manipal be phys 2011 12 quantumphysics

INTRODUCTION TO QUANTUM PHYSICS

Blackbody Radiation & Planks Hypothesis The Photoelectric Effect The Compton Effect Photons and Electromagnetic Waves The Quantum Particle The Double-Slit Experiment Revisited The Uncertainty Principle

TOPICS

Text Book PHYSICS for Scientists and Engineers with Modern Physics (6th ed) By Serway & Jewett

1 BE-PHYSICS-INTRODUCTION TO QUANTUM PHYSICS-2011-12 MIT- MANIPAL

INTRODUCTION

Failure of classical mechanics

Brief summary of chapter 40 of the text book



BLACKBODY RADIATION & PLANKS HYPOTHESIS

Origin of thermal radiation the classical view point

Concept of oscillators

INTRODUCTION


Definition of a Black-Body

Black-Body Radation Laws

1- The Stefan-Boltzmann Law

2- The Wiens Displacement Law

3- The Rayleigh-Jeans Law

4- The Planck Law

Application for Black Body

Conclusion

Summary



Definition of a black body

An object that absorbs all incident radiation. A small hole cut into a cavity is the most popular and realistic example.

None of the

incident radiation

escapes.



The radiation is absorbed in the walls of the

cavity. This causes a heating of the cavity walls.

The oscillators in the cavity walls vibrate and

cavity walls re-radiate at wavelengths

corresponding to the temperature of the cavity,

producing standing waves in the cavity. Some of

the energy from these standing waves can leave

through the opening. The electromagnetic

radiation emitted by the black body is called

black-body radiation.



The black body is an ideal absorber of

incident radaition.

The emitted "thermal" radiation from a black

body characterizes the equilibrium

temperature of the black-body.

Emitted radiation from a blackbody does not

depend on the material of which the walls are

made.



Figure shows Intensity

of blackbody radiation

versus wavelength at

three temperatures.

At about 6000 K (not

shown in fig.), the peak

is in the center of the

visible wavelengths and

the object appears white.



Basic Laws of Radiation 1) All objects emit radiant energy.

2) Hotter objects emit more energy than colder

objects (per unit area). The total power of the emitted radiation increases with temperature.

This is Stefan-Boltzmann Law. 3) The peak of the wavelength distribution

shifts to shorter wavelengths as the black body temperature increases

This is Wiens Law.



Stefan Boltzmann Law.

P = Ae T4

P = power radiated from the surface of the object(W) T = temperature (K) = 5.670 x 10-8 W/m2K4 (Stefan-Boltzmann constant) A = surface area of the object (m2) e = emissivity of the surface (for a blackbody e =1).

Black-Body Radiation Laws (1)




Wiens Displacement Law.

lm T = constant = 2.898 10-3 m.K, or lm T

-1

Where lm = peak of the wavelength distribution in the black body emission spectrum. T- equilibrium temperature of the blackbody. (A black-body reaches thermal equilibrium when the incident radiation power is balanced by the power re-radiated).



SJ: P-SE 40.1 Thermal radiation from Different Objects.

Find the peak wavelength of the blackbody radiation emitted by each of the following.

A. The human body when the skin temperature is

35C

B. The tungsten filament of a light bulb, which operates at 2000 K

C. The Sun, which has a surface temperature of about 5800 K.




The Rayleigh-Jeans Law.

4B

Tck2)T,(I

=

This law tries to explain the distribution of energy from a black body. I (l,T) dl is the intensity or power per unit area emitted in the wavelength interval dl from a blackbody.



The failure has become known as the ultraviolet catastrophe.

*It agrees with experimental

measurements for long

wavelengths.

* It predicts an energy

output that diverges towards

infinity as wavelengths grow

smaller.

kB Boltzmann's constant T- equilibrium blackbody temperature c- velocity of light.

Rayleigh-Jeans Law




The Planck Law

1 -Tk

hc

e

1

5

2hc2= T) ,(

B

I

This law too explains the distribution of energy from a black body. I (l,T) dl is the intensity or power per unit area emitted in the wavelength interval dl from a blackbody. Only the extra quantity (compared to the Rayleigh-Jeans Law) coming here is the constant known as Planks constant introduced by Max Plank in this revolutionary theory.



The Planck Law gives a distribution that

peaks at a certain wavelength, the peak

shifts to shorter wavelengths for higher

temperatures, and the area under the

curve grows rapidly with increasing

temperature. In short the law fitted the

experimental data for all wavelength

regions and at all temperatures.

