Modern Physics Class Final

INTRODUCTION TO QUANTUM PHYSICS

• Blackbody Radiation & Plank’s

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

1MIT- MANIPAL

INTRODUCTION

Failure of classical mechanics

Brief summary of chapter 40 of the text book

2

INTRODUCTION TO QUANTUM PHYSICS

BLACKBODY RADIATION & PLANCK’S HYPOTHESIS

Origin of thermal radiation – the classical view point

Concept of oscillators

INTRODUCTION

3BE-PHYSICS-INTRODUCTION TO QUANTUM PHYSICS-

2010-11MIT- MANIPAL

Definition of a Black-Body

Black-Body Radation Laws

1- The Stefan-Boltzmann Law

2- The Wien‘s Displacement Law

3- The Rayleigh-Jeans Law

4- The Planck Law

Application for Black Body

Conclusion

Summary

4


A body which absorbs all the electromagnetic radiation falling on it is called a black body.

Black body’s are good absorbers of radiation are also good emitters.

A blackbody is one whose absorptivity is 100%.

A black body will emit radiation at the fastest rate. A black body has maximum emissive power at a particular temperature.

Blackbodies absorb and re-emit radiation in a characteristic pattern called a spectrum.Lamp black surface may be considered as perfectly black for all practical purposes.


http://en.wikipedia.org/wiki/Electromagnetic_radiation

http://www.google.co.in/imgres?imgurl=http://library.thinkquest.org/C007571/images/block.jpg&imgrefurl=http://library.thinkquest.org/C007571/english/advance/background4.htm&h=198&w=213&sz=3&tbnid=-vcYvhVNgpx0mM:&tbnh=99&tbnw=106&prev=/images?q=black+body+radiation&hl=en&usg=__7_ZtQeLihF6y4_v_r1GpBdmWM8E=&ei=n3HVS4KhIY3c7AO73vm1Dw&sa=X&oi=image_result&resnum=7&ct=image&ved=0CBwQ9QEwBg




It consists of a double walled hollow metallic sphere with a narrow opening and lamp-blacked.

When the radiation enters into the body through the hole, it suffers multiple reflections inside the sphere and is completely absorbed.

This causes a heating of the cavity walls.

FIGURE 28.1 The opening to the cavity inside a hollow object is a good approximation of a black body.

Fery designed a black body

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.

7


1. The black body is an ideal absorber of incident radaition.

2. The emitted "thermal" radiation from a black body characterizes the equilibrium temperature of the black-body.

3. Emitted radiation from a blackbody does not depend on the material of which the walls are made.

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9

1. At the given temperature the intensity of radiation increases along with the wavelength and reaches the maximum value at a particular value of wavelength λmax. Then it decreases along with the increase of the wavelength.

2. At a given temperature, the intensity is not uniformly distributed in the radiation spectrum of a black body.


General characteristics of curves

Figure shows Intensity of blackbody radiation versus wavelength at three temperatures.

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3. The wavelength λmax at the maximum emission of intensity shifts towards the shorter wavelength as the temperature increases.

λmaxT= a constant This represents Wien’s displacement law.



4. For all wavelengths the energy emission increases along with the temperature.

5. The area under the each curve gives the amount of energy emitted at a given temperature.



12

At room temperature, black bodies emit mostly infrared wavelengths, but as the temperature increases past a few hundred degrees Celsius, black bodies start to emit visible wavelengths, appearing red, orange, yellow, white, and blue with increasing temperature.

By the time an object is white, it is emitting substantial ultraviolet radiation.

http://en.wikipedia.org/wiki/Infrared

http://en.wikipedia.org/wiki/Celsius

http://en.wikipedia.org/wiki/Ultraviolet_light

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 Wien’s Law. 13

BE-PHYSICS-INTRODUCTION TO QUANTUM PHYSICS-2010-11

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Stefan Boltzmann Law.

Stefan’s states that the power of the emitted radiation is directly proportional to the fourth power of its absolute temperature.Thus

P α T4

P = σ A e T4

P- power in watts radiated from the surface of an objectσ- is a constant called Stefan’s constant. σ= 5.670 x 10-8 W/m2.K4

A- the surface area of the object in square meters.e – the emissivity of the surfaceT – the surface temperature in Kelvins.

