sup 2.pdf

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httpslidepdfcomreaderfullsup-2pdf 120

Supplement to Chapter 2

REVIEW QUESTIONS

21 For a spatially uniform ideal gas how is the velocity distribution f (v) defined

22 Derive the relation p = 23EV for an ideal gas

23 In deriving the MaxwellndashBoltzmann distribution how are microstates and macrostates defined

24 For a given set of occupation numbers N kl how many microstates correspond to that macrostate25 What is the assumption of ldquoequal a priori probabilityrdquo

26 What is the function that must be maximized in the derivation of the MaxwellndashBoltzmann distribution

27 What are the constraints on the set of variables in the function to be maximized in the previousquestion

28 Describe Lagrangersquos procedure for finding the maximum of a functionF (xyz) on the constraint surface g(xyz) = 1

29 Using the results of Questions 4ndash8 derive the MaxwellndashBoltzmann distribution

210 What is the form of the MaxwellndashBoltzmann distribution

211 For ideal gas particles in an external potential U (r) that is equal to zero at some location ro what isthe equilibrium particle density

212 For a uniform ideal gas of density n calculate the normalization constant C in the Maxwell velocitydistribution f (v) = C exp(minus12mv2kT )

213 How is the speed distribution F (v) related to the velocity distribution f (v)

214 How are the average speed v and vrms defined

215 For an ideal gas at temperature T what is vrms

216 How is the mean free path defined

217 For a quantum mechanical system of N particles in a three-dimensional potential U (r) what is theHamiltonian operator

218 How are the single-particle energy eigenfunctions defined

219 For a system of N BosendashEinstein particles without interactions describe the form of the N -particle

energy eigenfunctions220 What are the allowed values for the quantum state occupation numbers N k for BosendashEinstein parti-cles

221 For a system of N FermindashDirac particles without interactions describe the form of the N -particleenergy eigenfunctions

222 What are the allowed values for the quantum state occupation numbers N k for FermindashDirac particles

223 In deriving the BosendashEinstein and FermindashDirac distribution functions how were microstates andmacrostates defined

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SUPPLEMENT TO CHAPTER 2 REVIEW QUESTIONS FOR CHAPTER 2 348

224 How many ways can ν fermions be arranged in K quantum states

225 How many FermindashDirac microstates are there for a macrostate defined by the bin occupation numbersν 1 ν 2 if the kth bin has K k states

226 Using the result of the previous question derive the FermindashDirac distribution function f FD(ε) = (eα+βε+1)minus1

227 What is the form of the BosendashEinstein distribution function f BE(ε)

228 For a particle with no external potential in three dimensions write the Schrodinger energy eigenvalueequation

229 For a particle in a cubic box of side L what are ldquoperiodic boundary conditionsrdquo

230 What are the eigenfunctions and eigenvalues of the Schrodinger equation for a free particle withperiodic boundary conditions

231 For a particle in a cubic periodic box of side L calculate the number of energy eigenstates with energyless than E

232 Under what condition do the quantum mechanical distribution functions reduce to the Maxwell dis-tribution

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SUPPLEMENT TO CHAPTER 2 EXERCISE 22 349

EXERCISES

Exercise 21 Draw a realistic picture of the situation in a typical noble gas say argon at STP by drawingtwo spheres with diameters proportional to the ldquointeraction diametersrdquo of argon atoms separated by a dis-tance that is proportional with the same proportionality constant to the typical distance between particlesnamely d = 1n13 On the same scale how large is the mean free path

Fig S21

Solution At STP the particle density is 27

times1025mminus3 Thus d = 33

times10minus9m The Handbook of Chem-

istry and Physics gives an approximate diameter for Ar based on crystal structure of 3 1 A = 31times 10minus10mThus the ratio of the atomic diameter to d is about one to ten

The mean free path of hard spherical particles is calculated approximately in Problem 28 and accuratelyin Exercises 215 through 217 It is given by ℓ = 1(

radic 2 πD2n) where D is the diameter of the particles and

n is the particle density For argon at STP ℓ = 87times 10minus8 In Fig S21 in which d is represented by twoinches ℓ would be about 44 feet (134 m)

Exercise 22 A two-dimensional ideal gas is a collection of noninteracting particles that move on a planeSuppose a two-dimensional ideal gas of N particles with a total energy E is confined by one-dimensionalldquowallsrdquo to an area A The pressure on a wall is defined as the force per unit length Show that p = EA

v

vdt

x

y

Fig S22 All the particles with velocity v that are in the shaded region will strike the wall within the

time interval dt

Solution Let f (vx vy) dvx dvy be the number of particles per unit area that have velocities in the rangedvx dvy Consider a length ℓ of the container wall which we assume is in the y direction A particle whosevelocity is in the range dvx dvy will strike the wall in that length during the time interval dt if it is withinthe shaded region shown in Fig S22 The area of that region is ℓvx dt and therefore the number of suchparticles is

dN = ℓvx dt f (vx vy) dvx dvy (S21)

The momentum delivered to the section of wall by the rebounding particle is 2mvx Thus the total rate at

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which momentum is being delivered to that section of the wall is

dP

dt equiv F = 2ℓm

991787 vxgt0

f (vx vy)v2x dvx dvy

= ℓm

991787 f (vx vy)v2x dvx dvy

= ℓ991787

f (vx vy)983131

m2

(v2x + v2y)983133

dvx dvy

= ℓEA

(S22)

Exercise 23 Consider two very dilute gases of slightly different densities and temperatures separated bya thin wall with a small hole in it of area A Calculate the rate at which particles and energy are transferredthrough the hole as a function of the temperature and density differences to first order in ∆T and ∆n (SeeFig S23)

nLLT n RR

T

Fig S23 nL T L nR and T R are the densities and temperatures of the dilute gases on the two sides

Solution The rate at which particles pass through the hole from left to right is

R(L rarr R) = AnL(m2πkT L)32991787 vxgt0

eminusmv22kT Lvx d 3v (S23)

Using spherical coordinates with the polar axis in the x direction

R(Lrarr

R) = AnL(m2πkT L)322π 991787 π2

o

cos θ sin θ dθ 991787 infin

o

eminusmv22kT Lv3 dv

= πAnL(m2πkT L)32 12(2kT Lm)2

= 1radic

2πAnL(kT Lm)12

= 1

4nLvLA

(S24)

where vL is the average speed of the particles on the left given in Eq (230) The rate of transfer of particlesfrom right to left is

R(R rarr L) = AnRT

12Rradic

2πmk(S25)

Assuming that nL = n + ∆n2 T L = T + ∆T 2 nR = n

minus∆n2 and T R = T

minus∆T 2 the net rate of

particle transfer from left to right isdN R

dt = R(L rarr R) minus R(R rarr L)

= Aradic

2πmk

983131(n + ∆n2)(T + ∆T 2)12

minus (n minus∆n2)(T minus ∆T 2)12983133

asymp Aradic 2πmk

983080T 12∆n +

1

2

n

T 12∆T

983081(S26)

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A particle of speed v that goes through the hole carries an amount of energy mv22 Therefore the rate atwhich energy is carried through the hole from left to right can be obtained from Eq (S23) by putting afactor of mv22 in the integrand

RE (L rarr R) = AnL(m2πkT L)32991787 vxgt0

eminusmv22kT L(mv22)vx d 3v

= πmAnL(m2πkT L)32991787 π2

o

cos θ sin θ dθ middot991787 infin

o

eminusmv22kT Lv5 dv

= 1

2radic

πmAnL(2kT Lm)32 (S27)

=9830802k3

πm

98308112AnLT 32L

with a similar formula for RE (R rarr L) The net rate of transfer of energy is

dE Rdt

=9830802k3

πm

98308112A983131

(n + ∆n2)(T + ∆T 2)32 minus (n minus∆n2)(T minus ∆T 2)32983133

= 9830802k3

πm98308112

A1048616T 32∆n + 32nT 12∆T 1048617 (S28)

pump

Fig S24 A series of diffusion chambers each with many small holes through which the gas diffuses