But for this to happen, Plank made two bold and

controversial assumptions concerning the nature of the

oscillators in the cavity walls.



1) The energy of an oscillator can have only

certain discrete values En En = nhf

where n is a positive integer called a

quantum number, f is the frequency of

oscillation, and h is a constant called Plancks

constant. Energy of the oscillator is

quantized. Each discrete energy value

corresponds to a different quantum state,

represented by the quantum number n.



2) The oscillators emit or absorb energy only

when making a transition from one quantum state

to another.

Difference in energy will be integral multiples of hf.

If it remains in one quantum state, no energy

is emitted or absorbed.

0

EN

ER

GY

hf

2 h

f

3 h f

4 h f

0

1

2

3

4

t o n =

n E

Figure shows allowed energy

levels for an oscillator with

frequency f, and the allowed

Transitions.



The results of Planck's law

the denominator [exp(hc/lkBT )] tends to infinity faster than the numerator (l-5), thus resolving the infrared catastrophe. i.e. I (l,T) 0 as l 0. for very large l,

From a fit between Planck's law and experimental data, Planck's constant was derived to be

h = 6.626 10-34 J.s

TkTk

hc1)Tk/hcexp(

B4

BB

- T) ,( I

-



Planck's restriction of the available

energies for radiation gets around the

ultraviolet catastrophe in the following

way: the short wavelength/high frequency

modes are now limited in the energy they

can have to either zero, or E hf; in the

calculation of the average energy, these

modes with high energy are cut off by the

Boltzmann factor exp(-E/kBT), i.e. these

modes are rarely excited and, therefore,

contribute nothing to the average energy in

the limit 0.



Summary

The characteristics of blackbody radiation cannot

be explained using classical concepts.

Plank introduced the quantum concept and

Planks constant when he assumed that atomic

oscillators existing only in discrete energy states

are responsible for this radiation.

In Planks model, radiation is emitted in

single quantized packets whenever an oscillator

makes a transition between discrete energy states.



SJ: P-SE 40.2 The Quantized Oscillator A 2.0- kg block is attached to a spring that has a force constant of k = 25 N/m. The

spring is stretched 0.40 m from its equilibrium position and released.

A. Find the total energy of the system and

the frequency of oscillation according to classical calculations.

B. Assuming that the energy is quantized, find the quantum number n for the system oscillating with this amplitude.



SJ: P-SE 40.2 The Quantized Oscillator contd. C. Suppose the oscillator makes a transition from the n = 5.4 x 1033 state to the state corresponding to n = 5.4 x 1033 -1. By how much does the energy of the oscillator change n this one-quantum change.



SJ: Section 40.1 P-1 The human eye is most sensitive to 560 nm light. What is the temperature of a black body that would radiate most intensely at this wavelength?

SJ: Section 40.1 P-3 A blackbody at 7500 K consists of an opening of diameter 0.050 mm, looking into an oven. Find the number of photons per second escaping the hole and having wavelengths between 500 nm and 501 nm.



SJ: Section 40.1 P-5 The radius of our Sun is 6.96 x 108 m, and its total power output s 3.77 x 1026 W. (a) Assuming that the Suns surface emits as a black body, calculate its surface temperature. (b) Using the result, find lmax for the Sun.

SJ: Section 40.1 P-7 Calculate the energy in electron volts, of a photon whose frequency is (a) 620 THz, (b) 3.10 GHz, (c) 46.0 MHz. (d) Determine the corresponding wavelengths for these photons and state the classification of each on the electromagnetic spectrum.



Assignment: Try to answer the questions in page no. 1313, chapter 40 of the text book.

SJ: Section 40.1 P-9. An FM radio transmitter has a power output of 150 kW and operates at a frequency of 99.7 MHz. How many photons per second does the transmitter emit?



Introduction

What is Photoelectric Effect

Apparatus for studying Photoelectric Effect

Experimental Observations

Classical Predictions

Clash between Classical predictions

& Observed Experimental results

Einsteins model of the Photoelectric Effect

Explanation for the observed features of PE

Application

Conclusion

Summary

THE PHOTOELECTRIC EFFECT


What is Photoelectric Effect?

Apparatus for studying Photoelectric Effect

T

T Evacuated

glass/ quartz tube

E Emitter Plate/

Photosensitive

material /Cathode

C Collector Plate /

Anode

V Voltmeter

A - Ammeter




1. When plate E is illuminated by light of suitable frequency, electrons are emitted from E and a current is detected in A.

