2m

d2ψ

dx2+ V0 = Eψ or

d2ψ

dx2− 2(V0 − E)m

~2ψ = 0

b) If we let:

k =

√2mE

~2and κ =

√2m(V0 − E)

~2

Then,d2ψ

dx2+ k2ψ = 0 ; 0 ≤ x ≤ L

d2ψ

dx2− k2ψ = 0 ; x ≥ L

Similar to the situation of the finite well discussed in the text, the aboveequations are thus satisfied by the following:

ψ(x) = A cos(kx) and ψ(x) = A sin(kx) ; 0 ≤ x ≤ L

ψ(x) = Be−κx ; x ≥ L

Once again, the negative argument in the exponential is to insure our wavefunction decays as x→∞.

31

c) If the potential is to be infinite at x = 0, then the wave function mustgo to zero at x = 0. Therefore, we must impose new boundary conditionsto insure the continuity of the wave function. Above we had both even andodd solutions for the wave function, so let’s examine their behavior at x = 0given the new conditions:

Even : ψ(0) = A cos(k0) = 0

Odd : ψ(0) = A sin(k0) = 0

We see immediately that the fist of the equations is not physically acceptable,so we must reject the even solutions, thus we are left with: ψ(x) = A sin(kx)for 0 ≤ x ≤ L. We may now proceed with our boundary conditions at x = L:

ψ(L) = A sin(kL) = Be−κL

Imposing the continuity of the first derivative:

Ak cos(kL) = −Bκe−κL

Divide eq. 2 by eq. 1:k cot(kL) = −κ

−cot(kL) =κ

k

We insert our κ as defined in part (b):

−cot(kL) =

√2mV0

~2k2− 2mE

~2k2

Substitute k into the second term of the RHS:

−cot(kL) =

√2mV0

~2k2− 1

Let θ = kL:

−cot(θ) =

√2mV0L2

~2θ2− 1

We can transform this into a form similar to that of eq. 2.31 if:

θ02 =

2mV0L2

~2


Our equation to solve then becomes:

−cot(θ) =

√θ0

2

θ2− 1

This can be solved graphically for the values of θ for which the LHS and RHSexpressions intersect. From here we obtain the corresponding values of k:

k =θ

L

And solve eq. 2.24 for E:

E =k2~2

2m

Note that the value for θ02 which we derived above is slightly different from

that of eq. 2.30. This occurs because our potential well for this problem isactually oriented from x = 0 to x = L, as opposed to x = −L/2 to x = L/2in the example of section 2.3. This in turn gave us a different argument of thecotangent function and thus necessitated we choose a different θ0

2 in orderto keep the same clean form of eq. 2.31 to solve numerically. Procedurally,the process we just followed is identical to that of section 2.3.

14. Equation (2.37) states: ψ(x, t) = Aeikxe−iωt. We first evaluate theleft-hand side of (2.42):

− ~2

2m

d2ψ

dx2= − ~2

2m(−Ak2eikxe−iωt)[

−~2

2m

∂2

∂x2+ V (x, t)

]ψ(x, t) = A

~2k2

2meikxe−iωt + V (x, t)eikxe−iωt

Now for the right-hand side:

i~∂ψ(x, t)

∂t= ~ωAeikxe−iωt

If we assume the free particle situation, V=0. And, since ~k = p and p2/2m =KE, the left hand side becomes:

~2k2

2mAeikxe−iωt = Eψ(x, t)

33

Similarly, for the right-hand side, ~ω = E and:

~ωAeikxe−iωt = Eψ(x, t)

Therefore, the time-dependent Schrodinger equation does in fact hold forthetraveling wave.

15. The infinite potential well represents a situation where the potentialenergy does not evolve with time. The time-dependent solutions must thensatisfy eq. 2.45:

Ψ(x, t) = uE(x)e−iωt where w =E

~From eq. 2.20, the normalized spatial wave functions of a particle in theinfinte well are given by:

ψ(x) =

√2

Lsin(nπxL

)The corresponding energies are given by eq. 2.17:

E =n2h2

8mL2

We may then calculate ω from eqs. 2.39 and 2.17:

ω =E

~=

n2πh

4mL2

Our total wave function is then:

Ψ(x, t) =

√2

Lsin(nπxL

)exp

(−in

2πht

4mL2

)

3

Operators and Waves -Solutions

1. We may obtain the momentum by multiplying the momentum operatoras given in equation 3.2

p = −i~ ddx

times the wave function given in the problem

pψ(x) = −i~ ddxAeiαx = −i2α~Aeiαx

pψ(x) = α~Aeiαx

Our wavefunction is therefore seen to be an eigenfunction of the momentumoperator corresponding to the eigenvalue α~. Thus, a measurement of themomentum of the electron in this state would yield a value of α~.

The kinetic energy may be found via the relation:

KE =p2

2m

The square of the momentum operator that appears here in the numeratorwill result in the operator begin applied twice, therefore:

p2 = (−i~)2 d2

dx2= −~2 d

2

dx2

We substitute this expression into the expression of the kinetic energy:

KE =p2

2m= − ~2

2m

d2

dx2

35

36 3. OPERATORS AND WAVES - SOLUTIONS

and operate on the wavefunction:

− ~2

2m

d2ψ

dx2= − ~2

2m

d2

dx2Aeiαx

=~2

2mAα2eiαx ⇒ KE =

~2α2

2m

2. From (3.2):

p = −i~ ddx

p φ1(x) = i~k cos kx

p φ2(x) = −i~k sin kxThese wavefunctions are not eigenfunctions of the momentum operator.

3. Our task is to find ψ = Aφ1(x) +Bφ2(x) such that:

pψ = pψ

Where p is the eigenvalue of the momentum operator. So, let’s try operatingon the following wavefunction:

ψ = A cos(kx) +B sin(kx)

pψ = −i~dψdx

= −i~A d

dxcos(kx)− i~B d

dxsin(kx)

= i~k Asin(kx)− i~k cos(kx)

This choice of ψ is not generally an eigenfunction of p, as we can not factorout our original ψ = A cos(kx) +B sin(kx). Let us now try the following:

ψ = cos(kx) + i sin(kx)

pψ = −i~k(−sin(kx))− i2~k(cos(kx)) = ~k cos(kx) + i~k sin(kx)

= ~k(cos(kx) + i sin(kx))

Thus, we have found an eigenfunction of the momentum operator whichcorresponds to the eigenvalue ~k. Note, that using the Euler formula:

eikx = cos(kx) + i sin(kx)

37

Our wavefunction may be written:

ψ = eikx

Using this form of the wavefunction and the result of Problem 1, we maythus see that the wavefunction is an eigenfunction of the momentum opera-tor corresponding to the eigenvalue ~k.

4a. In general, from (3.3):

p ψ(x) = pψ(x)⇒ i~d

dxψ(x) = p ψ(x)

b. We may solve the differential equation in (a) by first dividing both sidesby −i~:

ψ(x) = Aeiphx

Using (2.2), p = ~k and:ψ(x) = Aeikx

c. Using the Euler equation:

ψ(x) = Acos kx+ iAsin kx

Since ψ(0) = ψ(a):ψ(0) = A

ψ(a) = Acos ka

cos ka = 1

ka = 2π

k =2πn

a

Once again, from (2.2):

p = ~k =2π~na

5. Notice that if the energy is less than the minimum value of V (x), the[V (x)−E] portion of the RHS is always positive. Therefore, if this equation


holds, the second derivative of the wavefunction will have the same sign asthe wavefunction itself, and the function will increase or decrease in valuemonotonically.

6. The first two excited states should correspond to a†ψ0 and (a†)2ψ0, re-spectively. Therefore, using (3.22) and (3.31):

a†ψ =

√2mω

~A0xe

−(mω/2~)x2

The next excited state may be found by applying the a† operator again:

a†(a†ψ) = A0mω

~

(2x2e−(mω/2~)x2 − ~

mωe−(mω/2~)x2

)

7. Sections 3.3.1 and 3.3.2 provide the initial setup for this problem. Wemust solve the appropriate Schrodinger equation for each region, then useboundary conditions to find our desired ratios. In the first region, whereV = 0, we have the following from section 3.3.1:

d2ψ1

dx2+ k1

2ψ1 = 0

and given by eq. 3.32:

k1 =

√2mE

~2

This yields solutions:ψ1 = Aeik1x +Be−ik1x

Region 2 produces equations 3.39 through 3.40:

d2ψ2

dx2− k2

2ψ2 = 0 where k2 =

√2m(V0 − E)

~2

With physically acceptable solutions:

ψ2(x) = De−k2x

Now, we impose continuity of the wave equations at x = 0 and of their firstderivatives at x = 0:

ψ1(0) = ψ2(0)⇒ A+B = D

39

ψ1′(0) = ψ2

′(0)⇒ ik1(A−B) = −k2D

Divide both side of the second equation by ik1 and our two equations become:

A+B = D (1)

A−B = − k2

ik1

D = ik2

k1

D (2)

Here, the fact that −1/i = i was used to bring i out of the denominator andcancel the negative. If we add these two equations, we solve for A in termsof D:

2A = D(1 + ik2

k1

)

A =1

2D(1 + i

k2

k1

) (3)

If we subtract (1) from (2), we can solve for B in terms of D:

2B = D(1− ik2

k1

)

B =1

2D(1− ik2

k1

) (4)

From here, we divide each (3) and (4) by D:

A

D=

1

2(1 + i

k2

k1

)

B

D=

1

2(1− ik2

k1

)

The reflection coefficient R is found by

R =|B|2 v1

|A|2 v1

=|B|2

|A|2

Using our ratios above:

R =(BD

)∗(BD

)

(AD

)∗(AD

)=

14(1− ik2

k1)(1 + ik2

k1)

14(1 + ik2

k1)(1− ik2

k1)

= 1


8. Since region 2 extends from x = 0 to infinity, the probability that theparticle penetrates into this region is:

P =

∫ ∞0

|ψ(x)|2 dx

Where, from (3.40), ψ2(x) = De−k2x and:

P =

∫ ∞0

D2e−2k2x dx

=D2

2k2

Substituting (3.39):

P =D2

2

(2m(V0 − E)

~2

)− 12

9. Following figure 3.6, we are given the general solutions for each region:

ψ1(x) = Aeik1x +Be−ik1x, x ≤ 0

ψ2(x) = Cek2x +De−k2x, 0 ≤ x ≤ L

ψ3(x) = Eeik1x, x ≥ 0

We impose the boundary conditions at x = 0 and x = L:

ψ1(0) = ψ2(0)

ψ1′(0) = ψ2

′(0)

ψ2(L) = ψ3(L)

ψ2′(L) = ψ3

′(L)

The first two yield the following system of equations:

A+B = C +D

ik1(A−B) = k2(C −D)

41

We divide the second of these by ik1 and add the result to the first to obtainan expression for A in terms of C and D:

2A = C +D − ik2

k1

(C −D) (1)

Now, let’s turn to the interface between regions 2 and 3. The boundaryconditions produce the following system:

Cek2L +De−k2L = Eek1L

k2Cek2L − k2De

−k2L = ik1Eek1L

Dividing the second of these equations by k2 reduces to the system to a moremanageable one:

Cek2L +De−k2L = Eek1L

Cek2L +De−k2L = ik1

k2

Eek1L

Add the two equations and we can solve for C in terms of E:

2Cek2L = Eek1L + ik1

k2

Eek1L

C =E

2

(1 + i

k1

k2

)e(ik1−k2)L (2)

Subtract those same equations to solve for D in terms of E:

2De−k2L = Eek1L − ik1

k2

Eek1L

D =E

2

(1− ik1

k2

)e(ik1+k2)L (3)

Insert (2) and (3) into (1):

2A =E

2

(1 + i

k1

k2

)e(ik1−k2)L +

E

2

(1− ik1

k2

)e(ik1+k2)L

− ik2

k1

E

2

[(1 + i

k1

k2

)e(ik1−k2)L +

(1− ik1

k2

)e(ik1+k2)L

]


Divide both sides by A and simplify:

4 =E

A

[(1 + i

k1

k2

)−(ik1

k2

− 1

)]eik1L−k2L +

E

A

[(1− ik1

k2

)+

(ik1

k2

+ 1

)]eik1L+k2L

=E

A

[(2 + i

k21 + k2

2

k1k2

)eik1L−k2L +

(2 + i

k22 − k2

1

k1k2

)eik1L+k2L

]Solve for E/A:

E

A= 4

[(2 + i

k21 + k2

2

k1k2

)eik1L−k2L +

(2 + i

k22 − k2

1

k1k2

)eik1L+k2L

]−1

11. The Heisenberg uncertainty principle states:

∆E∆t ≥ ~2

Therefore we solve for ∆E:

∆E =~

2∆t

=6.626× 10−34 J · s

2(4.0× 10−10 s)

= 8.28× 10−25 J

Now, converting to electron volts:

∆E = 8.28× 10−25 J · 1 eV

1.6× 10−19 J= 5.18× 10−6 eV

12. By the normalization condition:∫ ∞−∞|ψ(x)|2 dx = 1

Using the boundary of the problem:∫ 1

0

B2e−2x dx = 1

43

1

B2= −1

2

[1

e2− 1

]

B =√

2

[1− 1

e2

]− 12

13. Equation 3.48 gives the average value of an observable:

< Q >=

∫ψ∗(x)Qψ(x) dx

Where Q is the corresponding operator, in this case the kinetic energy oper-ator, which is given by:

p2

2m= − ~2

2m

d2

dx2

The integral becomes:

< KE > =

∫ 1

0

Be−x(− ~2

2m

d2

dx2

)Be−x dx

= − ~2

2mB2

∫ 1

0

e−2xdx

= − ~2

4mB2[e−2 − 1

]where B is the normalization constant found in problem 12.

4

The Hydrogen Atom -Solutions

1. From table 4.2, we see that:

P10(r) =

√1

a0

2r

a0

e−r/a0

where z=1 corresponds to hydrogen. Then the radial probability density isgiven by:

P 210(r) =

4r2

a30

e−2r/a0

2. Per equation 4.9, the radial probability density for the 1s state hydrogenis found by:

|P1s(r)|2 =

(2

√1

a0

r

a0

er/a0)2

=2r2

a30

e−2r/a0

The 1s radial wave function used here is listed in table 4.2 in the row corre-sponding to n=1 and l=0. Since hydrogen is the atom of interest, we used theappropriate atomic number Z=1. We shall maximize the above expressionby taking the first derivative and equating to zero:

d

dr|P1s(r)|2 =

4r

a30

e−2r/a0 +2r2

a30

(−2

a0

)e−2r/a0 = 0

=4r

a30

e−2r/a0 − 4r2

a40

e−2r/a0 = 0

45

46 4. THE HYDROGEN ATOM - SOLUTIONS

Divide both sides by e−2r/a0 and solve for r:

4r

a30

− 4r2

a40

= 0

4r

a30

(1− r

a0

)= 0

Therefore we have one solution at r = 0 and one at r = a0. Let’s test theintervals (0, a0) and (a0,∞) to reveal the nature of these points. Let r = a0/2for the former and r = 2a0 for the latter:

d

dr|P1s(r)|2

∣∣∣∣x=a0/2

=4

a30

(a0

2

)e−1 − 4

a30

(a2

0

4

)e−1

= e−1

(2

a20

− 1

a20

)> 0

d

dr|P1s(r)|2

∣∣∣∣x=2a0

=4

a30

(2a0) e−4 − 4

a30

(4a2

0

)e−4

=8

a20

e−4(1− 2) < 0

Since the first derivative of the radial probability density takes a positive val-ues for (0, a0) and negative values for (a0,∞), we may conclude that r = a0

is in fact the maximum value. This is consistent with the description of theBohr radius as the innermost orbital radius of the electron in the hydrogenatom.