Exercise 24 This exercise concerns a simplified model of the process of isotope separation by diffusionTwo chambers are separated by a thin wall containing many small holes as shown in Fig S24 In the leftchamber is a very dilute mixed ideal gas containing a density nA of particles of mass mA and a density nB

of particles of mass mB The right chamber is pumped so that its density can be approximated as zero (a)What is the ratio of the number of B particles to A particles in the gas that is being pumped out of the rightchamber (b) Assume that the gas being pumped out of the right chamber is being pumped into the leftchamber of a second diffusion stage similar to this one Assume that the process is repeated for 20 stagesthat nA = nB in the initial chamber and that mAmB = 08 What is nAnB in the final chamber

Solution (a) According to Eq (S23) the rate at which A particles pass through the wall is

dN Adt

= A

4 vAnA (S29)

where vA =radic

8kTπmA Similarly the rate at which B particles pass through the wall is

dN B

dt

= A

4

vBnB (S210)

The ratio of the number of B particles to that of A particles in the gas extracted from the right chamber is

dN Bdt

dN Adt =

vBnB

vAnA=983080mA

mB

98308112nB

nA(S211)

(b) In each stage the ratio nBnA is multiplied by the factor (mAmB)12 If nBnA is initially onethen after 20 stages

nB

nA=983080mA

mB

98308110= (8)10 = 0107 (S212)

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Exercise 25 In Section 22 we calculated the force per unit area exerted on the walls of a container filledwith an ideal gas by assuming that the walls were perfectly smooth so that each gas particle underwentsimple reflection when it hit the wall When we remember that the walls of a real container are made of atoms that are roughly the same size as the gas particles it becomes clear that the assumption that the wallsare smooth on an atomic scale is completely unrealistic Assuming that the piston shown in the figure is inequilibrium and taking as our system everything within the volume shown by the dashed line in Fig S25

use the momentum conservation law to derive the relation p = (23)EV without the unrealistic ldquosmoothwallrdquo assumption

F

x

Fig S25 A cylinder with a movable piston The system is everything within the dotted lines

Solution Our ldquosystemrdquo is everything that is contained in a cylindrical volume that includes the pistonplus a certain amount of the gas to the left of the piston At the left surface of the system particles carrymomentum into and out of the system Let us choose our x axis as shown in Fig S25 Then becausemomentum is a conserved quantity the x component of the momentum carried into the system by particlesentering from the left minus the x component of the momentum carried out of the system by particlesleaving at the left during a time interval ∆t must equal the momentum delivered to the system by the forcenamely F ∆t if the total momentum of the system is to remain constant If we take a small area ∆A on theimaginary surface that bounds the system on the left the number of particles entering the system through∆A during time interval ∆t that have velocities in the range d 3v centered at v is [f (v) d 3v]vx ∆A ∆tThe x component of momentum carried into the system by those particles is m[f (v) d 3v]v2x ∆A ∆t The x

momentum carried into the system by all the particles entering through ∆A is

m ∆A ∆t

991787 vxgt0

f (v)v2x d 3v (S213)

The total rate at which particles carry x momentum into the system at the left is

dP x(in)

dt = mA

991787 vxgt0

f (v)v2x d 3v (S214)

The particles that leave the system at the left carry with them negative x momentum since their values of vx must be negative The loss of negative P x counts as a gain of P x by the system It is easy to see that therate at which the system gains momentum by the loss of those particles is

dP x(out)dt

= mA991787 vxlt0

f (v)v2x d 3v (S215)

Combining the two effects we get

dP xdt

= mA

991787 f (v)v2x d 3v

= 23A

991787 f (v)

983080m

2 (v2x + v2y + v2z)

983081d 3v

= 23AEV

(S216)

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The force delivers negative P x to the system at a rate F Setting this equal to dP xdt gives the desiredresult Notice that in this analysis what goes on at the left surface of the piston is a process internal to thesystem and therefore of no relevance to the conservation of total momentum

Exercise 26 In computing the pressure by calculating the momentum carried by particles across animaginary surface we are ignoring another possible mode of momentum transfer across the surface Thatis that the particles on one side of the surface without moving across the surface can exert forces on theparticles on the other side due to the direct interparticle force In a dense system such as a liquid most of the pressure force across any surface is transmitted by interparticle forces rather than by particle transferIn fact in a solid where the particles are not free to move across the surface at all all the pressure force istransmitted in that way In order to estimate how much force is transmitted in a dilute gas by interparticleforces assume that the particles in a gas at STP are distributed randomly with a density n and have aninterparticle potential that depends on the separation of the particles r as

φ(r) = φoeminusr2a2 (S217)

where the range of the potential a = 2 A and φo = 1 eV Calculate the pressure transmitted by the in-terparticle force across a surface of area A and compare it with the kinetic pressure force (the momentumtransferred by moving particles)

Solution We choose as our surface dividing the gas into two parts the surface x = 0 We calculate thepotential energy at a point (minusxo 0 0) to the left of the surface due to all the particles that lie to the rightof the surface That is

φ(minusxo) =

991787 infin

o

dx

991787 infin

minusinfin

dy

991787 infin

minusinfin

dznφo exp983080minus(x + xo)2

a2 minus y2

a2 minus z2

a2

983081

= nφoπa2991787 infin

o

dx e(x+xo)2a2

(S218)

Letting u = (x + xo)a this can be written as

φ(minusxo) = nφoπa3991787 infin

xoa

eminusu2du (S219)

The force on a particle toward the left due to all the particles on the right is

F = minuspartφ(minusxo)

partxo= πnφoa2eminusx2

oa2 (S220)

The total force on all the particles on the left is the integral of F times the density of particles over all theregion on the left The y and z integrals give a factor of A the area of the surface separating the two halvesThus

F total = πn2φoa2A

991787 infin

o

eminusx2oa2dxo

= π32n2φoa3A

(S221)

The pressure exerted by the direct interparticle forces is

pint = F total

A = π32n2φoa3 (S222)

Putting in n = n(STP) = 27times 1025 m3 φo = 1 eV = 16times 10minus19 J and a = 2times 10minus10m we get pint =52times 103Nm2 But at STP pkin asymp 105Nm2 Thus pintpkin = 005

Exercise 27 The rate of a certain chemical reaction of the form A + A rarr A2 in an ideal gas is given by

R = αtimes983080 the density of pairs of particles with

cm kinetic energy greater than εo

983081 (S223)

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where R is the number of reactions per unit volume per second and α is some constant Write a one-dimensional integral expression for R as a function of particle density and temperature

Solution Consider a unit volume of the gas The number of particles with velocity in the range d 3v isf (v) d 3v where f (v) is the Maxwell velocity distribution function The number of pairs of particles wherethe first particle has velocity in d 3v and the second in d 3vprime is f (v)f (vprime) d 3v d 3vprime We must now transformfrom the absolute velocities of the two particles to center-of-mass and relative velocities defined by

V = (v + vprime)2 and u = v minus vprime (S224)

In terms of u and V the kinetic energy of the pair is

K = m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S225)

The term involving u is called the center-of-mass kinetic energy Thus

K cm = m

4 u2 (S226)

The significance of K cm is that K cm + φ(r) remains constant while the two particles interact where φ is theinteraction potential energy associated with the interaction force between the particles K cm is the energyavailable in the center-of-mass frame (where the two particles always have equal and opposite velocities) todo work against a repulsive two-particle interaction force

The number of pairs of particles within the unit volume with K cm gt εo is therefore

N = 1

2

991787 f (v)f (vprime)θ

983080m

4 u2 minus εo

983081d 3v d 3vprime (S227)

where θ is the unit step function The factor of 12 is necessary to compensate for the fact that the integralldquodouble countsrdquo the pairs of particles To see what is meant let us try to calculate the number of pairs of students in a class whose total ages add up to 39 years If ax is the age of student x (always an integer)then we might sum the quantity δ (ax + ay

minus39) where δ (0) = 1 and δ (n) = 0 for n

= 0 and x and y

independently take on the values of all the students in the class But this would count ldquoMary and Johnrdquoand ldquoJohn and Maryrdquo as two pairs although they are really only one