2. Photocurrent produced Vs potential difference applied graph shows that maximum kinetic energy of the emitted electrons,

Kmax = e Vs



3. Maximum kinetic energy of the

photoelectron is independent of light

intensity.

4. Electrons are emitted from the surface of the

emitter almost instantaneously.

5. No electrons are emitted if the incident light

frequency falls below a cutoff frequency.

6. Maximum kinetic energy of the

photoelectrons increases with increasing

light frequency.




1. If light is really a wave, it was thought that

if one shine light of any fixed wavelength, at

sufficient intensity on the emitter surface,

electrons should absorb energy continuously

from the em waves and electrons should be

ejected.

2. As the intensity of light is increased (made it

brighter and hence classically, a more

energetic wave), kinetic energy of the

emitted electrons should increase.

3. Measurable/ larger time interval between

incidence of light and ejection of

photoelectrons.



4. Ejection of photoelectron should not

depend on light frequency

5. Photoelectron kinetic energy should not

depend upon the frequency of the

incident light.

In short experimental results contradict classical predictions.



Einsteins Interpretation of em radiation

(A new theory of light)

Electromagnetic waves carry discrete energy packets (light quanta called photons now). The energy E, per packet depends on frequency f.

E = hf. More intense light corresponds to more photons, not higher energy photons. Each photon of energy E moves in vacuum at the speed of light c, where c = 3x 108 m/s. Each photon carries a momentum p = E/C.



Einsteins model of the photoelectric effect

A photon of the incident light gives all its

energy hf to a single electron (Absorption

of energy by the electrons is not a

continuous process as envisioned in the

wave model) and Kmax = hf -

is called the work function of the metal.

It is the minimum energy with which an

electron is bound in the metal.



All the observed features of photoelectric effect could be explained by Einsteins photoelectric equation.

1. Equation shows that Kmax depends only on

frequency of the incident light.

2. Almost instantaneous emission of photoelectrons

due to one -to one interaction between photons

and electrons.

3. Ejection of electrons depends on light frequency

since photons should have energy greater than

the work function in order to eject an electron.

4. The cutoff frequency fc is related to by fc = /h.

If the incident frequency f is less than fc , no

emission of photoelectrons.



Einstein predicted that a

graph of the maximum

kinetic energy Kmax Vs

frequency f would be a

straight line, given by the

linear relation,

Kmax = hf -

And indeed such a linear

relationship was observed.

And this work won Einstein his Nobel Prize in 1921



Application of photoelectric effect

Photomultiplier tube

Explain the device, theory, and its working



Summary

Einstein successfully extended Planks quantum hypothesis to explain photoelectric effect. In Einsteins model, light is viewed as a stream of particles, or photons, each having energy E = hf , where h is Planks constant and f is the frequency. The maximum kinetic energy Kmax of the ejected photoelectron is

Kmax = hf -

Where is the work function of the photocathode.



SJ: P-SE 40.3 The Photoelectric Effect for Sodium

A sodium surface is illuminated with light having

a wavelength of 300 nm. The work function for

sodium metal is 2.46 eV. Find

A. The maximum kinetic energy of the ejected

photoelectrons and

B. The cutoff wavelength for sodium.

SJ: Section 40.2 P-13. Molybdenum has a

work function of 4.2eV. (a) Find the cut off

wavelength and cut off frequency for the

photoelectric effect. (b) What is the stopping

potential if the incident light has wavelength of

180 nm?



SJ: Section 40.2 P-14. Electrons are ejected

from a metallic surface with speeds up to 4.60 x 105

m/s when light with a wavelength of 625 nm is used.

(a) What is the work function of the surface? (b)

What is the cut-off frequency for this surface?

SJ: Section 40.2 P-16. The stopping

potential for photoelectrons released from a

metal is 1.48 V larger compared to that in

another metal. If the threshold frequency for

the first metal is 40.0 % smaller than for the

second metal, determine the work function for

each metal.




Assignment: Try to answer the questions in page no. 1313, chapter 40 of the reference book.

SJ: Section 40.2 P-17. Two light sources are

used in a photoelectric experiment to determine

the work function for a metal surface. When

green light from a mercury lamp (l = 546.1 nm)

is used, a stopping potential of 0.376 V reduces

the photocurrent to zero. (a) Based on this

what is the work function of this metal? (b)

What stopping potential would be observed

when using the yellow light from a helium

discharge tube (l = 587.5 nm)?


