1237.quantum mechanics 1

The Birth of Quantum MechanicsAt the turn of the last century, there were several experimental observations which could not be explained by the established laws of classical physics and called for a radically different way of thinking.

This led to the development of Quantum Mechanics which is today regarded as the fundamental theory of Nature.

NEED:Some key events/observations that led to the development of quantum mechanics… Black body radiation spectrum (Planck, 1901)

Photoelectric effect (Einstein, 1905)

Model of the atom (Bohr’s 1926)

Quantum Theory of Spectra (Bohr, 1913)

Scattering of photons off electrons (Compton, 1922) Matter Waves (de Broglie 1925)

Model of atom

Rutherford model

Classical models

Bohr model of atom-Quantum model

Quantum theory of spectra (Origin of Characteristic x-rays )

hv

Hydrogen spectra

Black Body Radiation

A black body is an object that absorbs 100% of the radiation that hits it. Therefore it reflects no radiation and appears perfectly black.

Wien’s Displacement Law:

Stefan’s Law (1879): the total power P radiated from one square meter of black surface at temperature T goes as the fourth power of the absolute temperature:

4 8 4, 5.67 10 watts/sq.m./K .P T

KoA7102.898constant.Tmλ

Blackbody Approximation

A good approximation of a black body is a small hole leading to the inside of a hollow object

The hole acts as a perfect absorber

The nature of the radiation leaving the cavity through the hole depends only on the temperature of the cavity

Rayleigh-Jeans Law:

An early classical attempt to explain blackbody radiation was the Rayleigh-Jeans law

At long wavelengths, the law matched experimental results fairly well

4

2I , Bπck T

λ Tλ

Rayleigh-Jeans Law, cont.

At short wavelengths, there was a major disagreement between the Rayleigh-Jeans law and experiment

This mismatch became known as the ultraviolet catastrophe

Planck’s Quantum Theory

In 1900 Planck devised a theory of blackbody radiation which gave good agreement for all wavelengths. In this theory the molecules of a body cannot have arbitrary energies but instead are quantized - the energies can only have discrete values. The magnitude of these energies is given by the formula

E = nhv,

where n = 0,1,2,... is an integer, v is the frequency of vibration of the molecule, and h is a constant, now called Planck's constant:

h = 6.63 x 10- 34 Js .

Classical predictions of black body radiation

hv2hv

Quantum prediction

Planck’s Model, Graphs

Planck’s Wavelength Distribution Function Planck’s Wavelength Distribution Function Planck generated a theoretical expression for the

wavelength distribution

h = 6.626 x 10-34 J.s h is a fundamental constant of nature

2

5

2

1I ,

Bhc λk T

πhcλ T

λ e

The photoelectric effect

Another experiment which provides compelling proof for the photon nature of light is the photoelectric effect.

hvh-W0=1/2mv2

Classical view Quantum view

Photoelectric Effect Summary

Each metal has “Work Function” (W0) which is the minimum energy needed to free electron from atom.

Light comes in packets called Photons E = h v h=6.626 X 10-34 Joule sec

Maximum kinetic energy of released electrons K.E. = hv – W0

PHOTOELECTRIC EFFECT (cont)

The maximum KE of an emitted electron is then

maxK h W

Work function: minimum energy needed for electron to escape from metal (depends on material, but usually 2-5eV)

Planck constant: universal constant of nature

346.63 10 Jsh

Einstein

Millikan

Verified in detail through subsequent experiments by Millikan

Maximum KE of ejected electrons is independent of intensity, but dependent on ν

For ν<ν0 (i.e. for frequencies below a cut-off frequency) no electrons are emitted

There is no time lag. However, rate of ejection of electrons depends on light intensity.

Actual results:

E h

Einstein’s interpretation (1905):

Light comes in packets of energy (photons)

An electron absorbs a single photon to leave the material

h (- 0) = 1/2mv2

Numerical Questions related to photo electric effect

Q1. Photoelectric work function of a photo-sensitive material is 3 eV. Calculate its threshold frequency and threshold wavelength.

Hint:

h (- 0) = 1/2mv2

h - h0 = 1/2mv2

As h0 = W = 3 eV

hvW

Therefore 0 = W/h = 3 x 1.6 x 10 -19 /6.62 10 -34 Hz

And 0 = c/ 0 Ao

1/2mv2


Q 1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.10 percent of the incident photons produce photoelectrons, find the current in the cell.