We must now transform the variables of integration from d 3v d 3vprime to d 3u d 3V Whenever integra-tion variables are changed one must include the Jacobian determinant of the transformation Accordingto Eq (S224) ux and V x are functions of vx and vprimex and similarly for the y and z components Thetransformation formula is

dvx dvprimex rarr |J | dux dV x (S228)

This is the same transformation that is shown in the Mathematical Appendix to have |J | = 1 Also

f (v)f (vprime) = n2(m2πkT )3 exp983080minusm(v2 + vprime2)

2kT

983081 (S229)

where m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S230)

Therefore

N = 12n2(m2πkT )3

991787 eminusmV 2kT d 3V

991787 θ983080m

4 u2 minus εo

983081eminusmu24kT d 3u (S231)

The first integral can be done easily

I 1 =

991787 eminusmV 2kT d 3V = (πkTm)32 (S232)

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The second integral can be reduced to a one-dimensional integral by transforming to spherical coordinatesin u space

I 2 =

991787 θ983080m

4 u2 minus εo

983081eminusmu24kT d 3u

= 4π

991787 infin

uo

eminusmu24kT u2 du(S233)

where u = 2radic

εom Introducing a variable x = (m4kT )12u with a lower limit xo = (m4kT )12uo =radic εokT we can write I 2 as

I 2 = 32π983080kT

m

98308132 991787 infinxo

eminusx2x2 dx (S234)

Putting these expressions for I 1 and I 2 into Eq (S231) we get

R = 8αn2radic

2π

991787 infin

xo


If kT is much smaller than the activation energy for the reaction εo then xo is large and because of thefactor of exp(minusx2) in the integrand R is very small

Exercise 28 Consider an ideal gas composed mostly of A particles of mass mA but with a low density of B particles of mass mB If a steady force is applied to the B particles but not to the A particles (one couldimagine that only the B particles carry an electric charge and there is a weak electric field present) thenthe B particles will drift through the gas at a speed that is proportional to the force

v = microF (S236)

The proportionality constant micro is called the mobility constant Because of the drift there will be a flux of the B particles equal to

φ = nBv = micronBF (S237)

In the absence of any external force if the density of B particles is not uniform (let us say that it depends onthe coordinate z) then there will be a net flux of the B particles from the regions of high B particle density

to those of low density This diffusion of B particles will eventually bring about a uniform equilibrium stateIt is observed that the flux is proportional to the density gradient

φ = minusDdnB

dz (S238)

The constant of proportionality D is called the diffusion constant It is reasonable (and correct) to assumethat if there are both a force and a density gradient then the net flux is given by the sum of the two termsMake that assumption and use the equilibrium distribution [Eq (220)] in a gravitational field to derive arelationship between the mobility constant micro and the diffusion constant D

Solution For a gravitational field pointing in the negative z direction the force on a B particle is F z =minusmBg The flux of B particles is given by

φ = minusmicronBmBg minus D dnB

dz (S239)

At equilibrium the flux must vanish giving the following differential equation for the equilibrium distribution

microgmBnB = minusDdnB

dz (S240)

This can be written asd[log nB(z)]

dz = minusmicrogmb

D equiv minus1

h (S241)

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which has a solution

nB(z) = nB(0)eminuszh (S242)

with h = DmicrogmB But the MaxwellndashBoltzmann formula for the scale height is h = kTmBg Thus wesee that in order for these two formulas to agree the diffusion constant and the mobility constant must berelated by

D = microkT (S243)This formula was first derived by A Einstein

The next few exercises involve calculations using single-velocity beams A single-velocity beam is ashower of particles all with velocity vo but distributed randomly throughout space with density no Firstfor simplicity we will consider two-dimensional single-velocity beams

Exercise 29 In two dimensions a smooth hard wall has a normal n and moves with velocity u Asingle-velocity beam is being reflected from the wall (Fig S26) What is the force per unit length on thewall

n

u

u

vo

Fig S26 n is the normal to the wall u is the velocity of the wall

Solution We transform to the inertial frame in which the wall is at rest In that frame the beam particleshave velocity vo minus u The rate at which they strike a unit length of the wall is equal to their density timesthe inward normal component of their velocity That is

Rate per length = minusnon middot (vo minus u) (S244)

Each particle has a change in momentum equal to minus2m[n middot (vo minus u)]n equiv ∆p The momentum delivered toa unit length of the wall is equal to ∆p times the collision rate This is defined as the force per unit lengthon the wall

dF

dℓ = minus2mno[n middot (vo minus u)]2n (S245)

As one would expect for a smooth wall it is a normal force One should note that this equation is valid onlyif in the wall frame the particles are actually directed at the wall That is if

n middot (vo minus u) lt 0 (S246)

Exercise 210 A heavy smooth hard circular particle of radius R moves with velocity u through a single-velocity beam (Fig S24) What is the net force on the particle

Solution We let θ be the angle that a point on the surface of the circle makes with an axis drawn antiparallelto the vector vo minus u Then n(θ) middot (vo minus u) = minus|vo minus u| cos θ The restriction that n middot (vo minus u) lt 0 says that

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Θ

n

v-uo

Fig S27 The circle moves with velocity u through a single-velocity beam in two dimensions

we should integrate θ from minusπ2 to π2

F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ

= minus2mno|vo minus u|2R991787 π2minusπ2

cos2 θ n(θ) dθ

(S247)

Certainly the net force F is parallel to vo minus u Therefore it is sufficient to calculate

F middot (vo minus u) = 2mno|vo minus u|3R991787 π2minusπ2

cos3 θ dθ

= 83mRno|vo minus u|3(S248)

orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)

Exercise 211 A disk moves at a very low velocity u = ux through a very dilute two-dimensional idealgas Calculate the net force on the disk

Solution The collection of particles with velocities within the range d 2v centered at v can be consideredas a single-velocity beam with a density f (v) d 2v where f (v) is the two-dimensional Maxwell distributionfunction

f (vx vy) = n(m2πkT )eminusmv22kT (S250)

The net force on the disk due to those particles is by Eq (S249)

dF = 83mRn(m2πkT )eminusmv22kT |v minus u|(v minus u) d 2v (S251)

The net force due to all the gas particles is obtained by integrating dF over all velocities

F = 4m2Rn

3πkT

991787 eminusmv22kT |v minus u|(v minus u) d 2v (S252)

We make a transformation to the variable V = v minusu and use the following approximations which are validfor very small u

v2 = (V + u)2 asymp V 2 + 2V middot u (S253)

eminusαv2 asymp eminusαV 2eminus2αVmiddotu asymp eminusαV 2(1 minus 2αV middot u) (S254)

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Then

F = 4m2Rn

3πkT

991787 eminusmV 22kT (1minus m

kT V middot u)|V|V d 2V (S255)

The first term in the parentheses vanishes by symmetry Since u is in the x direction it is clear that the netforce must have only an x component The value of that component is

F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2

2kT V 4 cos2 θ dV dθ

= minus2Rnradic

2πmkT u

(S256)

Exercise 212 By reviewing the solution to Exercise 29 it is easy to see that Eq (S245) is also valid inthree dimensions with the trivial modification that dℓ must be replaced by an element of area dA on thewall surface and no is the three-dimensional density of particles in the beam Use this fact to redo Exercise211 for the case of a heavy hard spherical particle in three dimensions as shown in Fig S28

v-u

φd

dθ

o

Fig S28 In three dimensions a hard sphere moves with velocity u through a single-velocity beam

Solution We construct a system of spherical coordinates with the polar axis in a direction opposite to thevector vo minus u The area on the sphere associated with the range of spherical angles dθ and dφ is equal todA = R2 sin θdθdφ The force dF imparted to that area by the particles in the single-velocity beam is givenby

dF

R2 sin θdθdφ = minus2mno[n middot (vo minus u)]2n (S257)