4. Starting with equations 4.7 and 4.8,

dP = |ψ(r)|2 r2 sin(θ) dr dθ dψ

We construct our wavefunction by equation 4.4:

ψ(r, θ, φ) =Pnl(r)

rΘlm(θ) Φm(θ, φ)

Pnl(r), Θlm(θ), and Φm(θ, φ) for the 1s state may be found in tables 4.1 and4.2 in rows corresponding to n=1, l=0, and ml=0 (Note that ml=0 is the only

47

possibility for l=0 thus only one row for l=0 appears in table 4.1). Carryingon,

ψ(r, θ, φ) =2

r

√1

a0

(r

a0

)e−r/a0

1√2

1√2π

=1

a3/20

e−r/a01√π

We now integrate to find the probability:

P =

∫|ψ(r, θ, φ)|2 r2 sin(θ) dr dθ dφ

=1

π

1

a30

∫ a0

0

r2e−2r/a0 dr

∫ π

0

sin(θ) dθ

∫ 2π

0

dφ

=

∫ a0

0

4r2

a30

e−2r/a0 dr

Grouping 1/a0 with r and dr:

P = 2

∫ a0

0

(2r

a0

)2

e−2r/a0 d

(r

a0

)Substitute ρ = r

a0:

P = 2

∫ 1

0

ρ2e−2ρ dρ

This integral may be solved analytically using integration by parts:∫ 1

0

ρ2e−2ρ dρ = −1

2e−2ρρ2

∣∣∣∣10

−∫ 1

0

2ρ

(−1

2e−2ρ

)dρ

= −1

2e−2 +

(−1

2

)e−2ρρ

∣∣∣∣10

−∫ 1

0

(−1

2

)e−2ρ dρ

=1

4

(1− 5e−2

)

5. For a given n-state, the possible values of l are n-1, n-2, . . . 0. Therefore,


n=5 allows l=0,1,2,3,4. For any given l value, the possiblities for the mag-netic quantum number are ml = -l, -l+1, . . . 0, 1, . . . l. The results are asfollows:

l = 0 : m = 0

l = 1 : m = −1, 0, 1

l = 2 : m = −2,−1, 0, 1, 2

l = 3 : m = −3,−2,−1, 0, 1, 2, 3

l = 4 : m = −4,−3,−2,−1, 0, 1, 2, 3, 4

6. The radial wave function P5d(r) corresponds to n=5 and l=2, thereforewe calculate from (4.6):

v nodes = n− l − 1

= 5− 2− 1

= 2

7. Recall from equation 4.4 our construction of a wave function as a productof radial and angular parts:

ψ(r, θ, φ) =Pnl(r)

rΘlml(θ)Φ(φ)

Our strategy will be to express the given wave function in terms of radial andangular parts, then use their properties to check whether the wavefunctionis an eigenfunction of the Schrodinger equation and find the correspondingenergy. For our given wavefunction, we see that we may separate it in termsof dependence on θ and dependence on r as cos(θ) and r2e−Zr/2a0 . Notethat the presence of r2 is due to fact the radial part is to be divided by r ineq. 4.4 to construct the total wave function. We need not incorporate theconstant C as it will not affect whether the wave function is a solution of theSchrodinger equation. For the angular part, note that in table 4.1, cos(θ)

49

appears in the spherical harmonic corresponding to l = 1 and ml = 0.When we distribute the left-hand side of eq. 4.4, we get:

−~2

2m

d2

dr2Pnl(r) +

~2l(l + 1)

2mr2Pnl(r)−

1

4πε0

Ze2

rPnl(r)

Taking the second derivative of the radial wave function:

d2

dr2Pnl(r) = 2e−Zr/2a0 − 2rZ

a0

e−Zr/2a0 +Z2r2

4a0

e−Zr/2a0

Upon substitution and using l = 1, the left-hand side of eq. 4.4 becomes:

−~2

2m

d2

dr2Pnl(r) +

~2l(l + 1)

2mr2Pnl(r)−

1

4πε0

Ze2

rPnl(r)

= −~2

me−Zr/2a0 +

~2rZ

ma0

e−Zr/2a0 +~2Z2r2

8ma0

e−Zr/2a0 +~2

me−Zr/2a0 − Ze2r

4πε0e−Zr/2a0

=~2rZ

ma0

e−Zr/2a0 − Ze2r

4πε0e−Zr/2a0 +

~2Z2r2

8ma0

e−Zr/2a0

Recall from equation 1.20 that the Bohr radius a0 may expressed as follows:

a0 =4πε0~2

me2

If we substitute this expression for a0 into the first term of the previousequation, the first two terms cancel and equation 4.5 becomes:

~2Z2

8ma0

r2e−Zr/2a0 = EPnl(r)

The radial wave function is in fact an eigenfunction of the Schrodinger equa-tion with corresponding eigenvalue (energy):

E =~2Z2

8ma0


8.

Hydrogen E1 = −13.6 eV

E2 = −3.4 eV

E3 = −1.5 eV

Ne+9 E1 = −1360 eV

E2 = −340 eV

E3 = −151 eV

9. We shall start with eq. 4.5:(−~2

2m

d2

dr2+

~2l(l + 1)

2mr2− 1

4πε0

Ze2

r

)Pnl(r) = EPnl(r)

And 1.20:

a0 =4πε0~2

me2

The substitution as directed is ρ = r/a0, so we can directly substitute for F:r = a0ρ. We let Z=1 as is the case hydrogen and carry out the substitution:(

−~2

2m

d2

d(ρa0)2+

~2l(l + 1)

2m(ρa0)2− 1

4πε0

Ze2

ρa0

)Pnl(r) = EPnl(r)

Since a0 is a constant, we may bring it outside the derivative in the first termon the left-hand side and substitute eq. 1.20 for one a0 in each of the firsttwo terms.(− 1

a0

~2

2m

me2

4πε0~2

d2

dρ2+

1

a0

~2l(l + 1)

2m(ρa0)2

me2

4πε0~2− 1

a0

Ze2

ρa0

)Pnl(r) = EPnl(r)

After making the appropriate cancellations and factoring we have:

e2

a0

1

4πε0

(−1

2

d2

dρ2+

1

2

l(l + 1)

ρ2− 1

ρ

)Pnl(r) = EPnl(r)(

−1

2

d2

dρ2+l(l + 1)

2ρ2− 1

ρ

)Pnl(r) =

E

(1/4πε0)(e2/a0)Pnl(r)

51

10 a.) For l = 1, our expression for Vl(ρ) becomes:

V1(ρ) =1

ρ2− 1

ρ

b.) From problem 8, E3 = −1.5 eV . Per the problem statement, make thispositive and set equal to V1(ρ):

1

ρ2− 1

ρ= 1.5 eV

1.5ρ2 + ρ− 1 = 0

ρ = −1.22, ρ = .55

c.) We find from table 4.2:

P31(r) =

√1

a0

4√

2

27√

3

(r

a0

)2(1− r

6a0

)e−r/3a0

Now, substitute ρ = r/a0 and evaluate the derivative:

dP31

dρ=

√1

a0

4√

2

27√

3

[2ρe−ρ/3 − 1

2ρ2e−ρ/3 − 1

3ρ2e−ρ/3 +

ρ2

18e−ρ/3

]Using sample values ρ = −1 and ρ = 1:

dP31

dρ/P31

∣∣∣∣ρ=−1

= −53

21

dP31

dρ/P31

∣∣∣∣ρ=1

=4

3

11. Equation 4.15 gives the following:[~2

2m∇2 + V (r, t)

]Ψ = i~

∂Ψ

∂t


Given that Ψ(r, t) = φ(r)T (t) and letting H equal the operator within squarebrackets:

T (t)Hφ(r) = i~φ(r)T (t)

dt

Divide through by Ψ(r, t) = φ(r)T (t):

1

T (t)i~dT (t)

dt=

Hφ(r)

φ(r)

Setting the left-hand side equal to the separation constant E:

1

T (t)i~dT (t)

dt= E

dT (t)

T (t)= −iE

~dt

Integrate both sides:

ln T (t) = −iE~t

Let ω = E/~:

T (t) = e−iωt

Using the fact that φ(r) is a solution to the time-independent case, the time-dependent solution may be written as:

Ψ(r, t) = φ(r)e−iωt

12. The transition integral for the general case is given by∫φ∗n1l1m1

l

(z

a0

)φn2l2m2

ldV

We are interested in the transition 3d1 → 2p1, so the transition integralbecomes: ∫

φ∗3,2,1

(z

a0

)φ2,1,1 dV

53

From here on, we assume that the radial distance will be given in termsrelative to a0, so we take a0 = 1 and constructs the wavefunctions by tables4.1 and 4.2.

φ3,2,1 =P3,2(r)

rY2,1(θ, φ)

=2√

2

81√

15r2e−r/3

(−√

15

8πsin(θ)cos(θ)eiφ

)

φ2,1,1 =P2,1(r)

rY1,1(θ, φ)

=1

2√

6re−r/2

(−√

3

8πsin(θ)eiφ

)

Now, we take the complex conjugate of φ3,2,1, substitute z = rcos(θ), andproceed with the integration:∫

φ∗3,2,1zφ2,1,1 dV =1

648π

∫r4e−5r/6sin2(θ)cos2(θ) dV

=1

648π

∫ ∞0

∫ π

0

∫ 2π

0

r6e−5r/6sin3(θ)cos2(θ) dr dθ dφ

The integration may be separated as follows:

1

648π

∫ ∞0

r6e−5r/6 dr

∫ π

0

sin3(θ)cos2(θ) dθ

∫ 2π

0

dφ =1

648π

40310784

15625

4

152π

= 2.123

13. In order to find the transition coefficient for hydrogen, we will use theresults found in Appendix FF (online) along with eqs. 4.20 and 4.21. WeAccording to the selection rules, summarized in table 4.3, ∆ml = ±1 for x-and y-polarized light. Therefore, we must calculate the transition integralsfor both the 2p1 → 1s0 and 2p−1 → 1s0 cases. As directed in Appendix FF,we use the following in place of the z operator in eq. 4.21:

x =1

2(r+ + r−)

y =1

2i(r+ − r−)


We first solve the case for x-polarized light.

I12 = 2(|φ∗1s0xφ2p1|2 + |φ∗1s0xφ2p−1|2

)Which gives us the following when we substitute the above expression for x:

φ∗1s0xφ2p1 =1

2(φ∗1s0r+φ2p1 + φ∗1s0r−φ2p1)

φ∗1s0xφ2p−1 =1

2(φ∗1s0r+φ2p−1 + φ∗1s0r−φ2p−1)

Since φ∗1s0r+φ2p1 and φ∗1s0r−φ2p−1 will equal zero, we may use the results givenin Appendix FF for the remaining terms, giving us:

I12 = 2

(1

6R2i +

1

6R2i

)=

2

3R2i

Using the results of eqs. 4.23 and 4.26 into eq. 4.22:

A21 =6.078× 1015

(121.6)3

1.109

6= 6.25× 108 per atom per second

We may follow the same procedure for y-polarized light. First, the transitionintegral:

I12 = 2(|φ∗1s0yφ2p1|2 + |φ∗1s0yφ2p−1 |2

)Using the operator substitution:

φ∗1s0yφ2p1 =1

2i(φ∗1s0r+φ2p1 − φ∗1s0r−φ2p1)

φ∗1s0yφ2p−1 =1

2i(φ∗1s0r+φ2p−1 − φ∗1s0r−φ2p−1)

Once again, the nonzero terms lead to:

I12 = 2

(−1

6R2i +−1

6R2i

)=−2

3R2i

55

The result for y-polarized light is then:

A21 =6.078× 1015

(121.6)3

−1.109

6= −6.25× 108 per atom per second

14. The p-state corresponds to l = 1. Per the selection rules, the 4p electronmay decay to the following states: 3s, 3d, 2s, 1s.

15. a) Let’s imagine that the angular momentum vector l points from theorigin and lies on the surface of a cone, then the radius of the circle at thebase of the cone is:

r = |l|sinθAnd, for a change of the azimuthal angle dφ, the distance that the tip of theangular momentum vector moves is given by:

d|l| = |l|sinθ dφ

b) We start with the relationship of the torque to angular to momentum:

d|l|dt

= |τ |

This equation together with eq. 4.40 immediately leads to the equation:

d|l|dt

=e

2m|l×B|

The definition of the vector product then gives:

d|l|dt

=e

2m|l|B sin(θ)

Substituting d|l| = |l|sinθ dφ, we may then find our Larmor frequencies:

|l|sin(θ)dφ

dt=

e

2m|l|B sin(θ)

ωL =dφ

dt=eB

2m


16. The f-state corresponds to l = 3, therefore the possible values of ml

and ms are:ml = −3,−2,−1, 0, 1, 2, 3

ms = −1

2and

1

2

17. The total angular momentum of the 4f electron will contain contribu-tion form both the electron’s spin and its orbital angular momentum. Section4.3.4 shows that the total angular momentum may take on the values:

J = j1 + j2, j1 + j2 − 1, . . . |j2 − j1|

We let j1 be our orbital angular momentum and j2 the spin. The electronoccupies an f-state, therefore j1 = 3, and the intrinsic spin of the electronmeans j2 = 1/2. We may now list the possible values for J:

J =7

2,

5

2

The spin-orbit coupling energy is then found by:

Es−o =ζ~2

2l =

3ζ~2

2for j = l +

1

2

Es−o =−ζ~2

2(l + 1) = −2ζ~2 for j = l − 1

2

18. We take into consideration two angular momenta: the orbital angu-lar momentum and the spin. The 2p electron has l = 1 and spin s = 1/2.Therefore, the possible values of the combined angular momentum will be:

j =3

2,

1

2

From (4.57) and the equations that follow:

g3/2 =4

3

57

g1/2 =2

3The splitting of the m-levels is given by:

∆E = gjµBBmj

where µB = 9.2732 × 10−24J T−1. For j = 3/2, the possible m-values arem = −3/2, −1/2, 1/2, 3/2 while the possibilities for j = 1/2 are m = ±1/2.Now, to calculate the splitting of energies:

j =3

2: m = −3

2∆E = −5.796× 10−4 eV

m = −12

∆E = −1.932× 10−4 eV

m = 12

∆E = −1.932× 10−4 eV

m = 32

∆E = 5.796× 10−4 eV

j =1

2: m = −1

2∆E = −9.660× 10−5 eV

m = 12

∆E = 9.660× 10−5 eV

19. Using the two equations following eq. 4.50, the separation in energiesdue to spin-orbit coupling is:

∆E =ζ~2

2l −(−ζ~2

2

)(l + 1) =

(ζ~2

2

)(2l + 1)

Since we are using the atomic system of units, ~ = 1 and

∆ =ζ

2(2l + 1) (1)

We then let l=1 (due to the occupation of a p-orbital):

∆E =3

2ζ = 7.9× 10−6

ζ = 5.27× 10−6

The spin-orbit constant ζ for the levels neon may be calculated directly byeq. 4.48, however we may use equation 4.51 to set up a relationship between


ζHe and ζNe. Since He+ and Ne9+ are both hydrogenic atoms (containingonly one electron), the quantity ⟨( a0

Zr

)3⟩

should be equal in both cases as the grouping of a0 and z with r will accountfor the scaling of the radial distance with changing nuclear charge. We nowuse the appropriate atomic numbers of helium and neon to generate a ratiobetween ζHe and ζNe.:

ζNeζHe

=104

24= 625⇒ ζNe = 625ζHe = 3.29× 10−3

For the separation between the 3p3/2 and 3p1/2 levels, we use equation (1)above:

∆E =3.29× 10−3

2(2l + 1) = 4.94× 10−3

A comparison with the separation of the corresponding helium states revealsthe vastly increased impact of spin-orbit coupling in the case of neon.