Hint:

.483.0

10483.0

106.1sec1002.3

.sec1002.3.100

10302.010.0

10.0

.1030212.0103.3

109375.0

103.3106.1

105.1

103.3

105.1

103.3

105.1

103.3

105.1

5.1

.

.103.310400

/1031062.6

6

19112

112116

116116

19

13133233

9

834

A

Amp

Ccelltheincurrentricphotoelect

s

photonsofpercent

seV

eVsn

eV

eVs

eV

Js

eV

skgm

eV

Wn

Vmlightincidenttheofpower

thetoingcorrespondemittedphotonsofno

eVEm

smJshcE


Q Light from the sun arrives at the earth , an average of 1.5 x 1011 m away, at the rate of 1.4 x 103 W/m2 of area perpendicular to the direction of light. Assume that sun light is monochromatic with a frequency of 5.0 x 1014 Hz, how many photons fall per second on each square meter of the earth surface directly facing the sun?

Hint:

.sec/102.4100422.0100.51062.6

104.1 21231434

3

photonsn

PtnhE

COMPTON SCATTERING (cont)

Compton’s explanation:“billiard ball” collisions between particles of light (X-ray photons) and electrons in the material

Classical picture:oscillating electromagnetic field causes oscillations in positions of charged particles, which re-radiate in all directions at same frequency and wavelengthas incident radiation.

Change in wavelength of scattered light is completely unexpectedclassically

θ

ep

pBefore After

Electron

Incoming photon

p

scattered photon

scattered electron

Oscillating electronIncident light wave Emitted light wave

Experiment: Outgoing photon has longer wavelength

Recoil electron carries some momentum and KE

Incoming photon has momentum, p, and wavelength

This experiment really shows photon momentum!

Electron at rest

Compton Scattering

Pincoming photon + 0 = Poutgoing photon + Pelectron

hc

hvE h

p

Energy of a photon

Momentum of a photon

Conservation of energy Conservation of momentum

1/ 22 2 2 2 4e e eh m c h p c m c ˆ

e

h p i p p

1 cos

1 cos 0e

c

h

m c

12 Compton wavelength 2.4 10 mce

h

m c

From this Compton derived the change in wavelength

θ

ep

pBefore After

Electron

Incoming photon

p

scattered photon

scattered electron

COMPTON SCATTERING (cont)

Numerical Questions related to Compton effect

Q1 An x-ray photon of frequency 3 x 10 19 Hz collides with an electron and is scattered through an angle of 900. Find its new frequency.

Hint:

1 cos

1 cos 0e

c

h

m c

The nature of light…….

The birth of quantum mechanics is intimately linked with the theories and discoveries relating to the nature of light

Is the nature of light that of a wave or a particle???

Light has a dual nature

Wave (electromagnetic) - Interference - Diffraction Particle (photons) - Photoelectric effect - Compton effect

Wave - Particle Duality for light

The connecting link – Planck’s constant_______________________________

Dual Nature

Radiation

Matter

hE

p

h

SUMMARY OF PHOTON PROPERTIESSUMMARY OF PHOTON PROPERTIES

E h

h hp

c

E p k2

h

2k

Energy and frequency

Also have relation between momentum and wavelength

2 2 2 2 4E p c m c

c

Relation between particle and wave properties of light

Relativistic formula relatingenergy and momentum

E pcFor light and

Also commonly write these as

2 angular frequency

wavevector

hbar

WAVE-PARTICLE DUALITY OF LIGHTIn 1924 Einstein wrote:- “ There are therefore now two theories of light, both indispensable, and … without any logical connection.”

Evidence for wave-nature of light• Diffraction and interferenceEvidence for particle-nature of light• Photoelectric effect• Compton effect

•Light exhibits diffraction and interference phenomena that are only explicable in terms of wave properties

•Light is always detected as packets (photons); if we look, we never observe half a photon

•Number of photons proportional to energy density (i.e. to square of electromagnetic field strength)

We have seen that light comes in discrete units (photons) with particle properties (energy and momentum) that are related to thewave-like properties of frequency and wavelength.

MATTER WAVES

h

p

In 1923 Prince Louis de Broglie postulated that ordinary matter can havewave-like properties, with the wavelength λ related to momentum p in the same way as for light

de Broglie wavelength

de Broglie relation346.63 10 Jsh

Planck’s constant

Prediction: We should see diffraction and interference of matter waves

De Broglie

NB wavelength depends on momentum, not on the physical size of the particle

ACT: De Broglie WavelengthA stone is dropped from the top of a building.