Again as in two dimensions n middot (vo minus u) = minus|vo minus u| cos θ The restriction that n middot (vo minus u) lt 0 restricts θto the range 0 lt θ lt π2 Therefore the total force is given by the integral

F = minus2mno|vo minus u|2R2991787 2πo

dφ

991787 π2o

dθ cos2 θ sin θ n(θ φ) (S258)

By viewing the process in the rest frame of the sphere it is obvious that the net force will be in the directionof vo minus u Thus we need only calculate

F middot (vo minus u) = minus2mno|vo minus u|3R2991787 2πo

dφ

991787 π2o

cos3 θ sin θ dθ

= minus4πmno|vo minus u|3R2983131cos4 θ

4

983133π2o

= πmnoR2|vo minus u|3

(S259)

which implies thatF = πmnoR2|vo minus u|(vo minus u) (S260)

a result that differs only slightly from the two-dimensional case

Exercise 213 What is the mobility constant of a hard sphere in a very dilute ideal gas

Solution The mobility constant is the constant micro in the equation u = microF relating the steady-state velocityof an object moving through a fluid to the force being exerted on the object But for steady-state motion the

8102019 sup 2pdf



external force must balance the average force that the fluid particles are exerting on the object Thereforewhat we have to calculate is the average force exerted by the particles of a very dilute ideal gas on a spheremoving through the gas at velocity u

We view the gas as being composed of a very large number of single-velocity beams The collectionof all the gas particles with velocities in the element d 3v forms a single-velocity beam of density f (v) d 3vThat beam exerts an average force on the sphere of

dF = πmR2f (v) d 3v |v minus u|(v minus u) (S261)

The total force on the sphere is

F = πmR2991787

f (v)|v minus u|(v minus u) d 3v (S262)

It is clear that the net force will be opposed to the velocity of the sphere u so that it is sufficient to calculate

F middot u = πmR2n(m2πkT )32991787

eminusmv22kT |v minus u|(v minus u) middot u d 3v (S263)

Transforming the variable of integration to the relative velocity V = v minus u using the approximations

eminusm(V+u)22kT

asympeminusmV 22kT (1

minusmV

middotukT ) (S264)

and introducing spherical coordinates with the polar axis antiparallel to u we can write the integral for F middotu

as

F middot u =2π2mR2n(m2πkT )32991787 πo

sin θ dθ

times991787 infin

o

V 2dV eminusmV 22kT 1048616

1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)

Because the integralint πo cos θ sin θdθ = 0 the first term in the parentheses vanishes Thus F middotu can be written

in terms of the integrals 991787 πo

cos2 θ sin θ dθ = 23 (S266)

and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)

when this is done one getsF = minus8

3(2πmkT )12R2nu (S268)

Note that from this and Einsteinrsquos relation [Eq (S243)] we can get the diffusion constant for hard spheresin a very dilute ideal gas

D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω

Fig S29 At what rate do collisions occur within the volume ω

8102019 sup 2pdf



v-v2 1

Fig S210 In the frame of the first beam the particles in the other beam have velocity v2 minus v1 The

length of the cylinder shown is |v2minusv1| dt Its cross-sectional area is πd 2

Exercise 214 Two single-velocity beams are passing through the same volume ω They have densities n1and n2 and velocities v1 and v2 The particles in both beams are hard spheres of diameter d At what ratedo collisions occur within the volume ω (See Fig S29)

Solution We observe the situation within the inertial frame of the first beam Then there is a densityn1 of stationary particles and a density n2 of particles with velocity v2 minus v1 The number of stationaryparticles in ω is n1ω The probability that one of those stationary particles will be hit within a time intervaldt is equal to the probability that the center of one of the moving particles occupies the cylinder of volumeπd 2

|v2

minusv1

|dt shown in Fig S210 That probability is n2πd 2

|v2

minusv1

|dt Thus the number of collisions

within ω during time interval dt isdN = ωn1n2πd 2|v2 minus v1| dt (S270)

and the rate of collisions within ω is

Rcoll = ωn1n2πd 2|v2 minus v1| (S271)

Notice that the rate is symmetric with respect to exchange of v1 and v2 which shows that we would haveobtained the same result by making the calculation in the rest frame of the second beam

Exercise 215 Within a volume ω in an ideal gas of spherical particles of diameter d at what rate docollisions occur

Solution Within ω the rate of collisions between pairs of particles of which one particle has velocity inthe range d 3v1 and the other has velocity in the range d 3v2 is

Rv1v2 = πωd 2f (v1) d 3v1 f (v2) d 3v2 |v1 minus v2| (S272)

By integrating v1 and v2 over their full ranges we obtain twice the total collision rate within ω because wecount collisions between particles of velocities v and vprime when v1 = v and v2 = vprime and also when v1 = vprime

and v2 = v Thus the total collision rate is

R = 12πωd 2991787

f (v1)f (v2)|v1 minus v2| d 3v1 d 3v2 (S273)

We transform to center-of-mass and relative velocities u = v1 minus v2 and V = (v1 + v2)2

f (v1)f (v2) = n2(m2πkT )3 exp[minusm(v21 + v22)2kT ]

= n2(m2πkT )3 exp983131minus983080m

4 u2 + mV 2

983081983087kT

983133 (S274)

With this transformation

R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V

991787 eminusmu24kT |u| d 3u

= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf



Exercise 216 In an ideal gas of density n at temperature T what is the total distance traveled by all theparticles within a volume ω during one second

Solution The number of particles is nω Their average speed is v =radic

8kTπm Therefore the totaldistance traveled by all particles together is

s = nωv = nωradic 8kTπm (S276)

Exercise 217 Given the results of the previous two exercises what is the average distance betweencollisions of a particle in an ideal gas

Fig S211 Each line represents the absolute distances traveled by one particle The dots are points

where the particle collided with another particle

Solution Let us imagine that Fig S211 shows a graph of the absolute distances traveled by each of theparticles in a volume ω in an ideal gas during one second For each particle a dot has been drawn at thelocation of each collision involving that particle Note that each collision creates two dots on the diagrambecause it involves two particles The total length of the lines is s = nω

radic 8kTπm The total number of

dots is 2R = 4ωd 2n2radic

πkTm Therefore the average distance between dots on the lines (that is the meanfree path of the particles) is

ℓ =

s

2R =

1

radic 2πd 2n (S277)

Exercise 218 An ideal gas of A particles contains a low density of B particles The diffusion constantfor the B particles defined in Eq (S238) is D Assume that the initial density of the B particles n(x 0)depends only on the variable x Derive a differential equation the diffusion equation for n(x t)

Solution Consider a cylinder of unit cross-sectional area whose axis is parallel to the x axis Let N (xo t)be the total number of B particles within the cylinder to the left of the plane x = xo By the meaning of the B particle flux it is clear that

partN (x t)

partt = minusJ (x t) = D

partn(x t)

partx (S278)

where J is the flux at position x at time t But by the definition of the particle density it is also true thatn(x t) = partN (x t)partx Differentiating Eq (S278) with respect to x and using the fact that part 2Npartxpartt =partnpartt gives the desired equation

partn

partt = D

part 2n

partx2 (S279)

Exercise 219 A Brownian particle is a particle in a fluid that is large enough to be observed in anordinary microscope (which means that it must be much larger and heavier than the fluid molecules) but isstill small enough that its gradual drift due to the impulses delivered to it by rebounding fluid particles isobservable in the same microscope The thermal motion of a truly macroscopic object is much too small to

8102019 sup 2pdf



be observable The resultant random drifting motion of such a particle is called Brownian motion Considera dilute mixture sometimes called a suspension of identical Brownian particles in water Assume that theparticles in water are known to have a diffusion constant D A particle begins at time 0 with x coordinate0 What is the probability distribution for its x coordinate at time t