Introduction

What is Compton Effect

Schematic diagram of Comptons apparatus



Explanation for Compton Effect

Derivation of the Compton Shift Equation.

Conclusion

Summary

THE COMPTON EFFECT


Introduction

Relativistic formula relating energy and momentum of a particle

2 2 2 2 4E p c m c

Here E is the total energy of the particle

Relativistically E = m c2

P and m are the momentum and mass of the particle.

Relativistic expression for the momentum of a particle

P = m v

where m = mass of the particle, v = speed of the particle & c = speed of light in vacuum

Expressions for relativistic momentum and relativistic kinetic energy of a particle

THE COMPTON EFFECT


For light, m = 0, E= pc . Also c = l f

h

=fhf

=c

Ep=

2

2

c

v - 1

1 and

And finally, the relativistic kinetic energy of a particle is K = ( -1) m c2.

SUMMARY OF PHOTON PROPERTIES

Energy, frequency, and wavelength, E = hf = hc / l

Also we have relation between momentum and wavelength of a photon as follows

Relation between particle and wave properties of light

THE COMPTON EFFECT


Compton (1923)

measured intensity of

scattered X-rays from

solid target (scattering of

X-rays from electrons), as

function of wavelength

for different angles. In

such a scattering, a shift

in wavelength for the

scattered X-rays takes

place, which is known as

Compton Effect.

scattered beam

What is Compton Effect ?

THE COMPTON EFFECT



Oscillating electromagnetic waves of frequency f0

incident on electrons should have two effects. (a)

oscillating electromagnetic field causes oscillations in

electrons, which re-radiate in all directions (b)

radiation pressure should cause the electrons to

accelerate in the direction of propagation of the waves.

Because different electrons will move at different

speeds after the interaction, depending on the amount

of energy absorbed from em waves, for a particular

angle of incidence of the incoming radiation, the

scattered wave frequency should show a distribution of

Doppler- shifted values.

THE COMPTON EFFECT


Schematic diagram of Comptons apparatus

Here X- ray photons are

scattered through 90

from a carbon target.

The wavelength is

measured with a

rotating crystal

spectrometer using

Braggs law. Intensity

of the scattered X-rays

are measured using the

ionization chamber.

Explain the experimental details and results

THE COMPTON EFFECT



Contrary to the classical predictions where X-rays are treated as waves, in Compton experiment, at a given angle, only one frequency for scattered radiation is seen. This is shown in the figure, scattered x-ray intensity versus wavelength for Compton scattering at = 0, 45, 90, and 135. Compton could explain the experimental result by taking a billiard ball type collisions between particles of light (X-ray photons) and electrons in the material.

THE COMPTON EFFECT


The graphs for three nonzero angles show two peaks,

one at l0 and one at l > l0 . The shifted peak at l is

caused by the scattering of X-rays from free electrons.

Shift in wavelength was predicted by Compton to

depend on scattering angle as

This is known as Compton shift equation, and the

factor is called the Compton wavelength.

) cos - 1(cm

h - '

e

0

cm

h

e nm 0.00243 = cm

h

e

Prediction were in excellent agreement with the experimental results.

THE COMPTON EFFECT


Derivation of the Compton Shift Equation

Photon is treated as a particle having energy E = hf = hc/l and zero rest energy. They collide elastically with free electrons initially at rest as shown in figure.

In the scattering

process, the total

energy and total

linear momentum of

the system must be

conserved.

THE COMPTON EFFECT


K+'

hc=

hce

0

Applying the law of conservation of energy to the process gives

where

hc/ l0 = E0 is the energy of the incident photon,

hc/ l = E is the energy of the scattered photon, and Ke is the kinetic energy of the recoiling electron.

Substituting for Ke we get

me

(40.12)2)( c1- +'hc

=hc

0

THE COMPTON EFFECT


Applying law of conservation of momentum to this collision, both in x and y components of momentum are conserved independently.