5

Many-Electron Atoms -Solutions

1. An electron has an intrinsic spin s = 1/2, therefore the possible val-ues of ms are ±1/2. The 4f state of the electron has an orbital angularmomentum quantum number l = 3 and therefore may have values ml =−3,−2,−1, 0, 1, 2, 3.

In the case of two electrons occupying the 4f state, we first see from ourresults above, that there are 7 possible ml quantum numbers 2 ms quantumnumbers. There are thus 14 possible states in which to put the first electron.Once we assign first electron, there are now 13 possible states. Because theorder in which placed them does not matter in our configuration, we thendivide the product by two:

# of distinct states =14 · 13

2= 91

2. If we interchange the coordinates of (5.5), we are left with:

Ψ =1√2

[φa(2)φb(1)− φa(1)φb(2)]

= 11√2

[φa(1)φb(2)− φa(2)φb(1)]

This expression is identical to (5.5) but with the opposite sign.

59

60 5. MANY-ELECTRON ATOMS - SOLUTIONS

3. Using the form of eq. 5.6, we attribute each row of the determinantto a particular state and each column to a particular electron occupying acorresponding state (row). Our system contains these electrons, thereforeN=3 in our coefficient and

Φ =

√1

3

∣∣∣∣∣∣ψ10α(1) ψ10α(2) ψ10α(3)ψ10β(1) ψ10β(2) ψ10β(3)ψ20α(1) ψ20α(2) ψ20α(3)

∣∣∣∣∣∣4. For each of these elements, we shall fill their shells in the following order:1s, 2s, 2p, 3s, 3p, 4s, 4p, and so on. The number of electrons in a neutralatom of any element will equal the number of protons in its nucleus, thereforewe use the atomic number of each element as the number of electrons we useto fill these shells:

Flourine(Z = 9) : 1s2 2s2 2p5

Magnesium(Z = 12) : 1s2 2s2 2p6 3s2 [Ne] 3s2

Silicon(Z = 14) : 1s2 2s2 2p6 3s2 3p2 [Ne] 3s2 3p2

Potassium(Z = 19) : 1s2 2s2 2p6 3s2 3p6 4s1 [Ar] 4s1

Cobalt(Z = 27) : 1s2 2s2 2p6 3s2 3p6 4s2 3d7 [Ar] 4s2 3d7

An alternate (shorter) way to notate these, as I’ve done above to the right,is to use the symbol of the inert gas that occurs closest before a particularelement to denote the filled shells through complete rows of the periodic tableand merely denote the filling of the remaining shells.

5. a.) The elements with a p4 ground configuration will reside in the samecolumn of the periodic table: O, S, Se, Te, Po

b.) The elements with d5 ground configuration are located in the fifth col-umn of the transition metals on the periodic table: Mn, Te, Re, Ns

6. The ground state configuration of carbon is 1s22s22p2. The next twolow-lying configurations are 1s22s22p3s and 1s22s22p3p.

7. The nitrogen atom will have the following ground configuration: 1s2 2s2 2p3.

61

Since the 1s and 2s orbitals have less energy than the 2p orbital, the nexthigher configurations are 1s2 2s2 2p2 3s and 1s2 2s2 2p2 3p.

8. ns2: Here are two electrons with l=0, therefore, L=0. Their spins pro-duce S=0,1. Since L+S must be even for two equivalent electrons, the onlypossibility is 1S (S=0, L=0).

nd2: Two electrons with l=2 may produce L=4,3,2,1,0. Again, the addi-tion of their spin angular momenta produce S=0,1. Imposing the conditionthat L+S must be even, we have the following configurations: 1G (L=4,S=0), 3F (L=3, S=1), 1D (L=2, S=0), 3P (L=1, S=1), and 1S (S=0, L=0).

4f 2: These two electrons occupy f-states (l=3), thus may contribute toL=6,5,4,3,2,1,0. As has been the case above, S=0,1. Imposing the condi-tion that L+S must be even, we have the following LS terms: 1I (L=6,S=0), 3H (L=5, S=1), 1G (L=4, S=0), 3F (L=3, S=1), 1D (L=2, S=0),3P (L=1, S=1), and 1S (S=0, L=0).

Let’s apply Hund’s rules to the 4f configuration. Of these terms, the onescorresponding to the maximum value of S are 3H, 3F , and 3P . Of these, theone with maximum value of L (3H) should lie lowest.

9. We shall add up the contributions from the various ML and MS valuespossible for each LS-term:

1I(L = 6, S = 0) : (13 values ML)× (1 value MS) = 133H(L = 5, S = 1) : (11 values ML)× (3 values MS) = 331G(L = 4, S = 0) : (9 values ML)× (1 value MS) = 93F (L = 3, S = 1) : (7 values ML)× (3 values MS) = 211D(L = 2, S = 0) : (5 values ML)× (1 value MS) = 53P (L = 1, S = 1) : (3 values ML)× (3 values MS) = 91S(L = 0, S = 0) : (1 values ML)× (1 value MS) = 1

The sum of states is 91, which is precisely the result obtained in problem 1.

10. The 2p electron has l = 1 while the 3d electron has l = 2 thereforethe possible values of L are L = 1, 2, 3. The spins may produce S = 0, 1.


The possible configurations are then:

1P, 3P, 1D, 3D, 1F, 3F

11. It is important to take note here that we have a case of non-equvalentelectrons for the [Xe] 4f 5d configuration and one of equivalent electronswith [Xe] 4f 2. For [Xe] 4f 5d, one electron has l=3 and the other l=2.Since there is no additional constraint on the values of L and S except forthe rules of addition of angular momentum, L may take L=5,4,3,2, or1 andS=0,1.

For 4f 2, we must impose the condition that L+S be even. The two l=3electrons could form L=6,5,4,3,2,1,0 and S=0,1. The allowed LS terms arethen (L=6, S=0), (L=5, S=1), (L=4, S=0), (L=3, S=1), (L=2, S=2), (L=1,S=1), and (L=0,S=0).

12. The 3H term corresponds to S = 1, L = 5. The possible values forJ are then J = 4, 5, 6. The 1F term corresponds to S = 1, L = 3, whichproduces J = 2, 3, 4. The states have equal S, therefore the higher value ofL lies lowest which would correspond to the 3H term. Since the 4f subshellcan hold 14 electrons, it is less than half-filled and the minimum value of Jlies lowest (J = 4). Therefore, the L, S, and J values for the lowest state ofthe 4f 2 configuration are L = 5, S = 1, and J = 4.

13. As described in section 5.5.2, the nuclear charge increases as we goacross a row in the periodic table, thus decreasing the average value of theelectrons’ distance from the nucleus. This causes an increase in the Coulombinteraction, thus we expect sulfur (S) to possess the greatest Coulomb inter-action of these three elements.

14. The spin-orbit interaction of S should be largest due to its having thehighest nuclear charge and the fact that the average value of the distancefrom the nucleus to the 3p electron will be less for S than for the other twoelements. This reduces the screening of the magnetic field that the electronfeels from the nucleus.

63

15. Recall from equation 4.51 that the strength of the spin-orbit interac-tion increases with atomic number as Z4. Since the spin-orbit interaction ismainly responsible for the breakdown of LS coupling, the O atom has thesmallest spin-orbit interaction and the states of O corresponds most nearlycorrespond to pure LS terms.

16. The average values of r found from the applet are:

La+ 4f : 1.14a0 5d : 2.88a0

Ce2+ 4f : 1.00a0 5d : 2.34a0

Pr3+ 4f : .94a0 5d : 2.08a0

As we see, the average distances for 4f are indeed smaller than for 5d.

17. To find the total energy, we simply click on neon in the periodic ta-ble interface, then the red arrow on the bottom row of controls to run thecalculation. Then, click on the ”Averages” tab to display the kinetic, po-tential, and total energies. We see a value of -128.55 for the total energy.We now remove a 2p electron by selecting the 2p shell via the incrementalobjects in the bottom row of controls, then clicking the (-e) button. Verifyyou have removed the electron from the correct shell as the configuration inthe second to last row of controls should now read 1s2 2s2 2p5. Now, run theapplet again for the neon ion and finding the energy under the ”Averages”tab, you should find the energy to be approximately 127.82. The bindingenergy is the difference of these two values:

|−128.55− (−127.82)| ≈ .73

Going back to the neutral neon atom, the third column to the left on the pageobtained after striking the Averages tab gives a 2p single electron energy of0.85. This is very close to the difference in total energies of the neutral Neatom and the ion.

18.

Si : < 1r3>= 2.05a0 ζ = 130.243 cm−1

P : < 1r3>= 3.31a0 ζ = 229.512 cm−1

S : < 1r3>= 4.84a0 ζ = 363.99 cm−1


The decreasing distance between electrons and their respective nuclei de-creases the screening of the magnetic field felt by the electrons due to theapparent motion of the nucleus.

19. We can calculate the Slater integrals for Si, P, and S by first select-ing the appropriate element in the periodic table of the Hartree-Fock applet,clicking the red arrow to run the calculation, then clicking the ”Other” tab.Now, the second box should be changed to ”2”, then continuing across: ”3”,”p”, ”3”, ”p”. The red arrow on the right runs the calculation. We obtainthe following values.

F 2(3p, 3p)Element au cm−1

Si .166 36418P .197 43180S .220 48323

We see a definite increase along the row, constitent with the qualitativeresult obtained in problem 13.

20. The following transitions are permitted:

[Xe]4f5d : 3H → 3G, 3G→ 3H, 3G→ 3F, 3F → 3G, 3F → 3D,3D → 3F, 3P → 3D, 3P → 3S, 3S → 3P

[Xe]4f2 : According to these selection rules, no transition

is permitted between these LS terms.

21. Using fig. 5.5, the energy of 1S of the 1s 3s configuration is approxi-mately 185,000 cm−1 while 1P has an energy of approx. 170,000 cm−1. Theenergy of the emitted photon should equal the difference of the energy levels,thus:

∆E = 25, 000 cm−1

According to equation 5.12, 1 eV of photon energy corresponds to light ofwave number 8065.54 cm−1. The photon energy emitted in this transition

65

must equal:25, 000 cm−1

8065.54 cm−1/eV= 3.10 eV

Using the relation E = hc/λ,

λ =hc

E=

1240 eV · nm3.10 eV

= 400 nm

6

The Emergence of Masers andLasers - Solutions

1. As given in figure 6.7, the range of the 4F2 band is approximately17, 000 cm−1 to 19, 000 cm−1 . As these quantities are given in terms of1/λ, we may quickly obtain the corresponding frequencies in Hz by using therelation I.22:

c = λf ⇒ f =c

λ

Therefore, we may convert c to cm/s and simply multiply:

17× 103 cm−1(

3× 1010 cm

s

)= 5.1× 1014 Hz

19× 103 cm−1(

3× 1010 cm

s

)= 5.7× 1014 Hz

Similarly, for 4F1:

23× 103 cm−1(

3× 1010 cm

s

)= 6.9× 1014 Hz

27× 103 cm−1(

3× 1010 cm

s

)= 8.1× 1014 Hz

2. Using fig. 6.9, the wavelength of this transition is approximately 6248 A.

67

68 6. THE EMERGENCE OF MASERS AND LASERS - SOLUTIONS

3. The 2p core is given to be in the state 2P3/2. Using the following schemefor spectroscopic notation:

2s+1LJ

we take J = 3/2. Since the excited electron (3p) has orbital angular momen-tum l=1, the possible values for K are:

K = J + l, J + l − 1, . . . |J − l| = 5

2,3

2,1

2

The spin orbit interaction spits each level with a value of K into two levelscorresponding to the total angular momentum having the values K ± 1

2.

4. The possible values of F from S = 1/2 and I = 1/2, using the rulesof angular momentum addition, are F = 0, 1 . For F=1, the possible valuesof MF are -1, 0, 1 and the only possibility for F=0 is MF = 0.

5. The situation described in problem 4 gives I = 1/2 and the fact thatwe are dealing with hydrogen which means we have one electron, for whichS = 1/2. The possible values for F are then F = 0, for which MF = 0;and F = 1, for which MF = −1, 0, 1. Since the F=0 level is not split by amagnetic field, we only calculate gF for F = 1.Using eq. 6.17,

For F=1, gF =(1 + 1)− 1

2(1

2+ 1

2) + 1

2(1

2+ 1)

1 + 1=

9

8

Therefore, we consider the F=1 states via equation 6.5. Since gF is positive,the MF = 1 state will have a negative µz value and be drawn to regions oflow magnetic field. Consequently, the MF = −1 will have a positive µz valueand be drawn to regions of high magnetic field outside the magnetic trap.These atoms will be lost, decreasing the kinetic energy and temperature ofthe atoms caught in the trap.

6. The relation F = I + S leads to the following:

F · I =1

2(F2 + I2 − S2)

F · S =1

2(F2 + S2 − I2)

69

Substituting these into the expressions for Sz and Iz, and using the appro-priate eigenvalues for F 2, S2, and I2:

Sz →F (F + 1) + I(I + 1)− S(S + 1)

2F (F + 1)Fz

Iz →F (F + 1)− I(I + 1) + S(S + 1)

2F (F + 1)Fz

The eigenvalues for Fz will be MF~, therefore:

µz = −gSµBSz + gNµNIz

= ~MF

[−gSµB

F (F + 1) + I(I + 1)− S(S + 1)

2F (F + 1)+ gNµN

F (F + 1)− I(I + 1) + S(S + 1)

2F (F + 1)

]We see that the expression in square brackets may be replaced by −gFµB,therefore:

µz = −gFµBMF

7. Given S = 1/2 and I = 5/2, our possible values for F are F = 3 andF = 2. These levels split as follows:

F = 3 : MF = −3,−2,−1, 0, 1, 2, 3

F = 2 : MF = −2,−1, 0, 1, 2

As we see, the MF = −3 and MF = 3 states are not mixed with any stateswhere F = 2, thus will be “pure” states with straight lines. All other statesof F = 3 and all states where F = 2 will be mixed, thus will have curvedlines.

70 6. THE EMERGENCE OF MASERS AND LASERS - SOLUTIONS

7

Statistical Physics - Solutions

1. As described in the text, there are sixteen possible outcomes of flippinga coin four times. The possible outcomes can be classified according to thenumber of heads. We may use equation 7.2 (or 7.3) to calculate the statisti-cal weight associated with each outcome by letting N = 4 and n equal thenumber of heads obtained in each distribution:

HHHH :

(4

4

)=

4!

4!0!= 1

HHHT :

(4

3

)=

4!

3!1!= 4

HHTT :

(4

2

)=

4!

2!2!= 6

HTTT :

(4

1

)=

4!

1!3!= 4

TTTT :

(4

3

)=

4!