Solution Imagine a system of pure water with a large collection of the Brownian particles randomlydistributed over the plane x = 0 The initial density of Brownian particles would be

n(x 0) = C δ (x) (S280)

where C is some positive constant equal to the two-dimensional density of the particles in the yndashz planeAfter time zero the density of Brownian particles will satisfy the diffusion equation

partn(x t)

partt = D

part 2n(x t)

partx2 (S281)

We are now faced with a mathematical problem of finding a solution of the diffusion equation that satisfiesthe initial condition given by Eq (S280) In order to guess the form of the solution it is best to thinkabout the physics of the Brownian motion process The particle drifts a noticeable amount during a giventime interval because it is being knocked randomly left and right by individually unnoticeable amounts very

many times during the time interval Its final x displacement is the sum of a very large number of separate xdisplacements It is natural to guess that the probability distribution for an individual particlersquos x coordinatewill be a Gaussian function of the form predicted by the central limit theorem The density of particles attime t will certainly be proportional to the probability density for an individual particle Therefore we trya solution of the form

n(x t) = C radic

γ (t)π eminusγ (t)x2 (S282)

where γ (t) is as yet unknown The square root factor is necessary in order to maintain normalization Sincethe particles are conserved it is necessary that

991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)

Putting this form into the differential equation we get

(12γ minus12 minus γ 12x2)dγ

dt = minus4D( 12γ 32 minus γ 52x2)

= minus4D( 12γ minus12 minus γ 12x2)γ 2

which would be satisfied if dγdt

γ 2 = minusd(1γ )

dt = minus4D (S285)

or 1γ = 4Dt Thus the solution of Eq (S281) is

n(x t) = C radic

4πDteminusx24Dt (S286)

The probability density for a single particle differs from the particle density for the ensemble of Brownianparticles only by its normalization Therefore

P (x t) = 1radic


Historical note In the year 1828 the botanist Robert Brown detected in a suspension of fine pollengrains in water an irregular ldquoswarming motionrdquo He attributed it to a somewhat mystical primitive life forceA long period of controversy followed regarding the question of whether the Brownian motion was a genuineeffect or was only an artifact of his experiment due to vibrations of the microscope currents in the water

8102019 sup 2pdf



caused by evaporation etc By 1905 when Einsteinrsquos theoretical explanation and mathematical analysis of the effect appeared it had been confirmed unambiguously that an irregular diffusive motion with no apparentenergy source continued indefinitely within a closed system of small suspended particles Stimulated byEinsteinrsquos analysis a number of experimental physicists but most notably the French physicist Perrin madedetailed quantitative experimental studies of Brownian motion These studies had two important effects

1 They completely confirmed the fundamental concept which was controversial at that time that heat

was simply microscopic kinetic energy2 They allowed using Einsteinrsquos theoretical analysis a determination for the first time of the mass of

individual molecules and from that of Avogadrorsquos numberA more recent study (see Scientific American August 1991) has concluded that with Brownrsquos experi-

mental techniques the size and density of pollen grains and the microscopes available at the time of Brownrsquosexperiments Brown could not possibly have seen what is now called Brownian motion What Brown sawreally was an artifact of his experimental technique He has immortalized his name by a lucky fluke

Exercise 220 A harmonic oscillator potential is a potential of the form U (x) = 12kx2 A single particleof mass m in a harmonic oscillator potential has quantum mechanical energy levels

E n = hω(n + 12) (S288)

where n = 0 1 2 and ω =radic

km which is the angular frequency of vibration of a classical particle in thesame potential What is the energy and degeneracy of the ground state of a system of five noninteractingparticles in a harmonic oscillator potential in the cases that (a) the particles are spin-0 bosons (b) theparticles are spin-12 fermions (c) the particles are spin- 12 bosons and (d) the particles are spin-0 fermionsNote cases (c) and (d) are known to be impossible (See Fig S212)

Solution (a) For bosons any number of particles may occupy a given single-particle quantum state Thefive-particle ground state is obtained by putting all the particles in the single-particle state with quantumnumber n = 0 Thus E g = 52 hω The ground-state is clearly unique that is it has degeneracy one (b) Forspin-12 fermions each single-particle state can hold two particles of opposite spin The spin of the particlein the n = 2 state can be either up or down giving the five-particle ground state a degeneracy of two Theground-state energy is E g = 2(hω2)+2(3hω2)+(5hω2) = 13hω2 (c) This has the same energy as case(a) but the number of spin-up particles can vary from 0 to 5 giving the ground state a degeneracy of 6 (d)

Only one particle can go into each single-particle state Thus E g = 25hω2 and the state is nondegenerate

n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)

Fig S212 The ground-state configurations in the four cases

Exercise 221 For particles in a one-dimensional hard-walled box of length π the single-particle energyeigenstates are un(x) =

radic 2π sin(nx) where n = 1 2 If we assume that the particles are bosons and

ignore the problem of normalization what is the explicit form of the three-particle wave function associatedwith the occupation numbers N 1 = 2 N 2 = 1 N 3 = N 4 = middot middot middot = 0

Solution

ψ(x1 x2 x3) = Asumperm

u1(xi)u1(xj)u2(xk) (S289)

8102019 sup 2pdf



where A is a normalization constant and in the sum (ijk) take the values (123) (132) (312) (321)(231) and (213) The result is

ψ =A(sin(x1)sin(x2) sin(2x3) + sin(x1)sin(x3) sin(2x2)

+ sin(x2)sin(x3) sin(2x1)) (S290)

Exercise 222 For a particle of mass m in a two-dimensional periodic box of area L2 (a) how manysingle-particle quantum states are there with energies less than E (b) What is the density of single-particlequantum states as a function of energy (c) If m = me and L = 1 cm what is the numerical value of thedensity of single-particle quantum states

Solution (a) The Schrodinger energy equation for the system is

minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)

with the periodic boundary conditions

u(0 y) = u(L y) and u(x 0) = u(x L) (S292)

The solutions of this equation are plane waves of the form

u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)

and K 1 and K 2 are integers The corresponding energy eigenvalue is

E (k1 k2) = h2(k21 + k22)

2m (S295)

If we plot these allowed wave vectors on a two-dimensional k plane they form a square lattice with a spacingof 2πL (see Fig 25) For a given value of energy E the states with energy less than E all fall within acircle whose radius is k =

radic 2mEh The number of such states is

N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)

(b) The density of eigenstates isdN (E )

dE =

mL2

2πh2 (S297)

(c) For m = me and L = 10minus2m dN (E )dE = 13times 1033 statesjoule

Exercise 223 1200 particles are to be distributed among three energy states with energies ε1 = 1 eV

ε2 = 2 eV and ε3 = 3 eV Assume that the total energy is 2400 eV and that each possible microstateis equally probable (a) What is the probability distribution for the number of particles in state 1 if theparticles are distinguishable (b) What is the probability distribution for the number of particles in state 1if the particles are BosendashEinstein particles

Solution (a) For distinguishable particles a microstate is defined by giving the energy value of eachparticle That is E 1 E 2 E 1200 where E k can be ε1 ε2 or ε3 Let N 1 N 2 and N 3 represent the numberof particles in each energy state The number of microstates with given values of N 1 N 2 and N 3 is

K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf



The restriction that E = 2400 eV will be satisfied if and only if N 3 = N 1 But also N 1 + N 2 + N 3 = 1200Thus only N 1 is independent For any N 1 between 0 and 600

N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)

When we take the restrictions into account the number of microstates with a given value of N 1 is

K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)

To find the most likely value of N 1 we let

F (N 1) = log K (N 1) asymp1200 log(1200)minus 2N 1 log N 1

minus (1200minus 2N 1) log(1200minus 2N 1) (S2101)

and set dF (N 1)dN 1 = 0 getting

minus2log N 1 + 2 log(1200minus 2N 1) = 0 (S2102)

or (1200

minus2N 1)N 1 = 1 whose solution is N 1 = 12003 = 400

We will calculate P (N 1)P (400) = exp[F (N 1) minus F (400)] A bit of algebra and use of the expansionlog(1 + ε) asymp εminus ε22 will show that

F (400 + n) minus F (400) asymp minus3n2400 (S2103)

orP (400 + n)