(40.13) :component x em + 'h=

0h coscos v

(40.14) :component y 0 em - 'h= sinsin v

where h/l0 = p0 is the momentum of the incident photon h/l = p is the momentum of the scattered photon mev = Pe is the momentum of the scattered electron

THE COMPTON EFFECT


Rewriting the above equations as

(40.13) :component x ' p p-p cosecos0

(40.14) :component y ' e p sinsin p

Squaring and adding the above equations give

) (a p= p'+ p'p2- p 220

20 ecos

Rewriting equation 40.12 in terms of respective energy notations as

E0 = E + Ee - mc2

i.e., E0 - E + mc2 = Ee ---------(b)

THE COMPTON EFFECT


Square equation (b), substitute for

4222e

2e cmcpE

and for from equation (a). Write the resulting equation in terms of respective wavelengths, we get the Compton shift equation as

2ep

) cos - 1(cm

h - '

e

0

THE COMPTON EFFECT


Summary

X-rays are scattered at various angles by electrons in

a target. In such a scattering, a shift in wavelength is

observed for the scattered X-rays and the

phenomenon is known as Compton Effect. Classical

physics does not predict the correct behaviour in this

effect. If x-ray is treated as a photon, conservation of

energy and linear momentum applied to the photon-

electron collisions yields for the Compton shift:

Where me is the mass of the electron, c is the speed of

light, and is the scattering angle.

) cos - 1(cm

h - '

e

0

THE COMPTON EFFECT


SJ: P-SE 40.4 Compton scattering at 45

X-rays of wavelength l0 = 0.20 nm are scattered from

a block of material. The scattered X-rays are observed

at an angle of 45 to the incident beam. Calculate

their wavelength.

What if we move the detector so that scattered x-

rays are detected at an angle larger than 45? Does

the wavelength of the scattered x-rays increase or

decrease as the angle increase?

SJ: Section 40.3 P-21 Calculate the energy and momentum of a photon of wavelength 700 nm.

THE COMPTON EFFECT


SJ: Section 40.3 P-23. A 0.00160 nm photon scatters from a free electron. For what photon scattering angle does the recoiling electron have kinetic energy equal to the energy of the scattered photon?

SJ: Section 40.3 P-25. A 0.880 MeV photon is scattered by a free electron initially at rest such that the scattering angle of the scattered electron is equal to that of the scattered photon ( = ). (a) Determine the angles & . (b) Determine the energy and momentum of the scattered electron and photon.

THE COMPTON EFFECT


THE COMPTON EFFECT

Assignment: Try to answer the questions in page no. 1313, chapter 40 of the reference book.


Photons and Electromagnetic Waves

Evidence for wave-nature of light Diffraction Interference Evidence for particle-nature of light Photoelectric effect Compton effect

PHOTONS AND ELECTROMAGNETIC WAVES


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

Photoelectric effect and Compton effect can only be explained taking light as photons/ particle

This means true nature of light is not describable in terms of any single classical picture.

In short, the particle model and the wave model of light compliment each other.



The Wave Properties of Particles

We have seen that light comes in discrete units

(photons) with particle properties (energy and

momentum) that are related to the wave-like

properties of frequency and wavelength.

In his 1923 doctoral dissertation, Louis de Broglie

postulated that because photons have both wave

and particle characteristics, perhaps all forms of

matter have wave-like properties, with the

wavelength related to momentum p in the same

way as for light

De Broglie PHOTONS AND ELECTROMAGNETIC WAVES


de Broglie Hypothesis

h

pl

de Broglie wavelength

346.63 10 Jsh -

Plancks constant

and

frequency of the particle

hE=f

Energy of the particle



mv

hph ==

p = momentum of the particle,

p = m v for a non-relativistic particle

m = mass of the particle

v = velocity of the particle

V ev m 22

1

The electron accelerated through a potential difference of V has a non relativistic kinetic energy

m = mass, v = velocity

p = m v = V e m 2



Davisson -Germer experiment

& Electron Diffraction pattern

(Go through the details of the experiments)

These two experiments confirmed de- Broglie relationship p = h /l.

Subsequently it was found that atomic beams,

and beams of neutrons, also exhibit diffraction

when reflected from regular crystals. Thus de

Broglie's formula seems to apply to any kind of

matter.



Now the dual nature of matter and radiation is

an accepted fact. And it is stated in the principle

of complementarity. This states that wave and

particle models of either matter or radiation

compliment each other.

SJ: P-SE 40.5 The wavelength of an Electron Calculate the de- Broglie wavelength for an electron moving at 1.0 x 107 m/s.

SJ: P-SE 40.6 The Wavelength of a Rock A rock of mass 50 g is thrown with a speed of 40 m/s. What is its de Broglie wavelength?