0!4!= 1

2. We use (7.3) where N = 4 and n = 2:(4

2

)=

24

4= 6

71

72 7. STATISTICAL PHYSICS - SOLUTIONS

3. We may set up equation 7.7 for each energy level:

n2

g2

=n2

2=N

Zeε2/kBT (1)

n1

g1

=n1

1=N

Zeε1/kBT (2)

Divide (1) by (2):n2

n1

= 2e−(ε2−ε1)/kBT

We are given ε2 − ε1 = 0.025 eV , T = 298K, and we use kB = 8.617 ×10−5eV/K:

n2

n1

= 2 exp

(−0.025 eV

(8.617× 10−5 eV/K)(298 K)

)= .755

4 Applying (7.20) to (7.21), we want to show:

4π

(2πmkBT )3/2

∫ ∞0

p2e−p2/2mkBTdp = 1 (1)

The integral above is of the form I2(a) from equation (7.18) if we let u = p,a = 1/2mkBT , and n = 2. The value of this integral is given in the text:

I2(a) =1

4a

(πa

)1/2

=

√π

2(mkBT )3/2

By substituting this into the left-hand side of (1), we see that the normal-ization condition is in fact met.

5. F1(u) is defined in conjunction with Fig. 7.2:

F1(u) =4√πu2e−u

2

73

We maximize F1(u) by setting the derivative of F1(u) with respect to u tozero:

dF1

du=

4√π

(2ue−u

2 − 2u3e−u2)

= 0

2ue−u2

(1− u2) = 0

u = 0,±1

We test the intervals (0, 1) and (1,∞):

F ′1

(1

2

)> 0 and F ′1(2) < 0

Thus, F1(u) has a maximum at u = 1.

6. The average value of v2 may be found by:

⟨v2⟩

=

∫ ∞0

v2P (v) dv

=

∫ ∞0

4πv4

(m

2πkBT

)3/2

e−mv2/2kBT dV

Let a = m/2kBT . The above expression may then be expressed as:

⟨v2⟩

=4√πa3/2

∫ ∞0

v4e−av2

dv

This integral may be evaluated using (F.2) in the appendix. Let n=4:

I4(a) =6√π

8a5/2

⟨v2⟩

=24

a=

48kBT

m

7. To find the average value of v2, we calculate the integral:

(v2)av =

∫ ∞0

v2P (v) dv


Use equation 7.22 for P (v)dv:

(v2)av =

∫ ∞0

4πv4

(m

2πkBT

)3/2

e−mv2/2kBT dv

We may use eq. F.2 in Appendix F to find I4(1) = 3√π/8, thus (v2)av =

3kBT/m.

8. The escape speed may be found by equating the gravitational potentialenergy to the kinetic energy of the molecule:

GMm

r=

1

2mv2

v =

√2GM

r

where G is teh gravitational constant and M is the mass of the planet. There-fore, the probability that the molcule will escape the planet’s gravitationalfield will be equal to:

P =

∫ ∞√

2GM/r

P (v)dv

Where P (v)dv is given by (7.22).

9. To find εrms, we need to first find (ε2)av, then take the square root.Analogous to example 7.3,

(ε2)av =

∫ ∞0

ε2P (ε) dε

Using equation 7.26:

(ε2)av =2√π

∫ ∞0

ε5/2

(kBT )3/2eε/kBT dε

Per equation 7.27, u = ε/kBT and we may write the above equation in termsof u:

(ε2)av =2√π

(kBT )2

∫ ∞0

u5/2e−u du

75

At this stage, we can work the integral according to the procedure expressedin eqs. F.8-F.14:

(ε2)av =2√π

(kBT )2Γ(7/2)

=2√π

(kBT )2 Γ(7/2)

=2√π

(kBT )2 5

2Γ(5/2)

=2√π

(kBT )2 5

2

3

2Γ(3/2)

=2√π

(kBT )2 5

2

3

2

1

2Γ(1/2)

=15

4(kBT )2

We now take the square root of this last quantity:

εrms =

√15

2kBT

10. The average kinetic energy for each degree of freedom is given by:

εav =1

2kBT

=1

2(8.617× 10−5 eV/K)(298 K)

= .013 eV

The energy necessary to excite the hydrogen atom to its first excited state is10.2 eV, much greater than that provided here.

11. According to the equipartition of energy (end of section 7.3.1), eachparticle has an average kinetic energy of kBT/2 for each degree of freedom.For three degrees of freedom:

εav =3

2kBT


Therefore, total kinetic energy of a mole of gas should be:

ε =3

2kBTNA =

3

2(1.381×10−23J/K)(298K)(6.022×1023particles) = 3717 J

12. The specific heat is a measure of the energy requred to raise the tem-perature of an object. Using the equipartition of energy,

εavT

=1

2kB

for each degree of freedom. Therefore, if we use units of Joules and Kelvin forenergy and temperature, respectively, each mole of gas will have the followingspecific heat:

Atoms :3

2kB(6.022× 1023 = 12.47 J/K

Diatomic particles :5

2kB(6.022× 1023 = 20.79 J/K

Polyatomic particles :6

2kB(6.022× 1023 = 24.95 J/K

13. Per equation 7.28,

u(f) =8πf 2

c3

(hf

ehf/kBT − 1

)We apply the product rule and quotient rule of differentiation:

du(f)

df=

16πf

c3

(hf

ehf/kBT − 1

)+

8πf 2

c3

(h(ehf/kBT − 1

)− h2f

kBTehf/kBT

(ehf/kBT − 1)2

)

Let’s set this expression equal to zero and multiply both sides by(ehf/kBT − 1

)2:

16πf 2

c3

(ehf/kBT − 1

)+

8πf 2

c3

(ehf/kBT − 1

)− 8πf 2

c3

(hf

kBTehf/kBT

)= 0

77

Multiply both sides by c3/8πhr2 and let x = hf/kBT , then simplify:

2 (ex − 1) + (ex − 1)− xex = 0

3ex − xex − 3 = 0⇒ (3− x)ex = 3

14. The total energy density may be found be integrating (7.28) over allfrequencies: ∫ ∞

0

u(f)df =

∫ ∞0

8πh

c3f 3 1

ehf/kBT − 1df

Let µ = hf/kBT . The integral becomes:

8πk4B

c3h3T 4

∫ ∞0

µ3

eµ − 1du

Using F.22: ∫ ∞0

u(f)df = .0612 J/m3

15. Equation 7.28 gives the energy density as a function of frequency:

u(f) =8πf 2

c3

(hf

ehf/kBT − 1

)We may derive an expression as a function of wavelength by using relationI.23 from the introductory chaper: f = c/λ.

u(λ) =8π(c/λ)2

c3

(h(c/λ)

ehc/λkBT − 1

)=

8π

λ3

(h

ehc/λkBT − 1

)

16. From (7.39)

I = σT 4

= 6.42× 107 W/m2


17. The most probable frequency of light emitted may be calculated fromeq. 7.37. Since we are interested in energy of the photons, we may thenmerely solve for the numerator:

E = hf = 2.8214kBT

= 1.41 eV

To find the probability of emitted photons being in each range, we dividethe integral of (7.41) over each range by the total intensity which is given by(7.39). First, we must calculate (via E = hf) the corresponding frequencyof each energy value:

1.79eV ⇒ 4.328× 1014s−1

1.81eV ⇒ 4.377× 1014s−1

3.09eV ⇒ 7.472× 1014s−1

3.11eV ⇒ 7.520× 1014s−1

Now to solve the integrals and calculate the probabilities:

P (1.79 ≤ E ≤) = .00818

P (3.09 ≤ E ≤) = .00296

18. From (7.45) and (7.46),

dE = −PdV

Assuming an isolated system,

dE = dE1 + dE2 = 0

Which leads todE1 = −dE2

Using our expression for dE:

−P1dV1 = P2dV2

79

Since the total volume is fixed, dV1 = −dV2, and:

P1 = P2

19. According to eq. 7.50, the work done on the gas is given by:

dW = −P dV

Per the ideal gas law, we solve for P :

P =nRT

V

We then integrate to find W and change sign to reflect work done by the gas:

W = nRT

∫ V3

V2

dV

V= nRT ln

(V3

V2

)Eq. 7.48 states dE = dQto + dWon. As found above, the work done by thegas is equal to nRT ln(V3/V2). If the energy remains constant, the amountof heat added to the gas must be equal to the work done by the gas and

dQto = nRT ln

(V3

V2

)

20. According to the First Law of Thermodynamics, the change in inter-nal energy is given by:

dE = dQ(to) + dWon

From 1 to 2, the change in internal energy is due only to the work done onthe gas. However, from 2. to 3, the temperature must remain constant by aflow of heat, therefore the work (dW = −PdV ) term has a lesser contribution.

21. We first solve eq. 7.66 for Tc:

1

2.612

(N

V

)=

(2πmkBTc

h2

)3/2


Tc =

(1

2.612

N

V

)2/3h2

2πmkB

Substitute our given density:

N

V= 5.0× 1014 atoms/cm3

Tc =(1.16× 109

) h2

πmkB=

7.31× 10−17

m

We may supply the appropriate atomic mass to calculate the critical tem-perature. This equation may appear a bit peculiar due to a quantity oftemperature appearing on the left-hand side and one of mass on right. Wecan take a closer look at the units to more clearly see this relationship. Forclarity, let’s use SI units of J, K, m, and kg for the necessary quantities:

Tc =

(1

2.612

N

V

)2/3h2

2πmkB⇒ [Tc] =

1

m2

J2 · s2

kg · J/K

=1

m

kg m/s2 · s2 ·Kkg

= K

23. Equation 7.76 gives the probability that an energy will fall betweenε and ε+ dε:

P (ε) dε =3

2ε−3/2F ε1/2 dε

1

e(ε−εF )/kBT + 1

Thus, we may substitute our values for εFand T then carry out the integrationnumerically:

At T = 295K :3

2

∫ 4.1

3.9

3−3/2ε1/21

e(ε−εF )/295kB) + 1dε = 5.9× 10−17

At T = 3000K :3

2

∫ 4.1

3.9

3−3/2ε1/21

e(ε−εF )/3000kB) + 1dε = 5.9×10−17 = 2.41×10−3

It is seen here that the electron is much more likely to be found with anenergy between 3.9 ecV and 4.1 eV at 3000K.

81

24. If we take ne to be the number of conduction electrons per elemen-tary unit of a substance, the total number of conduction electrons in onemole of the substance is given by ncNA. The number of conduction electronsper unit mass will then be ncNA/M . Now, multiply by the mass density toleave the number of conduction electrons per volume:

N

V=ncρNA

M

25. We substitute our given quantities into the given formula, along withnc = 1 since gold is a monovalent metal:

N

V=

19.32 g/cm3(6.022× 1023atoms/mole)

197 g/mole

N

V= 5.90× 1022 electrons/cm3

26. From problem 25, N/V = 5.90 × 1022cm3. Therefore, since V = 1 cm3

here:N = 5.90× 1022cm3

27. (a) Using the formula from problem 24 and the fact that nc = 1 sincesodium is a monovalent metal:

N

V=ncρNA

M=

(.971 g/cm3)(6.022× 1023atoms/mole)

23.0 g/mole

= 2.54× 1022 electrons/cm3

(b) To find the Fermi energy, we use equation 7.73 as well as the quantitywe just calculated in part (a):

εF =h2

2m

[3

8π

N

V

]2/3

=(6.626× 10−34J · s)2

2(9.109× 10−31)

[3

8π(2.54× 1028 m−3)

]2/3

= 5.048× 10−19 J

= 3.155 eV


(c) The equation after example 7.6 relates the Fermi temperature to theFermi energy:

εF = kBTF ⇒ TF =εFkB

Now, we substitute the Boltzmann constant and the Fermi energy found inpart (c):

TF =3.155 eV

8.617× 10−−5 eV/K= 36, 616K

28. We may use a very similar process to that of problem 27, except wemust input the appropriate density and atomic weight as well as nc = 2 sinceMg is a divalent metal. Then, calculate N/V :

N

V=

2(1.7 g/cm3)(6.022× 1023 atoms/mole)

24.3 g/mole

= 8.42× 1022 e−

cm3

= 8.42× 1028 e−

m3

Calculate Fermi energy:

εF =h2

2m

[3

8π

N

V

]2/3

= 1.12× 10−18J

= 7.01 eV

Now, for the Fermi temperature:

TF =ε

kB= 81406 K

8

Electronic Structure of Solids -Solutions

1. The cesium choride lattice is depicted in Figure 8.5. If we take a lighter-shaded sphere to represent a cesium ion, we see that there are 8 chloride ionsthat serve as the nearest neighbors. Or, if the cesium ion serves as the centerpoint in a body-centered cubic lattice, the eight nearest neighbors are thecholoride ions at the corners.

Returning to Figure 8.5, recall that the designation of which shade cor-responds to which ion is arbitrary. Therefore, now alllow the darker spheresto be the cesium ions. By extrapolating the figure in one more direction, itcan be seen that cesium would have 6 next-nearest neighbors.

2. Face-centered cubic

3. We refer to Fig. 8.8(b) where we will use vectors to denote the loca-tions of the carbon atoms from a fixed origin, then use vector relationshipsto determine the bond angle. First, we select our origin to be located at pointA in the lattice. The other two point of interest are point B and the atomlocated in the center of the top face shown. Our primitive i and j will beoriented along the edges of the top face (starting at point A) and the k vectorwill point directly upward from point A. We may now construct the positionvectors to each point. Let a be the position vector from the origin to the topface-centered atom, while b points from point B to the origin (point A). It isour intent to construct a vector c pointing from the top face-centered atom

83

84 8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS

to point B. We will then find the angle between b and c Thus:

a =1

2i+

1

2j, b = −1

4i− 1

4j +

1

4k

Our strategy wil be to find the angle between b and c, so let’s first find c.By inspection we see that:

c = a + b

We substitute the expressions for a and b:

c = a + b =

(1

2− 1

4

)i+

(1

2− 1

4

)j − 1

4k

=1

4i+

1

4j − 1

4k

Now we use the dot (or scalar) product of b and c to find the angle betweenthe two:

b · c = |b||c| cos(θ)

For the left-hand side:

b · c = − 1

16− 1

16+

1

16= − 1

16

For the right-hand side:

|b||c| cos(θ) =

√3

16

√3

16cos(θ) =

3

16cos(θ)

Equating the two:

− 1

16=

3

16cos(θ)

cos(θ) = −1

3

θ = cos−1

(−1

3

)= 10928′

85

4. This distance from one lattice point to the lattice point directly acrossthe main diagonal is calculated by:

d =√a+ a+ a

=√

3a

5. For the simple cubic lattice, referring to figure 8.2, the nearest neigh-bor of any point will be located along the length of a side of the cube. Thus,the nearest neighbor distance is simply “a”. In the case of the face-centeredcubic lattice, Figure 8.6, for example, we may find the nearest neighbor dis-tance by finding the midpoint distance along the diagonal of a face. We maysimply find the diagonal distance via the Pythagorean theorem:

√a2 + a2 =

√2a

The midpoint, thus nearest neighbor distance, is then (√

2/2)a.

6. We find the number of atoms per unit volume:

.971 g/cm3

23.0 g/mole

.04mole/cm3(6.022× 1023atoms/mole

)= 2.54× 1022atoms/cm3

An fcc lattice has 4 atoms per cell, therefore the volume of each cell is:

11 atoms/cell

2.54× 1022 atoms/cm3/= 1.57× 10−22cm3/cell

The nearest neighbor distance is then:

√2(1.57× 10−22 cm3) = 7.63× 10−8 cm−3

7. Our strategy to find the center-to-center distance of the copper ions willbe to find the volume of one cell, from which we will find the length of an


edge of the cube, then use this to arrive at the face-centered nearest-neighbordistance. First, we calculate the number of atoms per unit volume:

8.96 g/cm3

63.5 g/mole= .141

mole

cm3

.141mole

cm3

(6.022× 1023 atoms

mole

)= 8.494× 1022 atoms

cm3

Now, we wish to know the volume per unit cell. A face-centered cubic latticehas four atoms per cell, therefore:

4 atoms/cell

8.494× 1022 atoms/cm3= 4.709× 10−23 cm

3

cell

One edge of the cube should now have length:(1.648× 10−22

)1/3= 3.611× 10−8cm

As given in problem 5, the nearest-neighbor distance is then (√

2/2)a, wherea is the length of an edge of the cube.