1. It decreases

2. It stays the same

3. It increases

What happens to the de Broglie wavelength of the stone as it falls?

p

h

hp

Speed, v, KE=mv2/2, and momentum, p=mv, increase.

Activity:Photon A has twice as much momentum as Photon B. Compare their energies.

• EA = EB

• EA = 2 EB

• EA = 4 EB

Electron A has twice as much momentum as Electron B. Compare their energies.

• EA = EB

• EA = 2 EB

• EA = 4 EB

m

pmvKE

22

1 22

hc

E phand so cpE

double p then quadruple E

double p then double E

21%

47%

33%

22%

40%

38%

MATTER WAVES

de Broglie suggested that Whole of the universe is composed of light (energy) and matter and both are inter-transferable physical quantities (on the basis of Einstein’s mass-energy equivalence relation). Therefore, if light (radiation) possesses the dual character, then matter must show the same, while in motion, as the nature loves symmetry.

E s t i m a t e s o m e d e B r o g l i e w a v e l e n g t h s

• W a v e l e n g t h o f e l e c t r o n w i t h 5 0 e V k i n e t i c e n e r g y2 2

1 02

1 . 7 1 0 m2 2 2e e e

p h hK

m m m K

• W a v e l e n g t h o f N i t r o g e n m o l e c u l e a t r o o m t e m p e r a t u r e

u

1 1

3, M a s s 2 8 m

2

2 . 8 1 0 m3

k TK

h

M k T

• W a v e l e n g t h o f R u b i d i u m ( 8 7 ) a t o m a t 5 0 n K

61 . 2 1 0 m3

h

M k T

Nature of de Broglie waves

Our traditional understanding of a wave….

Adding up waves of different frequencies

A “Wave Packet”

v

Vg

Wave Envelope

Wave Envelope The blue line represents the envelope

function This envelope can travel through space with

a different speed than the individual waves

Wave packet, phase velocity and group velocity ____________________________ The spread of wave packet

in wavelength depends on the required degree of localization in space – the central wavelength is given by

What is the velocity of the wave packet?

p

h

Wave packet, phase velocity and group velocity ________________________________ The velocities of the individual waves which

superpose to produce the wave packet representing the particle are different - the wave packet as a whole has a different velocity from the waves that comprise it

Phase velocity: The rate at which the phase of the wave propagates in space

Group velocity: The rate at which the envelope of the wave packet propagates

Speeds Associated with Wave Packet Speeds Associated with Wave Packet The phase speed of a wave in a wave packet is

given by

This is the rate of advance of a crest on a single wave The group speed is given by

This is the speed of the wave packet itself

phaseωv k

gdωv dk

Wave packet, phase velocity and group velocity ________________________________

Phase velocity

Group velocity

Here is the velocity of light and v is the velocity of the particle

v

cv p

2

vvg

c

Numerical questions related to de Broglie waves

Q An electron and proton have same velocity. Compare the wavelengths, phase velocities and group velocities.

Hint:

Group velocity: vg = v , so group velocity will be same for both.

electron proton= v =

Wavelengths: e = h/mev and p = h/mpv and me < mp so e > p

Phase velocity: vp =c2/v so if v is same then phase velocity will be same for both.


Q Find the de Broglie wavelength of a 40 keV electron used in certain electron microscope.

Hint: kinetic energy (T) =40 keV = p2/2m, then p = (2mT)1/2

= h/p


Q Find the de Broglie wavelength of a 1.0 mg grain blown by wind at a speed 20 m / sec.

Hint:

= h / m v = h (1-v2/c2)1/2/ m0 v


Q. 2

Find the de Broglie wavelength of an electron, whose speed is

(a) 1 x 108 m/sec (b) 2 x 108 m/sec.

Hint:

= h / m v = h (1-v2/c2)1/2/ m0 v


Q1. A photon and particle have the same de Borglie wavelength. Can anything be said how their linear momentum compare.

Hint: photon = particle

as = h / p therefore, if is same then p will be same for both photon and particle.


Q Find the KE of an electron whose de-Broglie wave length is same as that of a100 keV x-ray.

HINT: For an x-ray photon E = 100 keV

E=hv=100 keV

hc/ = 100 kev, from this can be determined

Now as given x-ray = electron

Therefore, =h/p implies p = h/

If momentum of the electron is known, then its KE = (½)mv2 = p2/2m

can be determined.