P (400) = eminus3n

2400 (S2104)

(b) For BosendashEinstein particles specifying N 1 N 2 and N 3 completely specifies the 1200-particle quantumstate But Eq (S299) gives N 2 and N 3 in terms of N 1 For any N 1 between 0 and 600 there is exactly onemicrostate and therefore all values of N 1 in that range are equally probable

Exercise 224 What is the maximum value of the function

F (x1 x2 x10) =10sumn=1

n x2n (S2105)

subject to the constraint10sumn=1

xn = 1 (S2106)

Solution Using Lagrangersquos method we look at the function

G =

10sumn=1

n x2n minus λ10sumn=1

xn (S2107)

The unconstrained maximum of G is given by

partG

partxn= 2nxn minus λ = 0 (S2108)

or xn = λ2n The value of λ is determined by the constraint equation

10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)

Exercise 225 In Exercise 222 it was shown that the number of single-particle quantum states of a two-dimensional quantum gas with energies within the interval dE is equal to (mL22πh2) dE (a) What is thespatial density of particles in a two-dimensional FermindashDirac gas as an explicit function of temperature andaffinity (b) For particles of electronic mass and T = 300 K what value of α will give a density of 1 particleper square Angstrom

Solution (a) For given values of α and β = 1kT the average number of particles in a single quantumstate of energy ε is (eα+βε + 1)minus1 The number of quantum states in energy interval dε is (mL22πh2) dεThus the average number of particles in all quantum states is

N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)

Changing the integration variable to x = eminusβε and dx = minusβeminusβε dε we can write N as

N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667

log(1 + eα) minus α1048669

(S2113)

The spatial density of particles is NL2 or

n = mkT

2πh21048667


(S2114)

(b) Setting n = 1020 particlesm2 and T = 300 K we getlog(1 + eα) minus α = 1821 (S2115)

The function on the left becomes large for large negative values of α If α is large and negative then eα isvery small and

log(1 + eα) minus α asymp eα minus α asymp minusα (S2116)

Therefore α asymp minus1821

Exercise 226 What is the density n as a function of T and α if the particles of the previous question arebosons

Solution For BosendashEinstein particles Eq (S2112) must be replaced by

N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2

983131minus log(eα minus x)

9831331o

= mkTL2

2πh21048667

α minus log(eα minus 1)1048669

(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf


SUPPLEMENT TO CHAPTER 2 REVIEW QUESTIONS FOR CHAPTER 2 348

224 How many ways can ν fermions be arranged in K quantum states

225 How many FermindashDirac microstates are there for a macrostate defined by the bin occupation numbersν 1 ν 2 if the kth bin has K k states

226 Using the result of the previous question derive the FermindashDirac distribution function f FD(ε) = (eα+βε+1)minus1

227 What is the form of the BosendashEinstein distribution function f BE(ε)

228 For a particle with no external potential in three dimensions write the Schrodinger energy eigenvalueequation

229 For a particle in a cubic box of side L what are ldquoperiodic boundary conditionsrdquo

230 What are the eigenfunctions and eigenvalues of the Schrodinger equation for a free particle withperiodic boundary conditions

231 For a particle in a cubic periodic box of side L calculate the number of energy eigenstates with energyless than E

232 Under what condition do the quantum mechanical distribution functions reduce to the Maxwell dis-tribution

8102019 sup 2pdf



EXERCISES


Fig S21









v

vdt

x

y


time interval dt




8102019 sup 2pdf




dP

dt equiv F = 2ℓm

991787 vxgt0


= ℓm


= ℓ991787

f (vx vy)983131

m2

(v2x + v2y)983133

dvx dvy

= ℓEA

(S22)


nLLT n RR

T






R(Lrarr

R) = AnL(m2πkT L)322π 991787 π2

o


o

eminusmv22kT Lv3 dv


= 1radic

2πAnL(kT Lm)12

= 1

4nLvLA

(S24)


R(R rarr L) = AnRT

12Rradic

2πmk(S25)






= Aradic

2πmk

983131(n + ∆n2)(T + ∆T 2)12


asymp Aradic 2πmk

983080T 12∆n +

1

2

n

T 12∆T

983081(S26)

8102019 sup 2pdf







o


o

eminusmv22kT Lv5 dv

= 1

2radic


=9830802k3

πm

98308112AnLT 32L


dE Rdt

=9830802k3

πm

98308112A983131


= 9830802k3

πm98308112

A1048616T 32∆n + 32nT 12∆T 1048617 (S28)

pump





dN Adt

= A

4 vAnA (S29)

where vA =radic


dN B

dt

= A

4

vBnB (S210)


dN Bdt

dN Adt =

vBnB

vAnA=983080mA

mB

98308112nB

nA(S211)


nB

nA=983080mA

mB

98308110= (8)10 = 0107 (S212)

8102019 sup 2pdf





F

x




m ∆A ∆t

991787 vxgt0

f (v)v2x d 3v (S213)


dP x(in)

dt = mA

991787 vxgt0

f (v)v2x d 3v (S214)


dP x(out)dt

= mA991787 vxlt0

f (v)v2x d 3v (S215)


dP xdt

= mA

991787 f (v)v2x d 3v

= 23A

991787 f (v)

983080m

2 (v2x + v2y + v2z)

983081d 3v

= 23AEV

(S216)

8102019 sup 2pdf








φ(minusxo) =

991787 infin

o

dx

991787 infin

minusinfin

dy

991787 infin

minusinfin


a2 minus y2

a2 minus z2

a2

983081


o

dx e(x+xo)2a2

(S218)



xoa

eminusu2du (S219)




oa2 (S220)



991787 infin

o

eminusx2oa2dxo

= π32n2φoa3A

(S221)


pint = F total

A = π32n2φoa3 (S222)





983081 (S223)

8102019 sup 2pdf







K = m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S225)


K cm = m

4 u2 (S226)



N = 1

2


983080m

4 u2 minus εo

983081d 3v d 3vprime (S227)



= 0 and x and y






2kT

983081 (S229)

where m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S230)

Therefore

N = 12n2(m2πkT )3


991787 θ983080m

4 u2 minus εo



I 1 =


8102019 sup 2pdf




I 2 =

991787 θ983080m

4 u2 minus εo


= 4π

991787 infin

uo


where u = 2radic


I 2 = 32π983080kT

m

98308132 991787 infinxo



R = 8αn2radic

2π

991787 infin

xo




v = microF (S236)





φ = minusDdnB

dz (S238)




dz (S239)



dz (S240)


dz = minusmicrogmb

D equiv minus1

h (S241)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf



EXERCISES


Fig S21









v

vdt

x

y


time interval dt




8102019 sup 2pdf




dP

dt equiv F = 2ℓm

991787 vxgt0


= ℓm


= ℓ991787

f (vx vy)983131

m2

(v2x + v2y)983133

dvx dvy

= ℓEA

(S22)


nLLT n RR

T






R(Lrarr

R) = AnL(m2πkT L)322π 991787 π2

o


o

eminusmv22kT Lv3 dv


= 1radic

2πAnL(kT Lm)12

= 1

4nLvLA

(S24)


R(R rarr L) = AnRT

12Rradic

2πmk(S25)






= Aradic

2πmk

983131(n + ∆n2)(T + ∆T 2)12


asymp Aradic 2πmk

983080T 12∆n +

1

2

n

T 12∆T

983081(S26)

8102019 sup 2pdf







o


o

eminusmv22kT Lv5 dv

= 1

2radic


=9830802k3

πm

98308112AnLT 32L


dE Rdt

=9830802k3

πm

98308112A983131


= 9830802k3

πm98308112

A1048616T 32∆n + 32nT 12∆T 1048617 (S28)

pump





dN Adt

= A

4 vAnA (S29)

where vA =radic


dN B

dt

= A

4

vBnB (S210)


dN Bdt

dN Adt =

vBnB

vAnA=983080mA

mB

98308112nB

nA(S211)


nB

nA=983080mA

mB

98308110= (8)10 = 0107 (S212)