√2

2a = 2.553× 10−8 cm

8. Using figure 8.6(b) as guide, we see the points at the centers the otherfaces may be expressed by teh coordinates:

ak +a

2(i+ j)

ai+a

2(j + k)

aj +a

2(i+ k)

These may be expressed in terms of the primitive vectors, respectively:

a

2(j + k) +

a

2(i+ k) = a1 + a2

a

2(i+ k) +

a

2(i+ j) = a2 + a3

87

a

2(j + k) +

a

2(i+ j) = a1 + a3

9. We shall begin by constructing the position vectors of each of the corners,starting with the corner directly above the origin indicated in Fig. 8.6 (b)and continuing clockwise:

l1 = ak

l2 = aj + ak

l3 = ai+ aj + ak

l4 = ai+ ak

The primitive vectors from eq. 8.4 are:

a1 =a

2

(j + k

), a2 =

a

2

(k + i

), a3 =

a

2

(i+ j

)Our strategy will be to solve these primitive vectors for i, j, and k andsubstitute into the lattice vectors. We first add a1 and a2:

a1 + a2 =a

2(i+ j) + ak

Then subtract a3:

a1 + a2 − a3 = ak

We now solve for k:

k =a1 + a2 − a3

a

We continue in such manner to find:

i =a2 + a3 − a1

a

j =a1 − a2 + a3

a


We finally carry out the substitution:

l1 = a1 + a2 − a3

l2 = a1 − a2 + a3 + a1 + a2 − a3

= 2a1

l3 = a2 + a3 − a1 + a1 − a2 + a3 + a1 + a2 − a3

= a1 + a2 + a3

l4 = a2 + a3 − a1 + a1 + a2 − a3

= 2a2

10. The four upper corners may be expressed as:

l1 = ak

l2 = a(k + i)

l3 = a(j + k)

l4 = a(j + k + i)

Once again, in order, these correspond to:

l1 = a1 + a2 = ak

l2 = a1 + 2a2 + a3 = a(k + i)

l3 = 2a1 + a2 + a3 = a(j + k)

l4 = a1 + a2 + 2a3 = a(j + k + i)

11. The results from problems 4 & 5 are as follows:

simple cubic : d = a⇒ R =a

2

bcc : d =

√3

2a⇒ R =

√3

4a

fcc : d =2

2a⇒ R =

√2

4a

89

We now substitute these values of R into the packing fraction equation:

simple cubic : F = 143π(a3

)3

a3=π

6

bcc : F = 2

43π(√

3a4

)3

a3=

√3π

8

fcc : F = 4

43π(√

2a4

)3

a3=

√2π

6

12. The expression

b1 · a1 =2π(a2 × a3)

a1 · (a2 × a3)· a1

May be written as follows:

2πa1 · (a2 × a3)

a1 · (a2 × a3)= 2π

For the cases where i 6= j, consider:

b1 · a2 =2π(a2 × a3) · a2

a1 · (a2 × a3)

We must merely show that the numerator is equal to zero since the result ofthe denominator is a scalar. By vector identity:

a2 · (a2 × a3) = (a2 × a3) · a3 = 0

Now, for i = 1 and j = 3:

b1 · a3 =2π(a2 × a3) · a1

a1 · (a2 × a3)= 0

In like manner, when i = j the expression in the numberator of bi · aj willalways equal the denominator. Otherwise, if i 6= j there will be an aj × ajterm in the numerator.


13. Equation 8.2 gives the following bcc primitive vectors:

a1 =a

2

(j + k − i

), a2 =

a

2

(k + i− j

), a3 =

a

2

(i+ j − k

)We must use eqs. 8.18-8.20 to determine the primitive vectors of the corre-sponding reciprocal lattice. However, it would be of great benefit to calculatethe necessary cross-products first. We shall use the determinant method here:

a2 × a3 =a2

4

∣∣∣∣∣∣i j k1 −1 11 1 −1

∣∣∣∣∣∣ =a2

2j +

a2

2k

a3 × a1 =a2

4

∣∣∣∣∣∣i j k1 1 −1−1 1 1

∣∣∣∣∣∣ =a2

2i+

a2

2k

a1 × a2 =a2

4

∣∣∣∣∣∣i j k−1 1 11 −1 1

∣∣∣∣∣∣ =a2

2i+

a2

2j

Now we are ready to use eqs. 8.18-8.20 to calculate the primitive vectors ofthe reciprocal lattice:

b1 = 2π

a2

3

(j + k

)a2

4

(j + k − i

)·(j + k

) =4π

a

(j + k

)

b2 = 2π

a2

3

(i+ k

)a2

4

(j + k − i

)·(j + k

) =4π

a

(i+ k

)

b3 = 2π

a2

3

(i+ j

)a2

4

(j + k − i

)·(j + k

) =4π

a

(i+ j

)Compared to eq. 8.4, we see that this reciprocal lattice is nothing but anfcc-lattice with sides of length 8π/a.

91

14. From (8.30), (8.28), and (8.29):

n = l×m =l1m1(a1 × a1) + l1m2(a1 × a2) + l1m3(a1 × a3)

l2m1(a2 × a1) + l2m2(a2 × a2) + l2m3(a2 × a3)

l3m1(a3 × a1) + l3m2(a3 × a2) + l3m3(a3 × a3)

Immediately, we see that the terms including a cross product of one vectorwith itself will equal zero. Note that each of the remaining terms will have aform of a scalar times a direction of either b1, b2, or b3 as defined by (8.18)through (8.20). Thus, n will be parallel to a vector of the reciprocal lattice.

15. We shall once again use eqs. 8.18-8.20 to calculate the primitive vectorsof the reciprocal lattice. But, again we carry out the cross-products first:

a2 × a3 =a2

4

∣∣∣∣∣∣i j k1 0 11 1 0

∣∣∣∣∣∣ =a2

4

(−i+ j + k

)

a3 × a1 =a2

4

∣∣∣∣∣∣i j k1 1 00 1 1

∣∣∣∣∣∣ =a2

4

(i− j + k

)

a1 × a2 =a2

4

∣∣∣∣∣∣i j k0 1 11 0 1

∣∣∣∣∣∣ =a2

4

(i+ j − k

)Since the denominator of each of eqs. 8.18-8.20 is the same, we may go aheadand calculate it to substitute into each equation:

a1 · (a2 × a3) =a

2

(j + k

)· a

2

4

(−i+ j + k

)=a3

4

Now our reciprocal lattice primitive vectors are:

b1 =2π

a

(−i+ j + k

)=

4π

a

1

2

(−i+ j + k

)b2 =

2π

a

(i− j + k

)=

4π

a

1

2

(i− j + k

)


b3 =2π

a

(i+ j − k

)=

4π

a

1

2

(i+ j − k

)We factored out 1/2 so as to get us closer to the form of eq. 8.2. Notice thatthese reciprocal vectors are nothing but the primitive vectors of a bcc-latticewith sides of length 4π/a.

16. We use (8.18) through (8.20) to calculate the primitive vectors of thereciprocal lattic:

r1 = 2πb× c

a1 · (b× c)

=2π√3ai+

2π

aj

r2 = 2πc× a

a1 · (b× c)

= − 2π√3ai+

2π

aj

r3 = 2πc× a

a1 · (b× c)

=2π

ck

The volume will be equal to r1 · (r2 × r3):

r2 × r3 =

∣∣∣∣∣∣i j k−2π√

3a2πa

0

0 0 2πa

∣∣∣∣∣∣=

4π2

a2i+

4π2

√3a2

j

V =16π3

√3a3

17. Referring to figure 8.6(b), we may visualize constructing the same prim-itive vectors depicted here, but in a rotated coordinate system such that the

93

x-y plane is parallel to the planes depicted in figure 8.10(b). To continueour analog, let a of problem 16 be equivalent to a3 of figure 8.6(b) and inlike manner a1 → b. We then wish to find the analog of a2. The distancebetween points in our unit cell is simply:(√

3a

2

)2

+(a

2

)2

= 12

d = 1

Therefore, we see that the new, rotated “y-coordinate” of c should be halfthat of a and b (thus equal a/4). This means that the “z-coordinate” mustequal

√15a/4 to ensure that our distance between points is equal to a.

c =a

4j +

√15a

4k

18. Once again, we shall use eqs. 8.18-8.20 extensively as we want to showthat:

a1 = 2πb2 × b3

b1 · (b2 × b3)(1)

a2 = 2πb3 × b1

b1 · (b2 × b3)(2)

a1 = 2πb1 × b2

b1 · (b2 × b3)(3)

Where b1, b2, and b3 are given by eqs. 8.18-8.20. Let us calculate thecross-product b2 × b3 first since it will be used in each equation.

b2 × b3 = 4π2

(a3 × a1

a1 · (a2 × a3)

)×(

a1 × a2

a1 · (a2 × a3)

)We first make the important observation that the denominators of the termson each side of the cross-product results in a scalar quantity. Therefore, itcan be brought outside the vector product and we may carry out the cross-product. Let us proceed by writing just the operation in question:

(a3 × a1)× (a1 × a2)


The following vector relation, which can be found in most books that discussvector calculus, can be of immense help here:

A× (B×C) = B (A ·C)−C (A ·B)

We relate the vectors that occur in the relation above to the vectors involvedin our desired operation as follows:

A = a3 × a1

B = a1

C = a2

After the substitution:

(a3 × a1)× (a1 × a2) = a1 [(a3 × a1) · a2]− a2 [(a3 × a1) · a1]

= a1 [a1 · (a2 × a3)] + a2 [(a1 × a1) · a3]

To arrive at this first term in the second expression of the right-hand side,we used the following vector relation:

A · (B×C) = (A×B) ·C

Notice that the part of this term in brackets is now identical to the expressionfor the denominator. We also see that the second term is equal to zero sincewe have a vector in a cross-product with itself. Our result for b2×b3 is now:

b2 × b3 = 4π2a1

We can use the same method for the other relevant cross-products to obtainthem as well:

b3 × b1 = 4π2a2

b1 × b2 = 4π2a3

The denominators of (1), (2), and (3), which are identical may be also readilysolved using the result from the cross-products above:

b1 · (b2 × b3) = 2πa2 × a3

a1 · (a2 × a3)· 4π2a1

95

= 8π3

By substituting the results we have obtained into the right-hand side of (1),we see that the equation holds true as can be shown for (2) and (3).

19. Three lattice planes for the simple cubic lattice are depicted in Fig8.15. Of these, figure 8.15(b) represents a member of the family with Millerindices (110). Therefore, the plane is perpendicular to the reciprocal latticevector:

g = b1 + b2

Equation 8.33 gives the distance between planes:

d =2π

|g|

We may find the magnitude of g using the primitive reciprocal vectors foundin example 8.1:

g =2π

ai+

2π

aj ⇒ |g| =

√2

2π

a

Therefore our result:

d =a√2

=

√2a

2

20. This family of lattice planes will be perpendicular to the reciprocallattice vector:

g = 2b1 + b2

=4π

ai+

2π

aj

|g| =(

16π2

a2+

4π2

a2

)1/2

=2√

5π

a

From (8.33):

d =2π

|g|=

a√5


21. a) First assume g = g′:∫ a

0

e−igxeigxdx =

∫ a

0

dx = a

On the other hand, if g 6= g′ our integral is as follows:∫ a

0

e−ig′xeigxdx =

∫ a

0

eix(g−g′)dx

By using Euler formula, the integrand may be expressed as a combination ofsine and cosine functions:∫ a

0

cos(x(g − g′)) + i sin(x(g − g′))dx

Therefore, from equation 8.10 we see that the arguments of the cos and sinfunctions will be of the form:

2πn

ax

Therefore, the integral from 0 to a is over a full period. Such integrals of cosand sin functions are equal to zero.

b) Eq. 8.14 gives:

f(x) =∑g

Fgeigx

Multiply by e−ig′x:

e−ig′xf(x) =

∑g

Fge−ig′xeigx

Integrate from 0 to a:∫ a

0

e−ig′xf(x) dx =

∑g

Fg

∫ a

0

e−ig′xeigx dx

From part (a), we know that the integral on the right-hand side has a valueof zero unless g = g′, in which case the value of the integral is ”a”. Thus,the summation disappears as the only term that survives is g = g′:∫ a

0

e−ig′xf(x) dx = aFg′

97

22. The projection of l onto g is:

d = l · g

|g|

The projection of l′ onto g is:

d′ = l′ · g

|g|

The distance between the planes containing the points l and l′ should thenbe:

d′ − d =1

|g|(l′ − l) · g

=1

|g|(2π(N + 1)− 2πN)

=2π

|g|

23. From (8.48),

k =2πn

Na, k′ =

2πn′

Na

Using the wave functions from (8.51), the integrand of (8.52) then becomes:

ψk′(x)ψk(x) =1

L[eik−k

′x]

Now, we apply the Euler formula and evaluate the integral:∫ L

0

ψk′(x)ψk(x)dx =1

L

[sin

2π

L(n− n′)x+ i cos

2π

L(n− n′)x

]∣∣∣∣L0

Since n 6= n′, the difference will be equal to an integer and the real part ofthe integral will equal zero:

1

L[sin 2π(n− n′)] = 0


24. Bloch’s Theorem states:

Ψk(r + l) = eik·lΨk(r)

If we let l = N1a1, according to Bloch’s Theorem:

Ψk(r +N1a1) = eik·N1a1Ψk(r)

Now, we use the Born-von Karman boundary condition to replace the left-hand side:

Ψk(r) = eik·N1a1Ψk(r)⇒ eik·N1a1 = 1

25. We shall first carry out the substitutions then apply orthogonality con-ditions as directed. Our terms as given in the text are:

Eq. 8.55 : ψk(r) =1

V 1/2eik·r

Eq. 8.58 : V (r) =∑g

Vgeig·r

Our other two needed quantities may be determined from these:

ψ′k(r) =1

V 1/2eik′·r and ψ∗k(r) =

1

V 1/2e−ik·r

Substituting into the integral:

I∗k′, k =∑g

VgV

∫ei(g+k′−k)·r dV

We separate the exponential terms, so the the integral may be written:

I∗k′, k =∑g

VgV

∫ei(g+k′)·re−ik·r dV

We may also think of this as:∑g

VgV

∫ψ∗k(r)ψ(k′+g)(r) dV

99

Using the orthogonality condition of 8.52:

k = k′ + g′

which is equation 8.62.