THE UNCERTAINTY PRINCIPLE

In quantum mechanics no measurement can be 100% accurate as with every moving particle a wave packet is associated, hence it is impossible to locate the position and momentum of the particle with 100 % confidence.

Acc to Heisenberg “The product of uncertainty in position of a particle in certain direction and the component of linear momentum of particle in same direction can never be less than ħ/2 or h/4π”

i. e. Δx. Δpx ≥ ħ/2

Wave packet

Δx

THE UNCERTAINTY PRINCIPLE

The narrower is the wave packet more precisely the position of the particle can be determined, however the wavelength in the narrow packet can not be well defined, as there are not enough wavelengths to measure λ accurately. This means that as p=h/λ can not be determined accurately and vice -versa.

Δx is small but Δp is large

Δx is large but Δp is small

Measuring the position and momentum of an electron

Shine light on electron and detect reflected light using a microscope

Minimum uncertainty in position is given by the wavelength of the light

So to determine the position accurately, it is necessary to use light with a short wavelength

Measuring the position and momentum of an electron (cont’d)

By Planck’s law E = hc/λ, a photon with a short wavelength has a large energy

Thus, it would impart a large ‘kick’ to the electron

But to determine its momentum accurately, electron must only be given a small kick

This means using light of long wavelength!

Example of Baseball

A pitcher throws a 0.1-kg baseball at 40 m/s

So momentum is 0.1 x 40 = 4 kg m/s

Suppose the momentum is measured to an accuracy of 1 percent , i.e.,

Δp = 0.01 p = 4 x 10-2 kg m/s

Example of Baseball (cont’d)

The uncertainty in position is then

No wonder one does not observe the effects of the uncertainty principle in everyday life!

Numerical-THE UNCERTAINTY PRINCIPLE (Applications)

Q1. Non-Existence of electron with in the nucleus.

Hint:

let us consider an electron exist

with in nucleus therefore, maximum

Diameter

10-14 m Δx ~x = 10 -14 m

Now by uncertainty principle Δx .Δp ≥ ħ/2, Δp = ħ/2 Δx, Δp = 5.27 x 10 -21 Kgm/sec

As now actual linear momentum p can not be less than Δp (uncertainty in LM) so p ~ Δp Therefore, minimum energy possessed by the electron if it is to stay with in the nucleus is E = p 2/2m 96.4 MeV

Thus, to remain in side the nucleus the minimum energy of the electron should be 96.4 MeV, but it has been found experimentally that no electron can possess energy more than 5 MeV. Thus electron can not exist with in the nucleus.


Q Compare the uncertainties in the velocities of an electron and a proton confined in a 1.00 nm box.

Hint: On the basis of

uncertainty principle

1 x 10 -9 .Δpx ≥ ħ/2,

so Δpx = 0.527 x 10 -25 Kgm/sec for both

Now as we know that Δpx = me Δve , implies that Δve = 0.058 x 106 m/sec

Similarly, Δpx = mp Δvp , implies that Δvp = 0.058 x 102 m/sec

Therefore, Δve > Δvp

Δx= 1.00 nm box

electron proton


Q An electron has a speed of 500 m/sec, correct up to 0.01 %, with what minimum accuracy you can determine the electron.

Hint:

Δvx = (0.01/ 100) x 500 = 0.05 m/sec. 500

Δpx = m Δvx = 9.1 10 -31 x 0.05 = 4.55 10 -32 Kg m Sec-1

Therefore on the basis of uncertainty principle Δx .Δpx ≥ ħ/2

Δx = ħ/ (2 X Δpx) = 2.316 mm.

There is also an energy-time uncertainty relation

Transitions between energy levels of atoms are not perfectly sharp in frequency.

/ 2E t

n = 3

n = 2

n = 1

32E h

32

Inte

nsity

Frequency

32

HEISENBERG UNCERTAINTY PRINCIPLE

There is a corresponding ‘spread’ inthe emitted frequency

810 st

An electron in n = 3 will spontaneouslydecay to a lower level after a lifetimeof order

Wave Function

The wave function is often complex-valued The absolute square |ψ|2 = ψψ is always

real and positive The wave function contains within it all the

information that can be known about the particle

Properties of the wave function When squared, the wave function is a probability density (MaxBorn –

1926). The probability P(x) dx of a particle being between x and x+dx was given in the equation:

The probability of the particle being between x1 and x2 is given by

The wave function must also be normalized so that the probability of the particle being somewhere on the x axis is 1.