8102019 sup 2pdf





F

x




m ∆A ∆t

991787 vxgt0

f (v)v2x d 3v (S213)


dP x(in)

dt = mA

991787 vxgt0

f (v)v2x d 3v (S214)


dP x(out)dt

= mA991787 vxlt0

f (v)v2x d 3v (S215)


dP xdt

= mA

991787 f (v)v2x d 3v

= 23A

991787 f (v)

983080m

2 (v2x + v2y + v2z)

983081d 3v

= 23AEV

(S216)

8102019 sup 2pdf








φ(minusxo) =

991787 infin

o

dx

991787 infin

minusinfin

dy

991787 infin

minusinfin


a2 minus y2

a2 minus z2

a2

983081


o

dx e(x+xo)2a2

(S218)



xoa

eminusu2du (S219)




oa2 (S220)



991787 infin

o

eminusx2oa2dxo

= π32n2φoa3A

(S221)


pint = F total

A = π32n2φoa3 (S222)





983081 (S223)

8102019 sup 2pdf







K = m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S225)


K cm = m

4 u2 (S226)



N = 1

2


983080m

4 u2 minus εo

983081d 3v d 3vprime (S227)



= 0 and x and y






2kT

983081 (S229)

where m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S230)

Therefore

N = 12n2(m2πkT )3


991787 θ983080m

4 u2 minus εo



I 1 =


8102019 sup 2pdf




I 2 =

991787 θ983080m

4 u2 minus εo


= 4π

991787 infin

uo


where u = 2radic


I 2 = 32π983080kT

m

98308132 991787 infinxo



R = 8αn2radic

2π

991787 infin

xo




v = microF (S236)





φ = minusDdnB

dz (S238)




dz (S239)



dz (S240)


dz = minusmicrogmb

D equiv minus1

h (S241)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf




dP

dt equiv F = 2ℓm

991787 vxgt0


= ℓm


= ℓ991787

f (vx vy)983131

m2

(v2x + v2y)983133

dvx dvy

= ℓEA

(S22)


nLLT n RR

T






R(Lrarr

R) = AnL(m2πkT L)322π 991787 π2

o


o

eminusmv22kT Lv3 dv


= 1radic

2πAnL(kT Lm)12

= 1

4nLvLA

(S24)


R(R rarr L) = AnRT

12Rradic

2πmk(S25)






= Aradic

2πmk

983131(n + ∆n2)(T + ∆T 2)12


asymp Aradic 2πmk

983080T 12∆n +

1

2

n

T 12∆T

983081(S26)

8102019 sup 2pdf







o


o

eminusmv22kT Lv5 dv

= 1

2radic


=9830802k3

πm

98308112AnLT 32L


dE Rdt

=9830802k3

πm

98308112A983131


= 9830802k3

πm98308112

A1048616T 32∆n + 32nT 12∆T 1048617 (S28)

pump





dN Adt

= A

4 vAnA (S29)

where vA =radic


dN B

dt

= A

4

vBnB (S210)


dN Bdt

dN Adt =

vBnB

vAnA=983080mA

mB

98308112nB

nA(S211)


nB

nA=983080mA

mB

98308110= (8)10 = 0107 (S212)

8102019 sup 2pdf





F

x




m ∆A ∆t

991787 vxgt0

f (v)v2x d 3v (S213)


dP x(in)

dt = mA

991787 vxgt0

f (v)v2x d 3v (S214)


dP x(out)dt

= mA991787 vxlt0

f (v)v2x d 3v (S215)


dP xdt

= mA

991787 f (v)v2x d 3v

= 23A

991787 f (v)

983080m

2 (v2x + v2y + v2z)

983081d 3v

= 23AEV

(S216)

8102019 sup 2pdf








φ(minusxo) =

991787 infin

o

dx

991787 infin

minusinfin

dy

991787 infin

minusinfin


a2 minus y2

a2 minus z2

a2

983081


o

dx e(x+xo)2a2

(S218)



xoa

eminusu2du (S219)




oa2 (S220)



991787 infin

o

eminusx2oa2dxo

= π32n2φoa3A

(S221)


pint = F total

A = π32n2φoa3 (S222)





983081 (S223)

8102019 sup 2pdf







K = m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S225)


K cm = m

4 u2 (S226)



N = 1

2


983080m

4 u2 minus εo

983081d 3v d 3vprime (S227)



= 0 and x and y






2kT

983081 (S229)

where m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S230)

Therefore

N = 12n2(m2πkT )3


991787 θ983080m

4 u2 minus εo



I 1 =


8102019 sup 2pdf




I 2 =

991787 θ983080m

4 u2 minus εo


= 4π

991787 infin

uo


where u = 2radic


I 2 = 32π983080kT

m

98308132 991787 infinxo



R = 8αn2radic

2π

991787 infin

xo




v = microF (S236)





φ = minusDdnB

dz (S238)




dz (S239)



dz (S240)


dz = minusmicrogmb

D equiv minus1

h (S241)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf







o


o

eminusmv22kT Lv5 dv

= 1

2radic


=9830802k3

πm

98308112AnLT 32L


dE Rdt

=9830802k3

πm

98308112A983131


= 9830802k3

πm98308112

A1048616T 32∆n + 32nT 12∆T 1048617 (S28)

pump





dN Adt

= A

4 vAnA (S29)

where vA =radic


dN B

dt

= A

4

vBnB (S210)


dN Bdt

dN Adt =

vBnB

vAnA=983080mA

mB

98308112nB

nA(S211)


nB

nA=983080mA

mB

98308110= (8)10 = 0107 (S212)

8102019 sup 2pdf





F

x




m ∆A ∆t

991787 vxgt0

f (v)v2x d 3v (S213)


dP x(in)

dt = mA

991787 vxgt0

f (v)v2x d 3v (S214)


dP x(out)dt

= mA991787 vxlt0

f (v)v2x d 3v (S215)


dP xdt

= mA

991787 f (v)v2x d 3v

= 23A

991787 f (v)

983080m

2 (v2x + v2y + v2z)

983081d 3v

= 23AEV

(S216)

8102019 sup 2pdf








φ(minusxo) =

991787 infin

o

dx

991787 infin

minusinfin

dy

991787 infin

minusinfin


a2 minus y2

a2 minus z2

a2

983081


o

dx e(x+xo)2a2

(S218)



xoa

eminusu2du (S219)




oa2 (S220)



991787 infin

o

eminusx2oa2dxo

= π32n2φoa3A

(S221)


pint = F total

A = π32n2φoa3 (S222)





983081 (S223)

8102019 sup 2pdf







K = m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S225)


K cm = m

4 u2 (S226)



N = 1

2


983080m

4 u2 minus εo

983081d 3v d 3vprime (S227)



= 0 and x and y






2kT

983081 (S229)

where m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S230)

Therefore

N = 12n2(m2πkT )3


991787 θ983080m

4 u2 minus εo



I 1 =


8102019 sup 2pdf




I 2 =

991787 θ983080m

4 u2 minus εo


= 4π

991787 infin

uo


where u = 2radic


I 2 = 32π983080kT

m

98308132 991787 infinxo



R = 8αn2radic

2π

991787 infin

xo




v = microF (S236)





φ = minusDdnB

dz (S238)




dz (S239)



dz (S240)


dz = minusmicrogmb

D equiv minus1

h (S241)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf





F

x




m ∆A ∆t

991787 vxgt0

f (v)v2x d 3v (S213)


dP x(in)

dt = mA

991787 vxgt0

f (v)v2x d 3v (S214)


dP x(out)dt

= mA991787 vxlt0

f (v)v2x d 3v (S215)


dP xdt

= mA

991787 f (v)v2x d 3v

= 23A

991787 f (v)

983080m

2 (v2x + v2y + v2z)

983081d 3v

= 23AEV

(S216)