10

Semiconductor Lasers -Solutions

1. As with other radiative transitions we have dealt with, the photons emit-ted must have the same energy as the difference in energies between the twostates. Here, the two states are the bottom of the conduction band and thetop of the valence band, with an energy difference ∆ = 1.519 eV . Usingequation I.25:

E =hc

λ

λ =hc

E=

1240 eV · nm1.519 eV

= 816.3nm

2. From (10.2):

ε(k)− Ec =~2k2

2m0me

= 2.97× 10.20J

= 0.19 eV

3. As outlined in section 10.3.1, heterostructures may be formed using semi-conductors with similar lattice constants. Thus, the heterostructures grown

101

102 10. SEMICONDUCTOR LASERS - SOLUTIONS

on a substrate of InP must have the same lattice constant as InP. We’ll usethe linear interpolation procedure for each alloy:

a(InxGa1−xAs) = a(InP) = xa(InAs) + (1− x)a(GaAs)

5.869 = 6.058x+ 5.653(1− x)⇒ x = .53

This leads to an ally of composition: In.53Al.48As

5. By examining figure 2.6, we see that the graph of√

(θ20/θ

2)− 1 inter-sects the graph of tan θ and −cot θ three times as it decays. Therefore, ifwe insure that

√(θ2

0/θ2)− 1 reaches zero after it intersects once, but before

it does again, the well will have only one bound state. Per equation 2.30, itθ2

0 that includes dependence on V0, the depth of the well. So, we solve for θ0

where the whole expression equals zero:√θ2

0

θ2− 1 = 0⇒ θ2

0 = θ2

We keep only the positive solutions and impose the condition that θ < π/2as that is where the graph of −cot θ begins in the positive domain. If√

(θ20/θ

2)− 1 decays completely before this point, then the electron will nothave more than one ground state.

θ <π

2⇒ θ2

0 <(π

2

)2

We now solve eq. 2.30 for V0, using SI units at first:

θ20 =

m0V0L2

2~2<π

2⇒ V0 <

π~2

m0L2

π~2

m0L2=

(3.1416)(1.055× 10−34J · s)2

.067(9.109× 10−31kg)(10× 10−9m)2

= 5.729× 10−21J

We should now convert to eV:

5.729× 10−21J · 1 eV

1.6× 10−19eV/J= .036 eV

103

The well must be less than 0.036 eV deep to insure that no more than onebound state exists.

6. We take Al0.3Ga0.7As to be the material for the barrier and GaAs forthe well. Using figure 10.12, we see that EB

c ≈ 0.23 eV and EWc ≈ 0 eV .

Therefore, we may solve (10.11) numerically if we do a little simplification.First, the right-hand side:

κ

k=

√2m0mBEB

c

~2k2− 2m0mBE

~2k2

Using our expression for k, this becomes:

κ

k=

√2m0mBEB

c

~2k2− mB

mW

Now let θ = kL/2 and then:

θ20 =

mBm0EBc L

2

2~2

Substitute mB = .104, mW = .067, and L = .1× 10−9 m. (10.11) becomes:

tan θ =

√θ2

0

θ2− mB

mW

=

√15.66

θ2− 1.55

This produces intersections at θ = 1.242 and θ = 3.177 corresponding ton = 1 and n = 3. Note that we have followed the procedure from section 2.3very closely. Following suit, we also must solve for odd solutions:

−cot θ =

√15.66

θ2− 1.55

This intersection, corresponding to n = 2, occurs at θ = 2.317. We maynow find k (k = 2θ/L) and therefore the energy. The values found for θcorrespond to the following energies:

θ = 1.242⇒ E1 = 0.035 eV


θ = 2.317⇒ E2 = 0.122 eV

θ = 3.177⇒ E3 = 0.230 eV

7. We will use the following procedure for these matrix multiplication oper-ations: A1 A2 A3

B1 B2 B3

C1 C2 C3

D1

D2

D3

=

A1D1 + A2D2 + A3D3

B1D1 +B2D2 +B3D3

C1D1 + C2D2 + C3D3

1 2 0

1 1 21 3 1

110

=

324

1 0 1

1 2 11 1 3

101

=

224

2 1 1

1 0 11 1 0

121

=

523

8. [0 11 0

] [0 −ii 0

]=

[i 01 −i

][0 11 0

] [2 00 −2

]=

[0 −22 0

]1 2 0

1 1 21 3 1

0 1 11 0 11 1 0

=

5 1 34 3 26 2 4

9. We shall proceed left to right, first consolidating the 1/2k1 and 1/2k2

terms as we will reintroduce them after carrying out the matrix operations.[k1 + k2 k1 − k2

k1 − k2 k1 + k2

] [eik2L 0

0 e−ik2L

]=

[eik2L(k1 + k2) e−ik2L(k1 − k2)eik2L(k1 − k2) e−ik2L(k1 + k2)

]

105

Using this result in eq. 10.30:

T =1

4k1k2

[eik2L(k1 + k2) e−ik2L(k1 − k2)eik2L(k1 − k2) e−ik2L(k1 + k2)

] [k1 + k2 k2 − k1

k2 − k1 k1 + k2

]Carrying this through will give us following:

T =

[T11 T12

T21 T22

]where each element is as follows:

T11 =1

4k1k2

[eik2L(k1 + k2)2 − e−ik2L(k1 − k2)2

]T12 =

−1

4k1k2

[eik2L(k2

1 − k22)2 − e−ik2L(k2

1 − k22)]

T21 =1

4k1k2

[eik2L(k2

1 − k22)− e−ik2L(k2

1 − k22)]

T22 =−1

4k1k2

[eik2L(k1 − k2)2 − e−ik2L(k1 + k2)2

]We see that the expression we found for T11 does in fact agree with the firstpart of eq. 10.31, however the Euler formula was used in the second part tofurther simplify:

T11 =1

4k1k2

[eik2L(k1 + k2)2 − e−ik2L(k1 − k2)2

]=

1

4k1k2

[(cos k2L+ i sin k2L)(k2

1 + 2k1k2 + k22)− (cos k2L− i sin k2L)(k2

1 − 2k1k2 + k22)]

=1

2k1k2

[2k1k2 cos k2L+ i(k21 + k2

2)sin k2L]

The same may be done for the remaining three elements if desired.

10. Making the substitution κ2 = −ik2 (rearranging the given expressionfrom the problem) into the expression for sinh(κ2x):

sinh(κ2x) =e−ik2x − eik2x

2


Multiply by i:

i sinh(κ2) = ie−ik2x − eik2x

2

=eik2x − e−ik2x

2i= sin(k2)

11. Equations10.25 and 10.26 give the following:

A1 = T11A2 + T12B2

B1 = T21A2 + T22B2

The transmission amplitude of light incident from the right will be equal toB1/B2 and the corresponding reflection amplitude is equal to A2/B2. Asdirected, we set A1 = 0.

0 = A1 = T11A2 + T12B2 (1)

B1 = T21A2 + T22B2 (2)

Solving (1) for A2 and substituting into (2):

B1 =−T21T12B2

T11

+ T22B2 = B21

T11

(T11T22 − T21T12)

We recognize the quantity in parentheses as the determinant of the T-matrix,therefore:

t21 =B1

B2

=det T

T11

This equation 10.38. Now for 10.39, we solve eq. (1) above for A2/B2:

r21 =A2

B2

=−T12

T11

12. For the downward step, we need only modify (10.15) such that:

k2 =

√2m(E − (−V0)

~2

=

√2m(E + V0)

~2

107

Therefore,

T11 = T22 =

√2mE~2 −

√2m(E+V0)

~2

2√

2mE~2

=1

2

[1 +

√1 + V0/E

]T12 = T21 =

1

2

[1−

√1 + V0/E

]

13. As in problem 12, we may modify the treatment of the potential barrierdepicted in figure 10.17 to describe the situation in figure 10.28. Once again,

k2 =

√2m(E + V0)

~2

We then use (10.31) and (10.34) to express the transmission amplitude interms of k1 and k2:

t =2k1k2

2k1k2cos k2L+ i(k21 + k2

2)sin k2L

where:

k1 =

√2mE

~2

Making the substitutions:

2k1k2 =4m

~2

√E(E + V0)

k21 + k2

2 =2m(2E + V0)

~2

t =2√E(E + V0)[

2√E(E + V0) cos

√2m(E+V0)

~2 L+ i(2E + V0) sin√

2mE~2 L

]

14. We carry out the matrix multiplication in (10.43) to obtain:

T =1

t12t21

[eiβL r12e

−iβL

r12eiβL e−iβL

] [1 r21

r21 1

]


T11 =1

t12t21

[eiβL + r12r21e

−iβL]

T12 =1

t12t21

[r21e

iβL + r12e−iβL

]T21 =

1

t12t21

[r12e

iβL + r21e−iβL

]T22 =

1

t12t21

[r12r21e

iβL + e−iβL]

15. a) The total rate of change in the amount of water in the reservoirwill equal the rate of water entering minus the rate in which it is leaving.Letting V equal the volume of water in the reservoir:

dV

dt= Rf −RD

Where RD is the total rate at which water drains. Since the reservoir isdraining from two ports, we set up the following:

RD = Rd1 +Rd2

Therefore, our total rate equation is:

dV

dt= Rf − h(C1 + C2)

b) To find the steady state height, we set:

dV

dt= 0

And solve for h:

h =Rf

c1 + c2

11

Relativity I - Solutions

1. Einstein’s second postulate on the special theory of relativity states thatthe speed of light is independent of the motion of the source, therefore thevelocity of the light in the reference frame of the spaceship is c.

2 a.) Before applying the Lorentz transformation, calculate γ:

γ =1√

1− .22= 1.02

Now, applying (11.15):

x′ = 1.224 m

y′ = 0 m

z′ = 0 m

t′ = 3.672 m/c

b.) We apply the inverse transformations (11.17):

x = 2.00 m

y = 0 m

z = 0 m

t = 4.00 m/c

109

110 11. RELATIVITY I - SOLUTIONS

3. a) The inverse Lorentz transformations are given by eqs. 11.17. Therefore,we need to first calculate γ:

γ =11√

1−u2/c2

=1√

1− .252

= 1.033

Now we transform from S to S ′:

x = γ(x′ + ut′) = 1.033 (2m+ .25(4)m)

= 3.10 m

y = y′ = 0 m

z = z′ = 0 m

t = γ(t′ + ux′/c2) =1.033

c(4 s+ .5 s)

=4.65 s

c

b) We now use eqs. 11.15 and the results from part (a) to transform backto S ′:

x′ = γ(x− ut) = 1.033(3.10 m− .25(4.65) m)

= 2.0 m

y′ = y = 0 m

z′ = z = 0 m

t′ = γ(t− ux/c2) =1.033

c(4.39 s− .25(3.10) s)

=4.0 s

c

4. Using (11.23):

LMLR

=1

2=

√1− u2

c2

111

1

4= 1 =

u2

c2

u2

c2=

3

4

u =

√3

2c

5. a) We may use equation 11.23 to calculate the length as perceived inour reference frame:

LM =

√1− .5c

c2(1 m) = .86 m

b) Equation 11.26 will give us the time we record the object passing us.However, we need to calculate ∆tR, which is the time we would record if wewere traveling along with meter stick:

∆tR =d

v=

11 m

.5(3× 108 m/s)= 6.67× 10−9 s

Now, to transform to our frame of reference:

∆tM =6.67× 10−9 s√

1− (.5c)2

c2

= 7.70× 10−9 s

6. The length along the x-axis will be contracted, but that along the y-axis will not. Therefore, we apply the Lorentz contraction to the componentof the length along the x-axis:

1m(cos 30) =√

3/2 m

Using this length:

LM,x =√

1− (.90c)2/c2

(√3

2m

)= 0.52 m


Now, use the Pythagorean theorem where this contracted distance is one legof a right triangle and y = 1m (sin 30) will be the other:

d = (.52)2 + (.5)2

= .72m

7. The percentage of length contraction will be equal to:(1− LM

LR

)× 100%

Therefore, we may calculate LM/LR via eq. 11.23:

LMLR

=

√1− u2

c2=

√1− (3× 10−6c)2

c2

LMLR≈ 1⇒ 0% contraction

While the high speed of the jet is truly impressive, they still do not travel withenough velocity to display relativistic effects. The time dilation experiencedhere may be calculated by equation 11.26:

∆tM =∆tR

1√1−u2/c2

=1 yr√

1− (3×10−6c)2

c2

∆tM ≈ 1 yr

Once again, the relativistic time dilation has an extremely small effect, wellwithin the intrinsic error of any wristwatch.

8. Using (11.39), we let v equal the velocity of rocket A and u equal thevelocity of rocket B:

v′ =.8c− .6c

1− (.8c)(.6c)c2

= 0.39 c

113

9. a) By differentiating both sides of the equation given in the problemfor the position of the center of mass with respect to t, we may calculate thevelocity of the center of mass:

d

dt2mxCM =

d

dt(mxe +mxp)

dxCMdt

=1

2(ue + up)

where ue and up equal the velocities of the electtron and positron, respec-tively. We now substitute these quantities to obtain the velocity of the centerof mass:

ux =1

2(.95c+ .2c) = .575c

b) The lifetime of the produced particle, gives 2.0 × 10−8 s will be takenas ∆tR in eq. 11.26:

∆tM =2.0× 10−8s√

1− (.575c)2

c2

= 2.44× 10−8s

10 a.) The time as measured by an observer on Earth is found by dividingthe distance by the rate:

t = 8.7 years

b.) We may then use the time dilation equation (11.26) to solve for ∆tR:

∆tR =√

1− (.99)2 × 8.7 years

= 1.22 years

11. a) The equations of section 11.4.6 will give us the transformed fre-quencies, however we are given the wavelength of the sodium D2 line. Thisis a minor obstacle, as we can use equation I.23:

f0 =c

λ0

=3× 108 m/s

589.0× 10−9m= 5.09× 1014Hz


For an approaching light source:

f =

√1 + β

1− βf0

where β = v/c = 0.3.

f =

√1.3

0.7(5.09× 1014Hz) = 6.94× 1014Hz

We then use eq. I.23 to calculate the correspond wavelength:

λ =c

f=

3× 108m/s

6.94× 1014Hz= 432.2 nm

λ− λ0 = 432.2 nm− 589.0 nm = −156.8 nm

b) For a receding light source:

f =

√1− β1 + β

f0 =

√.7

1.3(5.09× 1014Hz) = 3.74× 1014Hz

λ =c

f=

3× 108m/s

3.74× 1014Hz= 802.1 nm

λ− λ0 = 802.1 nm− 589.0 nm = 213.1 nm

c) The equation adjacent to figure 11.12 allows us to calculate the Dopplershift due to transverse motion:

f =f0

γ

We calculate γ:

γ =1√

1− u2/c2=

1√1− (.3)2

= 1.05

115

And substitute:

f =5.09× 1014Hz

1.05= 4.86× 1014 Hz

λ =c

f=

3× 108m/s

4.86× 1014Hz= 617.2 nm

λ− λ0 = 617.2 nm− 589.0 nm = 28.2 nm

12. First, calculate β:

β =v

c= .4

Convert λ = 656.5 nm to frequency:

f =c

λ= 4.57× 1014 Hz

Now, use (11.41) to calculate the shifted frequency:

f = 2.99× 1014 Hz

Convert this back to wavelength:

λ = 1002.8 nm

Now for the relative shift:λ− λ0

λ0

= .53

13. We use eq. 11.41 for the receding source:

β =v

c= 0.80

f =

√1− .80

1 + .80× 100 MHz = 33 MHz

14. According to one of the results of example 11.5:

f

f0

=λ0

λ


We may then use (11.42) to calculate β and the necessary speeds:

yellow: β = .094, v = 0.094c

green: β = .21, v = 0.21c

blue: β = .33, v = 0.33c

15. Equation 11.16 gives the definition for γ:

γ =1√

1− u2/c2

However, we first solve eq. 11.38 for v:

v′ = γ(v − u) · γ(1 + uv′/c2)

We first simplify the right-hand side, then bring the terms containing v′ tothe left, and solve:

v′ = γ2

(v − u+

vuv′

c2− u2v′

c2

)v′

γ2− v′vu

c2+u2v′

c2= v − u

v′(

1

γ2− uv

c2+u2

c2

)= v − u

v′(

1− u2

c2− uv

c2+u2

c2

)= v − u

v′ =v − u1− uv

c2

17. Eq. 11.43 allows us to calculate the proper time interval. We applythis equation at each segment of the observer’s journey. First, the travelerremains stationary for 1 m of time:

(proper time)2 = 12 − 02

proper time = 1 m

117

Then, he travels 4 m in 5 m of time:


proper time = 3 m

Finally he travels 1 m in 2 m of time:


proper time =√

3 m

Therefore, the total proper time is 4 +√

3 m.