Wave Function

The probabilistic interpretation of the wave function The wave function is often complex-valued

The absolute square |ψ|2 = ψψ is always real and positive ψ* is the complete conjugate of ψ

It is proportional to the probability per unit volume of finding a particle at a given point at some instant

The wave function contains within it all the information that can be known about the particle was first suggested by Max Born

Erwin Schrödinger proposed a wave equation that describes the manner in which the wave function changes in space and time

This Schrödinger wave equation represents a key element in quantum mechanics

Wave Function of a Free Particle Because the particle must be somewhere along

the x axis, the sum of all the probabilities over all values of x must be 1

Any wave function satisfying this equation is said to be normalized

Normalization is simply a statement that the particle exists at some point in space

21abP ψ dx

Numerical problems related to probability and normalization

Q Find the value of normalization constant A for the wave function

Hint: Probability of finding the particle equal to 1. square of the wave function should be equal to 1.

Q2 The wave function of a particle is

(a) Find the value of A

(b) Find the probability that the particle be found between x=0 and x=

HINT: (a) for the calculation of A apply normalization condition on the wave function

(b) After substituting the value of A in the wave function find the probability by integrating with in the limits 0 to Pi/2

2/2xAxe

12

0

2/2

dx

x xAxe

2/2/cos2 xforxA

4/

Expectation Values

ψ is not a measurable quantity Measurable quantities of a particle can be derived from ψ The average position is called the expectation value of x and is defined as

The expectation value of any function g(x) for a normalized wave function:

Abbreviated Notation: ‹*| g |›

Operators

Momentum:

Energy:

Expectation Values

Expectation value

the expectation value of momentum involves the representation of momentum

as a quantum mechanical operator.

Where

is the operator for the x component of momentum.

Expectation value

Since the energy of a free particle is given by

and the expectation value for energy becomes

for a particle in one dimension.

Numerical problems related to Expectation values

Q Find the expectation value for the position x of a particle in a box L wide assuming it in the ground state.

Hint: For ground state n=1

L

xn

Lwheredxxx

L sin2

,0

2

L

Schrodinger’s Time dependent equation

Schrodinger equation is the fundamental equation of quantum mechanics in the same sense as the second law of motion of motion is the fundamental equation of classical mechanics

The wave function associated with a moving particle is

The total energy of the sub-atomic particle is E =KE+PE = p2/2m +V

pxEti

Ae

T i m e - i n d e p e n d e n t S c h r ö d i n g e r e q u a t i o n

S u p p o s e p o t e n t i a l i s i n d e p e n d e n t o f t i m e2 2

2i ( )

2V x

t m x

L H S i n v o l v e s o n l y v a r i a t i o n o f Ψ w i t h t

R H S i n v o l v e s o n l y v a r i a t i o n o f Ψw i t h x ( i . e . H a m i l t o n i a n o p e r a t o r d o e s n o t d e p e n d o n t )

L o o k f o r a s e p a r a t e d s o l u t i o n ( , ) ( ) ( )x t x T t

S u b s t i t u t e :

, ( )V x t V x

2 2

2( ) ( ) ( ) ( ) ( ) ( ) ( )

2x T t V x x T t i x T t

m x t

2 2

2 2( ) ( ) ( )

dx T t T t

x d x

2 2

2( )

2

d d TT V x T i

m d x d t

N . B . T o t a l n o t p a r t i a ld e r i v a t i v e s n o w

e t c

Eigenvalues and Eigenfunctions

The values of energy for which the steady state Schrodinder equation can be solved are called eigen values and the corresponding solutions (wave functions) are called as eigen functions.

These words are derived from German words;

EIGEN WERT and EIGEN FUNKTION

eigen means proper or characteristic and wert means value.

To obtain specific values for physical parameters, for example energy, you operate on the wave function with the quantum mechanical operator associated with that parameter

Eigen function and eigen value:

Eigen values Eigen functions

1

2

3

n

1

2

3

n

Numerical problems related to Eigen values and eigen function

Q An eigen function of the operator d2/dx2 is = e3x. Find the corresponding eigen value.

Hint: As we know that Eigen value equation is

here

i = e3x

and Qop = d2/dx2

so when we apply the operator on the wave function the corresponding eigen value comes

qi = 9.