SJ: P-SE 40.7 An Accelerated Charged Particle A particle of charge q and mass m has been accelerated from rest to a nonrelativistic speed through a potential difference of V. Find an expression for its de Broglie wavelength.

SJ: Section 40.5 P-35 (a) An electron has a kinetic energy of 3.0 eV. Find its wavelength. (b) Also find the wavelength of a photon having the same wavelength.



SJ: Section 40.5 P-38 In the Davisson-Germer experiment, 54.0 eV electrons were diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at = 50.0,

what was the lattice spacing a between the vertical rows of atoms in the figure?



What is a Quantum Particle?

How to represent a quantum particle?

Wave packet Phase velocity Group velocity

Double Slit Experiment

Conclusion

Summary

THE QUANTUM PARTICLE


What is a Quantum Particle?

Quantum particle is a model by which

particles having dual nature are represented.

We must choose one appropriate behavior for

the quantum particle (particle or wave) in

order to understand a particular behavior.



How to represent a quantum particle?

To represent a quantum wave, we have to

combine the essential features of both an ideal

particle and an ideal wave.

An essential feature of a particle is that it is

localized in space. But an ideal wave is

infinitely long (unlocalized) as shown in figure

below.



Now to build a localized entity from an infinitely long wave, waves of same amplitude, but slightly different frequencies are superposed. The result of superposition of two such waves are shown below.



If we add up large number of waves in a similar way, the small localized region of space where constructive interference takes place is called a wavepacket, which represents a particle.

In the figure, large number of waves are Combined. The result is a wave packet,

which represents a

particle.



Mathematical Representation of a wave packet

superposition of two waves of equal amplitude, but

with slightly different frequencies, f1 & f2 and

wavelengths, traveling in the same direction are

considered. The waves are written as

( )t -xk cos A y1 11

= ( )t -xk cos A y2 22

=and

The resultant wave is, y = y1 + y2

t2

+ -x

2

k + k cost

2

-x

2

k cos 2A =y )()]([ 2121

Amplitude varies with t and x



Where k = k1 k2 and = 1 2.

The resulting wave oscillates with the average

frequency, and its amplitude envelope (in

square brackets, shown by the blue dotted

curve in figure) varies according to the

difference frequency.



This envelope can travel through space with a different speed than the individual waves. This speed is called the group speed or the speed of the wave packet (the group of waves)

k

k

g

2

2 speed, group The v

For a superposition of large number of waves to form a wave packet, this ratio is

dk

dg v



A realistic wave (one of finite extent in space) is

characterized by two different speeds. The phase

speed, the speed with which wave crest moves,

which is given by

& the group speed, the speed with which the

envelope (energy) moves. This is given by

In general these two speeds are not the same.



fp v

dk

dg v

Relation between group speed and phase speed

ppp

+dk

dk=

dk

)kv(d =

dk

d=But

g

p==.,e.i

==,havewe

phase

phase

k k

f k

g = p l d

dp

Substituting for k in terms of l, we get



Relation between group speed and particle speed

h

E 2f 2 ==

h

p 2=

ph

2=

2= kand

dp

dE=

dp h

2

dEh

2

=dk

d=v

g

For a classical particle moving with speed u, the kinetic energy E is given by

u=m

p =

dp

dE or

m2

dp p 2= dE and

m2

p

2

1 E

2

=mu= 2

velocity particle the v,dp

dE

dk

d i.e., g ===


speed


ie., we should identify the group speed with

the particle speed, speed with which the

energy moves.

To represent a realistic wave packet, confined to

a finite region in space, we need the

superposition of large number of harmonic waves

with a range of k values.




SJ: Section 40.6 P-43 Consider a freely moving

quantum particle with mass m and speed u. Its

energy is E= K= mu2/2. Determine the phase

speed of the quantum wave representing the

particle and show that it is different from the

speed at which the particle transports mass and

energy.


The slit separation d is much greater than the individual slit

widths and much less than the distance between the slit and

the detector. The electron detector is movable along the y

direction in the drawing and so can detect electrons

diffracted at different values of . The detector acts like the

viewing screen of Youngs double-slit experiment with

light as learned in interference of light.

Electron interference

Experimental details

And the discussion of

the results

THE DOUBLESLIT EXPERIMENT REVISITED


Photograph of a double-slit interference pattern produced by electrons.