18. The plots will be constructed by making the slopes of the x′-axis and ct′-axis equal to .2 and 5, respectively. We may also find the intersections of thex′-coordinate lines and x-axis by setting t = 0 in the Lorentz transformation:

x′ = γx⇒ x =x′

γ

Since u = .2c:

γ =1√

1− .22= 1.02

19. We use a similar procedure as for problem 18, however here β = .25and our x′-axis and ct′-axis will have slopes equal to .25 and 4, respectively.

20. Let x0 = ct, x1 = x, x2 = y, x3 = z. Our calculation of γ remainsvalid, γ = 1.02:

x′0 = 3.67 m

x′1 = 1.22 m

x′2 = 0 m

x′3 = 0 m

21. The inverse transformations for x and y are trivial, so we shall focuson x′ → x and t′ → t. First, solve the top expression of (11.15) for x:

x =x′

γ+ ut (1)


Now, solve the bottom expression of (11.15) for t:

t =t′

γ+ux

c2(2)

Substitute (2) into (1):

x =x′

γ+u

γt′ +

u2x

c2

x = γ(x′ + ct′)

Now, substitute (1) into (2):

t =t′

γ

ux′

γc2+u2t

c2

t = γ(t′ +ux′

c2)

22. Per equation 11.60, tσ = gσµtµ, and eq. 11.58, tµ = gµνtν , the g matrices

will serve to raise or lower the indices of a vector. We now use the givenexpression Λν

µ = gµρΛρσg

σν to operate on vν :

gσνvν = vσ

Now, the Λρσ will serve to transform to the ρ′ indices:

Λρσg

σνvν = Λρσv

σ = v′ρ

We now use gµρ to lower the index:

gµρ Λρσ g

σν vν = gµρv′ρ = v′ρ

We have now shown that v′µ = Λνµvν , which is similar to the contravariant

vector’s transformation in form but requires an alternately defined transfor-mation matrix Λν

µ.

23. First recall that gµρ and gσν will have the form:1 0 0 00 −1 0 00 0 −1 00 0 0 −1

119

The first operation, gµρΛρσ:

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

γ −βγ 0 0−βγ γ 0 0

0 0 1 00 0 0 1

=

γ −βγ 0 0βγ −γ 0 00 0 −1 00 0 0 −1

Now, gµρΛ

ρσg

σν :γ −βγ 0 0βγ −γ 0 00 0 −1 00 0 0 −1

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

=

γ βγ 0 0βγ γ 0 00 0 1 00 0 0 1

24. According to (11.61), we should be able to rewrite:

gµνxµxν = gµνΛρµΛσ

νx′ρx′σ

Recognizing the coefficients of x′ρx′σ in this expression and that given in the

problem:gµνΛρ

µΛσν = gρσ

12

Relativity II - Solutions

1. The components of the four-velocity are given by the relations betweeneqs. 12.3 and 12.4. Since the electron is traveling in the x-direction, we notethat:

dy

dt=dz

dt= 0

Therefore, our relevant components are:

v0 = γc and v1 = γdx

dt

We then calculate γ:

γ =1√

1− u2/c2=

1√1− (.2c)2

c2

= 1.02

Giving us:

v0 = 1.02c

v1 = 1.02(.2c) = .204c

b. The four-momentum as given in eq. 12.6 is simply the product of theelectron’s mass and each component of its four-velocity:

p0 = 1.02mec

p1 = .204mec

121

122 12. RELATIVITY II - SOLUTIONS

2. From (12.13):

K =

(1√

1− .22− 1

).511 MeV

= .011 MeV

3. We may use equations 12.12 and 12.13 which give expressions for therest energy and kinetic energy, respectively. Setting R=KE:

γmc2 = 2(γ − 1)mc2

γ = 2γ − 2⇒ γ = 2

Using eq. 12.3 for γ:

γ =1√

1− u2/c2= 2

√(1− v2/c2) =

1

2(1− v2

c2

)=

1

4

v2

c2=

3

4

v =

√3

2c

In order for the kinetic evergy to be equal to the rest energy, the particlemust be traveling at a speed of (

√3/2)c.

4. Setting (12.9) equal to two times (12.12):

γmc2 = 2mc2

γ = 2

123

Now solve for v from the definition of γ:

1√1−

(vc

)2= 2

v =c√2

5. We first evaluate the derivatives of f(x) called for in the Taylor series:

f ′(x) =1

2

1

(1− x)3/2

f ′′(x) =3

4

1

(1− x)5/2

We now evaluate at x=0 and substitute into the given form of the Taylorseries:

f(x) = 1 +1

2x =

3

8x2 + . . .

Now, let x = v2/c2:

1√1− u2/c2

= 1 +1

2

v2

c2=

3

8

v4

c4

This result is identical to eq. 12.10.

6. At rest, the kinetic energy is equal to zero. Moving at .99c,

γ = 7.09

Using (12.13):

KE = (7.09− 1).511 MeV = 3.11 MeV

Once the electron has been accelerated to .999c,

γ = 22.37

KE = (22.31− 1).511 MeV = 10.92 MeV


7. We shall use eq. 12.13 and let m equal the electron’s mass. However,it will be most convenient to use the mass given in Appendix A in units ofMeV/c2.

100 MeV = (γ − 1)

(.511

MeV

c2

)c2

100 = .511γ − .511

γ = 196.7

We now find the electron’s speed from eq. 12.3:

γ = 196.7 =1√

1− u2/c2

38688 =1

1− v2

c2

1− v2

c2= 2.58× 10−5

.99997 =v2

c2

v ≈ .99999c

The electron would have to be traveling toughly the spped of light possess100 MeV of kinetic energy.

8a.) Equation (12.9) allows us to calculate the spped of the proton fromthe total energy:

E = γmc2

1500 MeV = γ(938.3 MeV )

γ = 1.60 From the definition of γ:

v = .78c

125

b.) Now that we know the magnitude of the velocity, we may find themomentum from (12.7):

|p| = 1.60(938.3 MeV/c2)(.78c)

= 1171.9 MeV/c

Alternatively, one may use (12.14) which leads to:

p =

√E2

c2−m2c2

9. a) For the laboratory frame, recall that the target particle is at rest,therefore has zero momentum. We go ahead and carry out the four-vectoraddition as called for in the numerator of the expression for s:

|pA + pB|2

c2=

[(EA + EB)2 − (pA + pB)2]

c2

Note that we have separated the energy and momentum components of thefour-momentum. Since particle B is at rest, we realize that EB must equalthe rest energy of the particle and pB = 0:

s =[(EA +mBc

2)2 − |pA|2]

c2

In the center of mass frame, we take the colliding particles to have equal andopposite momenta. Therefore, we begin as we did for the laboratory frame,however we can no longer take the energy of particle B to be equal to its restenergy and we allow the sum of the momenta to be zero.

s =[(EA + EB)2 − (pA + pB)2]

c2=

(EA + EB)2

c2

b) We begin by directly substituting the sums of the four-vectors into thesum of the Mandelstam variables:

s+ t+ u =1

c2

[(EA + EB)2

c2− (pA + pB)2 +

(EA − EC)2

c2− (pA − pC)2

+(EA − ED)2

c2− (pA − pD)2

]


By conservation of energy and momentum:

EA − ED = EC − EB

EA − EC = ED − EBpA − pD = pC − pB

pA − pC = pD − pB

We substitute these quantities into our equation above:

s+ t+ u =1

c2

[(EA + EB)2

c2− (pA + pB)2 +

(ED − EB)2

c2− (pD − pB)2

+(EC − EB)2

c2− (pC − pB)2

]We use the fact that in the laboratory frame, pB = 0 and EB = mBc

2 andsimplify:

s+ t+ u =2EAmB

c2− 2EDmB

c2− 2ECmB

c2+m2

A + 3m2B +m2

C +m2D

=2mB

c2(EA − ED − EC) +m2

A + 3m2B +m2

C +m2D

Once again, from the conservation of energy, we know that EB = EA−ED−EC . Making this substitution and using EB = mBc

2:

s+ t+ u = −2m2B +m2

A + 3m2B +m2

C +m2D

= m2A +m2

B +m2C +m2

D

10. This is best solved using the center of mass frame, where the sumof momenta before and after the event is equal to zero. Stating conservationof energy and momentum:

Eρ = Eγ + Eπ

pπ = −pγSquaring the magnitude of the momenta and using (12.14):

p2π = p2

γ ⇒E2π

c2−m2

πc2 =

(Eγc

)2

127

The right-hand side may be replaced by Eγ = Eρ − Eπ:

E2π

c2−m2

pπc2 =

1

c2[E2

ρ + E2π − 2EρEπ]

−m2pπc

2 =m2ρc

4

c2− 2mρEπc

2

c2

2mρEπ = 612388

Eπ = 397 MeV

We may then use (12.9) to solve for the speed of the pion:

397 MeV = γmπc2

γ = 2.85 =1√

1− u2/c2

u = .94c

11. The particle collision here involes two particles with equal mass andspeed traveling in the opposite direction. Therefore, using the conservationof momentum:

pA + pB = pC = 0

Thus, we are in the center of mass frame. An important result here is thatthe resulting particle is at rest. We may then use the conservation energywhere the energy of the resulting particle is its rest energy and the energyof the colliding particles are given by eq. 12.9. We let m be the mass of theeach of the colliding particles and mC be the mass of the resulting particle:

EA + EB = EC ⇒ 2γmc2 = mCc2

mC = 2γm

Using equation 12.3:

mC = 2m√

1− (2/3)2= 2.68m

We see that since the mass of the resulting particle is greater than the sumof the colliding particles, some of their kinetic energy was converted into the


resulting particle’s rest-energy.

12. Since the electron and positron have the same mass and the same kineticenergies, their total energies must also be equal. Similarly, their momentawill be equal and opposite:

pp + pe = pµ1 + pµ2 = 0

Ep + Ee = 2Eµ

We now sum the energies of the electron and positron:

Ep + Ee = 2(10 GeV +mec2)

= 2(1000 MeV + .511 MeV )

= 2001.022 MeV

Each muon will possess 1000.511 MeV . Its kinetic energy will then be:

KE = 1000.511 MeV − 105.7 MeV

= 894.811 MeV

We may calculate its speed via (12.23):

894.86 MeV = (γ − 1)× 105.7 MeV

γ = 9.466

v = .99c

13. We consider the center of mass frame where the proton and antiprotonhave equal and opposite momenta. Since the proton and antiproton haveequal masses, their speeds must also be equal. Therefore, when we imposeconservation of energy and use eq. 12.9:

Ep + Ep = EX

where EX is the energy of the resulting particle. As stated above, we arein the center of mass frame, where the resulting particle is at rest and itsenergy is given by eq. 12.12:

2γmpc2 = mXc

2

129

2γ(938 MeV ) = 9700 MeV

γ = 5.17

We may then find the kinetic energies of the colliding particles by equation12.13:

KE = (γ − 1)mpc2 = 3912 MeV

Their velocities may be calculated by eq. 12.3:

γ = 5.17 =1√

1− u2/c2

1− v2

c2= .0374

v = .981c

The proton and antiproton will each have a speed of .981c in the center ofmass frame.

14a.) We start from the center of mass frame where we take the pion to beat rest. As it decays into the muon and neutrino, the following must be true:

pπ = pµ + pν = 0

Eπ = Eµ + Eν

p2µ = p2

ν

E2µ

c2−m2

µc2 =

E2ν

c2

Make the substitution Eν = Eπ − Eµ:

E2µ −m2

µc4 = E2

π + E2µ − 2E2

πE2µ

−m2µc

4 = m2πc

4 − 2m2πc

4Eµ

Eµ = 109.8 MeV

Now, we may find the speed of the muon via (12.13):

109.8 MeV = (γ − 1)mµc2


γ = 2.039

v = .87c

b.) Using the given mean lifetime of the muon and the speed found in part(a):

vt = .87(3× 108 m/s)(2.2× 10−6s)

= 574 m

15. First, we test the relation:

αiαj + αjαi = 2δijI

The Dirac-Pauli representation is given in eq. 12.42 and we merely substitute:

αiαj =

[0 σi

σi 0

] [0 σj

σj 0

]=

[σiσj 0

0 σiσj

]

αjαi =

[0 σj

σj 0

] [0 σi

σi 0

]=

[σjσi 0

0 σjσi

]αiαj + αjαi =

[σiσj + σjσi 0

0 σiσj + σjσi

]= (σiσj + σjσi)I

It can be shown that the Pauli matrices satisfy the following condition:

σiσj =

I if i = j−σjσi if i 6= j

Therefore, we see that:

σiσj + σjσi =

2 if i = j0 if i 6= j

Or, more concisely, σiσj + σjσi = 2δijI. Thus, αiαj + αjαi = 2δijI .

Now, we turn our attention to the relation: β2 = I. From eq. 12.42, weget the definition for β:

β =

[I 00 I

]

131

We carry out the multiplication:[I 00 I

] [I 00 I

]=

[I2 00 I2

]Since I2 = I, β2 = I.

17. Muons and electrons are both Dirac particles. The Feynman diagramsfor muon-electron scattering are equivalent to the Feynman diagram shownin Fig. 12.4 with one of the incoming and outgoing lines corresponding toan electron and the other corresponding to a muon.

13

Particle Physics - Solutions

1. a) This process can not occur as it would violate the conservation ofbaryon number and lepton number. The baryon number is one of the left-hand side but zero on the right. Likewise, the electron lepton number of theleft-hand side is zero while Le = −1 on the right.

b) This process can not occur due to violation of energy conservation. TheΣ has a rest energy of 1192 MeV while the Λ and the π have a total restenergy of 1250 MeV.

c) Conservations of charge, energy, baryon number, and lepton number areall observed, therefore the process may occur. Furthermore, the process con-serves strangeness and proceeds by the strong interaction.

d) Here, the conservation of muon lepton number is violated as the totalnumber on the left is 0, but is -1 on the right.

e) K− has a strange quantum number S=-1 where the π−π system hasS=0. Since charge, energy, lepton number, and baryon number are all con-served, this process proceeds via the weak interaction.

f) The ρ0, being a boson, will have a symmetric wave function. The spin of aρ0 will have a value of 1, therefore its total angular momentum will be J=1.By the conservation of angular momentum, J=1 for the right-hand side, thusthe spinless pions would have an antisymmetric total wave function. Alsobeing bosons, this is forbidden.

133

134 13. PARTICLE PHYSICS - SOLUTIONS

g) Strangeness is not conserved while the other conservation laws are, there-fore the process must occur via the weak interaction.

2.a.) This process can occur.b.) This process can occur only if the neutrino and antineutrino are of theelectron type.c.) This process is forbidden due to violation of lepton number conservation.d.) This process can occur.e.) This process can occur.f.) This process violates lepton number conservation and therefore may notproceed.g.) Strangeness is not conserved, therefore must proceed via weak interac-tion.