Numerical problems related to Eigen values and eigen function

Q Determine the eigen energy values and eigen functions of a particle trapped in a one dimensional box.

Hint:

2

222

2sin

2

mL

nEand

L

xn

L nn

S U M M A R YS U M M A R Y

T i m e - d e p e n d e n t S c h r ö d i n g e r e q u a t i o n

2 2

2,

2V x t i

m x t

2 *, , , ,P x t d x x t d x x t x t d x

P r o b a b i l i t y i n t e r p r e t a t i o n a n d n o r m a l i z a t i o n

T i m e - i n d e p e n d e n t S c h r ö d i n g e r e q u a t i o n

2 2

2( ) ( ) ( )

2

dV x x E x

m d x

/, ( ) ( ) ( ) i E tx t U x T t U x e

/, ( ) ( ) ( ) i E tx t x T t x e

C o n d i t i o n s o n w a v e f u n c t i o ns i n g l e - v a l u e d , c o n t i n u o u s , n o r m a l i z a b l e ,c o n t i n u o u s f i r s t d e r i v a t i v e

2, , 1d x P x t d x x t

Particle in box:

Particle in a Box

Particle in a Box A particle is confined to a one-

dimensional region of space

The “box” is one- dimensional

The particle is bouncing elastically back and forth between two impenetrable walls separated by L

Potential Energy for a Particle in a Box As long as the particle is inside

the box, the potential energy does not depend on its location

We can choose this energy value to be zero

The energy is infinitely large if the particle is outside the box

This ensures that the wave function is zero outside the box

Wave Function for the Particle in a Box Since the walls are impenetrable, there is zero

probability of finding the particle outside the box

ψ(x) = 0 for x < 0 and x > L

The wave function must also be 0 at the walls ψ(0) = 0 and ψ(L) = 0

Wave Function of a Particle in a Box – Mathematical The wave function can be expressed as a real, sinusoidal

function

Applying the boundary conditions and using the de Broglie wavelength

2( ) sin

πxψ x A

λ

( ) sinnπx

ψ x AL

The Particle in a Box

n is the quantum number (n= 1, 2, 3,....),L is the 'length' of the (one dimensional) molecular box, m is the mass of the particle (electron), and h is Planck's constant.

and

Energy Level Diagram – Particle in a Box The lowest allowed energy

corresponds to the ground state

En = n2E1 are called excited states

E = 0 is not an allowed state

The particle can never be at rest

Graphical Representations for a Particle in a Box

Expectation value for position and linear momentum of the particle in a one dimensional box:

As wave function or eigen function of the particle trapped in a box is known, expectation value of position and momentum can be evaluated from it as:

02

,

22

22222

2

2

.

00

2sin10

2sin2

2

2sin0 21cossin

02

20

20

*0

*

0*2

Ln

Ln

p

ismentummoaverageandforthandbackmovespariclethethus

Ln

mL

nmmEPm

pEthatknowweAs

resultStrange

L

Lxn

Li

L

Lxn

nL

Ln

Liptherefore

axxa

axdxaxSince

dxL

xnCosLL

xnSinL

niL

p

dxL

xnSinLxi

LL

xnSinL

dxxi

LdxpLp

ascomplexlittleispcalculatetobut

dxxLxL

xnSinL

Numerical questions related to Particle in a box:

Q Find the energy of an electron contained in a box of dimension 1 Ao such that its wave function has 5 nodes in the box.

Ans. The energy of en electron is

2

222

2

m

nEn


Q A particle is described by the wave function

8/1:

2/1,0int

;0

10;3)(

Ans

ervalinfoundbecanparticlethethatyprobabilittheFind

Elsewhere

xxx


Q Think the nucleus as a box with the size of 10-14 m across. Compute the lowest energy of a neutron confined in the nucleus.

Hint:

considering the nucleus as a cubical box of size 10-14 m.

so that x = y = z = 10-14 m

1

10

2

14

2

2

2

2

2

222

zyx

zyx

z

z

y

y

x

x

nnnand

mhere

nnn

mEn


Q Show that wave functions for two different states are ortho-normal for a particle in a one dimensional box.

Hint: the wave functions are said to be ortho-normal

if

nmfordx

whennormalizedtosaidarefunctionswavetheand

nmfordx

n

L

m

n

L

m

1

0

0

*

0

*

1237.quantum mechanics 1

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