2pd

h sin



The minimum occurs when

Electron wavelength is given by

For small angle ,

/2 sind l

/p h l

This experiment proves the dual nature of electrons.

The electrons are detected as particles at a localized

spot at some instant of time, but the probability of

arrival at that spot s determined by finding the

intensity of two interfering waves.

If slit 2 is blocked half the time, keeping slit 1

open, and slit 1 blocked for remaining half the time,

keeping 2 open, the accumulated pattern of

counts/ min is shown by blue curve. That is

interference pattern is lost and the result is simply

the sum of the individual results.



Results of the two-slit electron diffraction experiment with each slit closed half the time (blue). The result with both slits open (interference pattern) is shown in brown.



The observed interference pattern when both the

slits are open, suggests that each particle goes

through both slits at once. We are forced to

conclude that an electron interacts with both the

slits simultaneously shedding its localized

behaviour.

If we try to find out which slit the particle goes

through the interference pattern vanishes! Means,

if we know which path the particle takes, we lose

the fringes. We can only say that the electron

passes through both the slits.



SJ: Section 40.7 P-46 Electrons are incident on a

pair of narrow slits 0.060 m apart. The bright

bands in the interference pattern are separated

by 0.40 mm on a screen 20.0 cm from the slits.

Determine the potential difference through which

the electrons were accelerated to give this

pattern.

THE UNCERTAINTY PRINCIPLE


Quantum theory predicts that, it is fundamentally

impossible to make simultaneous measurements

of a particles position & momentum with

infinite accuracy. This is known as Heisenberg

uncertainty principle. The uncertainties arise from

the quantum structure of matter.

For a particle represented by a single wavelength

wave existing throughout space, l is precisely

known, and according to de- Broglie hypothesis, its p

is also known accurately. But the position of the

particle in this case becomes uncertain.



This means l = 0, p =0; but x =

In contrast, if a particle whose momentum is

uncertain (combination/ a range of wavelengths

are taken to form a wavepacket ), so that x is

small, but l is large. If x is made zero, l, &

thereby p will become .

In short ( x ) ( px) h / 4

Also ( E ) ( t) h / 4



SJ: P-SE 40.8 Locating an electron

The speed of an electron is measured to be

5.00 x 103 m/s to an accuracy of 0.0030%.

Find the minimum uncertainty in determining

the position of this electron.

SJ: P-SE 40.9 The Line Width of Atomic Emissions

The lifetime of an excited atom is given as 1.0 x 10-8 s.

Using the uncertainty principle, compute the line width

f produced by this finite lifetime?



SJ: Section 40.8 P-51 Use the uncertainty

principle to show that if an electron were

confined inside an atomic nucleus of diameter

2x 10-15 m, it would have to be moving

relativistically, while a proton confined to the

same nucleus can be moving

nonrelativistically.


SJ: Section 40.8 P-52 Find the minimum

kinetic energy of a proton confined within a

nucleus having a diameter of 1.0 x 1015 m.



QUESTIONS 1. Explain (a) Stefans law (b) Wiens displacement law

(c) Rayleigh-Jeans law. [1 EACH] 2. Sketch schematically the graph of wavelength vs intensity of

radiation from a blackbody. [1] 3. Explain Plancks radiation law. [2] 4. Write the assumptions made in Plancks hypothesis of

blackbody radiation. [2] 5. Explain photoelectric effect. [1] 6. What are the observations in the experiment on photoelectric

effect? [5] 7. What are the classical predictions about the photoelectric

effect? [3] 8. Explain Einsteins photoelectric equation. [2]



QUESTIONS 10.Which are the features of photoelectric effect-experiment

explained by Einsteins photoelectric equation? [2] 11.Sketch schematically the following graphs with reference to

the photoelectric effect: (a) photoelectric current vs applied voltage (b) kinetic energy of most-energetic electron vs frequency of incident light. [1EACH]

12.Explain compton effect. [2] 13.Explain the experiment on compton effect. [5] 14.Derive the compton shift equation. [5] 15.Explain the wave properties of the particles. [2] 16.Explain a wavepacket and represent it schematically. [2] 17.Explain (a) group speed (b) phase speed, of a wavepacket.

[1+1]



QUESTIONS 20.Show that the group speed of a wavepacket is equal to the

particle speed. [2] 21.Explain Heisenberg uncertainty principle. [1] 22.Write the equations for uncertainty in (a) position and

momentum (b) energy and time. [1]


manipal be phys 2011 12 quantumphysics

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