3. The rest energies may be found in tables 13.9 through 13.11.

a) Initial state: 140 + 938 = 1078 MeVFinal state: 940 + 135 = 1074 Mev

b) Initial state: 140 + 938 = 1116 MeVFinal state: 938 + 135 = 1073 Mev

c) Initial state: 140 + 938 = 1078 MeVFinal state: 494 + 1138 = 1632 Mev

The kinetic energy of this initial state must be greater than that of thefinal state to ensure conservation of energy.

4.b.) The νe has electron lepton number +1, while the e+ has electron leptonnumber of -1.c.) The sum of electron lepton numbers is 2 on the right-hand side but 0 on

135

the left-hand side.

5 To calculate the strangeness of these particles, we refer to tables 13.9-13.11 to find the quark content of each particle, then use eq. 13.26.

a) The Ω− has a quark content of sss, therefore using equation 13.26:

S = −[3] = −3

We repeat for the particles on the right-hand side to find that S(Λ) = −1and S(K−) = −1. Therefore ∆S = 1 and the process must occur via theweak interaction.

b) Using the same procedure as part (a), S(Σ) = −1, S(Λ) = −1, andS(γ) = 0. There is no change in strangeness, therefore the process may oc-cur via the electromagnetic, as we recognize the presence of the photon.

c) Using the same procedure as parts (a) and (b) above, S(Λ) = −1,S(p) = 0, and S(π−)=0. ∆S = 1 and the process must occur via the weakinteraction.

6.a.) ∆S = 1 This process may only occur via weak interaction.b.) No change in strangeness. May occur via strong, electromagnetic, orweak interaction.c.) ∆S = −1 This process may only occur via weak interaction.

7. a) This process may occur via the strong interaction as all conserva-tion laws are met.

b) The strangeness of the initial state is zero; however, the strangeness has avalue of -1 in the final state. The other conservation laws are observed, andtherefore this process must occur via the weak interaction.

c) This process is strictly forbidden as the baryon number of the initialstate is +1 while that of the final is zero. The Λ particle has baryon numberof +1, while the Σ antiparticle has a baryon number of -1.


d) This process is strictly forbiden as the muon lepton number of the an-tineutrino is -1 which is the total muon lepton number for the initial state.The final state has a muon lepton number zero. Note that the similar con-servation of electron lepton number is also broken.

e) Here again, lepton number conservation is violated. The electron neu-trino on the left-hand side has an electron lepton number of +1 whereas thepositron on the right-hand side has an electron lepton number of -1. As aconsequence, this process is strictly forbidden.

8. a.)Λ : uds p : uud π0 : uu, dd

b.)π− : du p : uud n : udd π0 : uu, dd

c.)π+ : ud p : uud K+ : us Σ+ : uus

9. The quark composition may be found in tables 13.9-13.11:

a)Ω− : sss Λ : uds K− : su

Here we have nothing but strange quarks in the initial state, but the finalstate contains one fewer s quark but does contain u, u, d quarks.

b)π− : du p : uud Λ : uds K : sd

There are no strange quarks in the inital state, but two in the final. Noticethat the antiquark in the left-hand meson is a u, while the antiquark thatappears in the meson that occurs on the right is an antidown quark.

c)p : uud K− : su Ξ− : dss K+ : us

There is no change in strangeness during this process. Even though thereis an additional strange quark on the right, there is also an s quark which

137

cancels the added strange quantum number.

10. a.)

π− : I = 1, I3 = −1 p : I = 12, I3 = 1

2

Possible Totals : I = 32, 1

2I3 = 3

2, −3

2, 1

2, −1

2

b.)

π− : I = 1, I3 = −1 n : I = 12, I3 = −1

2


2I3 = 3

2, −3

2, 1

2, −1

2

c.)

π+ : I = 1, I3 = 1 p : I = 12, I3 = 1

2


2I3 = 3

2, −3

2, 1

2, −1

2

d.)

n : I = 12, I3 = −1

2n : I = 1

2, I3 = −1

2

Possible Totals : I = 1, 0 I3 = 1, 0,−1

e.)

n : I = 12, I3 = −1

2p : I = 1

2, I3 = 1

2

Possible Totals : I = 1, 0 I3 = 1, 0,−1

11. If the wave has periodic boundary conditions at the two ends of theregion, then the length L must equal an integer multiple of the wavelength:

L = nλ⇒ λ =L

n

Using the de Broglie relation and this value of the wavelength:

p =h

λ=hn

L

Recall that ~ = h/2π therefore:

p =2π~nL


We can think of this relation in the following way. In momentum space, eachstate has a “length” of 2π~/L, therefore, the number of states in a differentiallength dp will be:

number of states =dp

2π~/L=

L

2π~dp (1)

If we now expand our region to include a cubic volume, extending to x = L,y = L, z = L from the origin, we now have three terms like (1) that form avolume element in a momentum space:

dN =L

2π~dpx

L

2π~dpy

L

2π~dpx

=V

(2π~)3d3p

12. a.)Λ : uds π− : du p : uud

b.)n : udd p : uud

c.)Ξ0 : uss Σ− : dds

d.)Ω− : sss Ξ0 : uss

13. We can use figure 13.29 as a guide to obtain the remaining weightsfrom the highest weight. Notice in figure 13.29(b) the highest weight is 3µ1,where µ1 is given by eq. 13.84. As we work along a diagonal such as the onewith the weights labeled in Fig. 13.29(b), we subtract an additional α1. Aswe work right-to-left along a row, we subtract (α1 + α2) for each step. Asfollows, we shall work from right to left across a row, and completing rowsfrom the top down to 3µ1 − 3α1. Expressions for α1 and α2 are given in eq.13.78:

3µ1 = 3

[1√3/3

]=

[3√3

]

139

3µ1 − α1 − α2 =

[2√3

]

3µ1 − 2α1 − 2α2 =

[1√3

]

3µ1 − 3α1 − 3α2 =

[0√3

]

3µ1 − α1 =

[5/2√3/2

]

3µ1 − 2α1 − α2 =

[3/2√3/2

]

3µ1 − 3α1 − 2α2 =

[5/2√3/2

]

3µ1 − 2α1 =

[20

]

3µ1 − 3α1 − α2 =

[10

]

3µ1 − 3α1 =

[3/2

−√

3/2

]

14

Nuclear Physics - Solutions

1. a) Recall from section 14.2 the nomenclature for various isotopes:

AZX

where A atomic mass number which is the sum of the number of neutronsand number of protons, while z is the atomic number which is the number ofprotons, and X is the element symbol. Therefore, the number of neutrons isequal to A minus Z:

37Li : 3 protons, 4 neutrons

6329Cu : 29 protons, 34 neutrons23892 U : 92 protons, 146 neutrons

2. There are 236 − 92 = 144 neutrons on the left side of the reaction,but on the right side there are only:

(90− 36) + (144− 56) = 142

Therefore, two neutrons are emitted.

3. Equation 14.3 provides an estimate of the radius of a nucleus:

R = 1.12A1/3 fm

141

142 14. NUCLEAR PHYSICS - SOLUTIONS

where A is the atomic mass number:

42He : R = 1.12(4)1/3 fm = 1.78fm168 He : R = 1.12(16)1/3 fm = 2.82fm5626Fe : R = 1.12(56)1/3 fm = 4.28fm20882 Fe : R = 1.12(208)1/3 fm = 6.64fm23793 Fe : R = 1.12(237)1/3 fm = 6.93fm

4. a.)Mass= 4.003B(2, 2) = 28.876 MeV

7.219 MeV/Nucleon

b.) Mass = 15.995B(8, 8) = 131.340 MeV

8.209 MeV/Nucleon

c.) Mass = 55.935B(30, 26) = 505.799 MeV

9.032 MeV/Nucleon

d.) Mass = 207.977

B(126, 82) = 1688.791 MeV

8.119 MeV/Nucleon

e.) Mass = 238.051

B(146, 92) = 1862.049 MeV

7.834 MeV/Nucleon

5. The semiempirical mass formula (14.6) will help us determine the bindingenergy of each nucleus, then the binding energy with one proton removed:a.)

B(2, 2)−B(2, 1) = 19.803 MeV

143

b.)

B(30, 26)−B(30, 25) = 9.230 MeV

c.)

B(126, 82)−B(126, 81) = 7.194 MeV

6. Once again, we use the semiempirical formula, but this time removingone neutron:a.)

B(2, 2)−B(1, 2) = 21.289 MeV

b.)

B(30, 26)−B(29, 26) = 12.056 MeV

c.)

B(126, 82)−B(125, 82) = 7.143 MeV

7. a.)

B(2, 2) = 20.952 MeV

b.)

B(8, 8) = 121.637 MeV

c.)

B(30, 26) = 487.170 MeV

d.)

B(126, 82) = 1624.748 MeV

e.)

B(146, 92) = 1806.076 MeV

8. The semi-empirical formula without the pairing term is as follows:

B(N,Z) = aA− bA2/3 − dZ2

A1/3− s(N − Z)2

A


If N=Z, we see that the fourth term immediately equals zero. Since we wantthe binding energy per nucleon, we can divide the entire equation by thenumber of nucleons, which will be equal to A:

B(N,Z)

A= a− ba−1/3 − dZ2

A4/3

Now we let Z=A/2:

Binding energy per nucleon = a− bA−1/3 − dA2

4A4/3

= a− bA−1/3 − d

4A2/3

Now, we differentiate with respect to A and set equal to zero:

dB

dA=

1

3A−4/3 − 1

2dA−1/3 = 0

Now solve for A:1

3A−1/3(bA−1 − 2

d) = 0

b

A=

2

d

A =2b

dThe critical points are A = 0 and A = 51.3. Per the verification calledfor in the problem, A/2 ≈ 26. Further testing of the value of the derivativein the intervals (0, 13) and (13,∞) further identifies A/2 = 26 as a maximum.

9. We use equation 14.13 first which relates the half-life to the propor-tionality constant λ:

t1/2 = 2 min = 120 s =ln2

λ

λ =ln2

120 s= .006 s−1

We now use the equation following eq. 14.12 and our initial condition R =1200 s−1 at t=0 to find N0:

R = 1200s−1 = .006N0e−.006(0) = (.006s−1)N0

145

N0 = 2× 105

Now we may use this same equation to find the transition rates at the sub-sequent times:

t = 4 min = 240 s : R = (.006s−1)(2× 105)e−.006(240)

R = 284 s−1

t = 6 min = 360 s : R = (.006s−1)(2× 105)e−.006(360)

R = 138 s−1

t = 8 min = 480 s : R = (.006s−1)(2× 105)e−.006(480)

R = 67 s−1

We see, just as expected, that the decay rate itself decays exponentially.

10. Use (14.13) to calculate λ, then (14.12) to calculate the fractional amountleft:

λ =ln 2

t1/2

= .0564yr−1

N

N0

= e−(.0564)(40)

= .105

11. Our strategy here should be to ultimately use the rate of decay equation(the one immediately following eq. 14.12) to find the number of 14

6 C atomsthat are responsible for the measured radiation in the sample. We may thenuse the atomic mass of carbon and Avogadro’s number to find the numberof total atoms in the one gram sample. Then will we have the necessary in-formation to calculate the proportion. First, since we are given the half-lifeof 14

6 C, we use eq. 14.13 to obtain the proportionality constant λ:

t1/2 =ln 2

λ⇒ λ =

ln 2

t1/2


λ =ln 2

5730 yr=

ln 2

3.012× 109 min= 2.30× 10−10 min−1

Now to bring this into the rate equation:

15.3 min−1 = (2.30× 10−10 min−1)N0 e−2.30×10−10 min−1

N0 = 6.65× 1010 atoms 146 C

We now calculate the total number of carbon atoms in the sample:

1 g · 1 mol

12 g· 6.022× 1023 atoms

1 mol= 5.018× 1022 atoms C

We now simply find our proportion:

6.65× 1010

5.018× 1022=

1.32× 10−12atoms 146 C

total atoms C

One check from Appendix B shows that this calculation agrees with the factthat the isotope of carbon containing 14 nucleons is not listed as a percent-age. This indicates its rarity and thus reliability for dating ancient relicssince these nuclear phenomena are readily measurable.

b) We use the value of λ calculated above as well as the value of N0 whichwould have been relevant at the time that the organism was alive. We con-vert 20,000 years to minutes and use our rate equation with this as the newtime:

20, 000 years = 1.051× 1010 minutes

R = (2.30× 10−10 min−1)(6.65× 1010 min) e−2.30×10−10 min−1(1.051×1010 min)

= 1.36 beta rays per minute

We see that there are many fewer counts in the ancient sample than in thecurrent one.

12. We complete the reactions and use the semiempirical formula whennecessary to calculate nuclear masses:a.)

20983 Bi→205

81 T l +42 He

147

∆E = 208.980u− 204.974u− 4.003u

= .003u

= 2.794 MeV

b.)23892 U →234

90 Th+42 He

∆E = 238.051u− 217959 MeV − 4.003u

= 54.868 MeV

c.)7736Kr →77

35 Br + e+ + νe

∆E = 71634.496 MeV − 71634.105 MeV − .511 Mev

= 2.881 MeV

d.)7735Br →77

34 Se+ e+ + νe

∆E = 71724.565 MeV − 71650.462 MeV − .511 Mev

= 73.592 MeV

13. We may calculate the release of energy of the 74Be nucleus by finding the

difference in binding energy between 74Be and 7

3He. Since the energy of 74He

is given in table 13.2 to be 39.25 MeV, we need just use the semi-empiricalformula (eq. 14.6) for 7

4Be:

B(3, 4) = 67a− b72/3 − 42d

71/3− s(−1)2

7− 0

= 97.899 MeV

We now subtract the two:

∆E = 97.899− 39.25 = 58.65 MeV

Thus the energy released is 58.65 MeV.


15. a) We begin by using equation 14.9 to calculate the nuclear massesof 92234U and 90230U :

m(92, 142) = 221710 MeV/c2

m(90, 140) = 217913 MeV/c2

Now, using the data from Appendix B, m(2, 2) = 3728.4 MeV/c2, and:

∆E = 221710 MeV − 217913 MeV − 3728.4 MeV = 68.44 MeV

b) The energy carried off by the alpha-particle will equal the difference inbinding energies of 92234U and 90230U . Using equation 14.6,

B(92, 142) = 1923.88 MeV

B(90, 140) = 1883.46 MeV

Thus, the alpha-particle carries off 40.42 MeV of energy.

17. We follow a very similar procedure to that used in section 14.5. However,we will use the following subscripts to represent the various particles in theprocess: D represents the deuteron, X∗ represents the excited nucleus, andp represents the proton. We begin with our statements of conservation ofmomentum:

pD = pp cosθ + Px

0 = pp sinθ + Py

Solving for Px and Py:Px = pD − pp cosθPx = −pp sinθ

The total kinetic energy before the collision is equal to the kinetic energy ofthe proton:

Ei =p2D

2mD

The total kinetic energy after the collision is:

Ef =p2p

2mp

+1

mX∗(pD − pp cosθ)2 +

1

2mX∗(pp sinθ)

2

149

Due to the conservation of energy, the energy absorbed by the nucleus willbe equal to the loss of kinetic energy:

E = Ei − Ef =p2D

2mD

−p2p

2mp

− 1

mX∗(p2D + p2

p − 2pDpp cosθ)

We now substitute the corresponding energies for the first two terms as wellas the following:

p2D

2= EDmD

p2p

2= Epmp

pDpD2

=√EDEpmDmp

Our expression for the energy of the excited nucleus due to deuteron strippingis now:

E = ED

(1− mD

mX∗

)− Ep

(1− mp

mX∗

)−

2√EDEpmDmp

mX∗cosθ

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