8102019 sup 2pdf








φ(minusxo) =

991787 infin

o

dx

991787 infin

minusinfin

dy

991787 infin

minusinfin


a2 minus y2

a2 minus z2

a2

983081


o

dx e(x+xo)2a2

(S218)



xoa

eminusu2du (S219)




oa2 (S220)



991787 infin

o

eminusx2oa2dxo

= π32n2φoa3A

(S221)


pint = F total

A = π32n2φoa3 (S222)





983081 (S223)

8102019 sup 2pdf







K = m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S225)


K cm = m

4 u2 (S226)



N = 1

2


983080m

4 u2 minus εo

983081d 3v d 3vprime (S227)



= 0 and x and y






2kT

983081 (S229)

where m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S230)

Therefore

N = 12n2(m2πkT )3


991787 θ983080m

4 u2 minus εo



I 1 =


8102019 sup 2pdf




I 2 =

991787 θ983080m

4 u2 minus εo


= 4π

991787 infin

uo


where u = 2radic


I 2 = 32π983080kT

m

98308132 991787 infinxo



R = 8αn2radic

2π

991787 infin

xo




v = microF (S236)





φ = minusDdnB

dz (S238)




dz (S239)



dz (S240)


dz = minusmicrogmb

D equiv minus1

h (S241)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf








φ(minusxo) =

991787 infin

o

dx

991787 infin

minusinfin

dy

991787 infin

minusinfin


a2 minus y2

a2 minus z2

a2

983081


o

dx e(x+xo)2a2

(S218)



xoa

eminusu2du (S219)




oa2 (S220)



991787 infin

o

eminusx2oa2dxo

= π32n2φoa3A

(S221)


pint = F total

A = π32n2φoa3 (S222)





983081 (S223)

8102019 sup 2pdf







K = m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S225)


K cm = m

4 u2 (S226)



N = 1

2


983080m

4 u2 minus εo

983081d 3v d 3vprime (S227)



= 0 and x and y






2kT

983081 (S229)

where m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S230)

Therefore

N = 12n2(m2πkT )3


991787 θ983080m

4 u2 minus εo



I 1 =


8102019 sup 2pdf




I 2 =

991787 θ983080m

4 u2 minus εo


= 4π

991787 infin

uo


where u = 2radic


I 2 = 32π983080kT

m

98308132 991787 infinxo



R = 8αn2radic

2π

991787 infin

xo




v = microF (S236)





φ = minusDdnB

dz (S238)




dz (S239)



dz (S240)


dz = minusmicrogmb

D equiv minus1

h (S241)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf







K = m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S225)


K cm = m

4 u2 (S226)



N = 1

2


983080m

4 u2 minus εo

983081d 3v d 3vprime (S227)



= 0 and x and y






2kT

983081 (S229)

where m

2 (v2 + vprime2) =

m

4 u2 + mV 2 (S230)

Therefore

N = 12n2(m2πkT )3


991787 θ983080m

4 u2 minus εo



I 1 =


8102019 sup 2pdf




I 2 =

991787 θ983080m

4 u2 minus εo


= 4π

991787 infin

uo


where u = 2radic


I 2 = 32π983080kT

m

98308132 991787 infinxo



R = 8αn2radic

2π

991787 infin

xo




v = microF (S236)





φ = minusDdnB

dz (S238)




dz (S239)



dz (S240)


dz = minusmicrogmb

D equiv minus1

h (S241)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf




I 2 =

991787 θ983080m

4 u2 minus εo


= 4π

991787 infin

uo


where u = 2radic


I 2 = 32π983080kT

m

98308132 991787 infinxo



R = 8αn2radic

2π

991787 infin

xo




v = microF (S236)





φ = minusDdnB

dz (S238)




dz (S239)



dz (S240)


dz = minusmicrogmb

D equiv minus1

h (S241)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf









n

u

u

vo





dF






8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf



Θ

n

v-uo



F = 991787 π2

minusπ2

dF

dℓ

dℓ

dθ

dθ


cos2 θ n(θ) dθ

(S247)



cos3 θ dθ


orF = 83mRno

|vo

minusu

|(vo

minusu) (S249)







F = 4m2Rn

3πkT





8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf



Then

F = 4m2Rn

3πkT




F x = minus4m3

Rnu3πk2T 2

991787 infino

991787 πminusπ

eminusmV 2


= minus2Rnradic

2πmkT u

(S256)


v-u

φd

dθ

o



dF




dφ

991787 π2o




dφ

991787 π2o

cos3 θ sin θ dθ


4

983133π2o


(S259)





8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf







F = πmR2991787






eminusm(V+u)22kT


minusmV

middotukT ) (S264)


as


sin θ dθ

times991787 infin

o


1 minus m

kT uV cos θ

1048617V 2u cos θ

(S265)




and 991787 infino

eminusmV 2

2kT V 5dV = (2kTm)3 (S267)


3(2πmkT )12R2nu (S268)


D = 83(2πm)12(kT )32R2n (S269)

v2

v1

ω


8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf



v-v2 1





|v2

minusv1


|v2

minusv1











R = 12πωd 2991787





4 u2 + mV 2

983081983087kT

983133 (S274)


R = 12πωd 2n2(m2πkT )3991787

eminusmV 2kT d 3V


= 2ωd 2n2radic

πkTm

(S275)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf














ℓ =

s

2R =

1




partN (x t)


partn(x t)

partx (S278)


partn

partt = D

part 2n

partx2 (S279)


8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf





n(x 0) = C δ (x) (S280)


partn(x t)

partt = D

part 2n(x t)

partx2 (S281)



n(x t) = C radic



991787 infin

minusinfin

n(x t) dx = C

991787 infin

minusinfin

δ (x) dx = C (S283)






γ 2 = minusd(1γ )

dt = minus4D (S285)


n(x t) = C radic



P (x t) = 1radic



8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf









E n = hω(n + 12) (S288)





n=0

n=1

n=2

n=3

n=4

(a) (b) (c) (d)





Solution



8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf





+ sin(x2)sin(x3) sin(2x1)) (S290)



minus h2

2m

983080part 2u

partx2 +

part 2u

party2

983081 = Eu (S291)


u(0 y) = u(L y) and u(x 0) = u(x L) (S292)


u(x y) = A exp[i(k1x + k2y)] (S293)

where

(k1 k2) =9830802π

L K 1

2π

L K 2

983081 (S294)


E (k1 k2) = h2(k21 + k22)

2m (S295)



N (E ) = πk2

(2πL)2 =

mEL2

2πh2 (S296)


dE =

mL2

2πh2 (S297)





K = 1200

N 1N 2N 3 (S298)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf




N 3 = N 1 and N 2 = 1200 minus 2N 1 (S299)


K (N 1) = 1200

(N 1)2(1200minus 2N 1) (S2100)






or (1200




orP (400 + n)

P (400) = eminus3n

2400 (S2104)



F (x1 x2 x10) =10sumn=1

n x2n (S2105)


xn = 1 (S2106)


G =

10sumn=1


xn (S2107)


partG



10sumn=1

xn = λ

2

10sumn=1

1

n = 1 (S2109)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

8102019 sup 2pdf



which gives

λ = 2983080 10sumn=1

1

n

983081minus1

= 068283 (S2110)

Then

F =10

sumn=1

n983080 λ

2n9830812

= λ2

4

10

sumn=1

1

n

= λ

2

= 0341417 (S2111)



N =

991787 infin

o

(mL22πh2) dε

eα+βε + 1

= mL2

2πh2991787 infino

eminusβε dε

eα + eminusβε

(S2112)


N = kTmL2

2πh2

991787 1o

dx

eα + x

= kTmL2

2πh2

983131log(eα + x)

9831331o

= kTmL2

2πh21048667


(S2113)


n = mkT

2πh21048667


(S2114)







N = mL2

2πh2

991787 infin

o

dε

eα+βε

minus1

= mkTL2

2πh2

991787 1

o

dx

eα minus x

= mkTL2

2πh2


9831331o

= mkTL2

2πh21048667


(S2117)

or

n = mkT

2πh21048667


(S2118)

sup 2.pdf

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