Black-Body Radiation Laws (1)BLACKBODY RADIATION & PLANCK’S HYPOTHESIS

We know,

I= P/A is the definition of intensity. For a black body e=1 exactly.

So we can write, Stefan’s law in the form: I= σ T4

Failed to predict the shapes of the curve and the shift of the peak with temperature described by Wien’s displacement law.

Black-Body Radiation Laws (2)

Wien’s Displacement Law.

lm T = constant = 2.898 × 10-3 m- K, or lm T-1

Where λm - peak of the wavelength distribution in the black body emission spectrum.

T- is the absolute temperature of the surface of the object emitting the radiation. 16


The peak of the wavelength distribution shifts to shorter wavelengths as the temperature increases.

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 35°C

Solution:T=273+35=308 K


2010-11MIT- MANIPAL


lm T = constant = 2.898 × 10-3 m- Klm = 2.898 × 10-3/Tlm = 9.4 μm

18

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

Solution:lm = 2.898 × 10-3 /Tlm = 2.898 × 10-3/2000

lm = 1.4 μm

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

lm = 2.898 × 10-3/5800=0.5 μm

Black-Body Radiation Laws (3)The Rayleigh-Jeans Law. The Rayleigh-Jeans law gives a distribution of energy for a particular wavelength λ as

4BTck2

)T,(Iλ

π=λ

19


Where kB – Boltzmann's constant (1.38 x 10-23 J/K).

T object temperature in kelvins.

c- velocity of light.

20

To describe the distribution of energy from a black body, it is useful to define I(λ,T)dλ to be the intensity, or power per unit area, emitted in the wavelength interval dλ.

dTck2

d)T,(I4

B

Where

I (,T) d is the intensity or power per unit area emitted in the wavelength interval d from a blackbody.

I (,T) d=P/A

Where λ=(λ1+ λ2)/2

d= 1~λ2

It agrees with experimental measurements for long wavelengths.

It predicts an energy output that diverges towards infinity as wavelengths grow smaller.

Rayleigh-Jeans Law

21


Limitation: It was found that Rayleigh-Jeans law holds good in the region of longer wavelength but not for shorter wavelengths.

Thus Wien’s law as well as Rayleigh-Jeans law do not agree with the experimental results throughout its spectrum.

€€

22

Ultraviolet catastropheThe failure has become known as the ultraviolet catastrophe.

4BTck2

)T,(Iλ

π=λ

According to the equation:When λ approaches zero, the function I(λ,T) approaches infinity. In contrast to this prediction, the experimental data plotted shows that as λ approaches zero, the function I(λ,T) approaches zero.This mismatch of theory and experiment was called Ultraviolet catastrophe.

(This ‘‘catastrophe“ – infinite energy-occurs as the wavelength approaches zero – ‘ultraviolet‘ was applied because ultaviolet wavelengths are short).

Black-Body Radiation Laws (4) The Planck Law

23


The distribution of energy from a black body at a particular wavelength λ is given as

1 -Tλk

hc

e

15λ

2hcπ2= T) ,(λI

B

Where kB – Boltzmann's constant (1.38 x 10-23 J/K). T- object temperature in Kelvins.c- velocity of lighth- Planck’s constant=6.626 x 10-34 J.s

Only the extra quantity (compared to the Rayleigh-Jeans Law) coming here is the constant known as Plank’s constant (h) introduced by Max Planck in this revolutionary theory.

The results of Planck's law

For very small λ,

• The denominator [exp(hc/λkBT)] tends

to infinity faster than the numerator (λ-

5), thus resolving the catastrophe.

i.e. I (l, T) 0 as l 0.

24


25

For very large λ:Show that at longer wavelengths, Planck’s radiation law reduces to the Rayleigh-Jeans law.

SJ: Section 40.1 P-10.

neglectedtermsorderhigher,smallxeargl(

x1.........!3

x

!2

xx1eUse

Tk

hcxTake

1 -Tλk

hc

e

15λ

2hcπ2= T) ,(λI

32x

B

B

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.Planck’s Assumptions

Planck assumed that the cavity radiation came from atomic oscillations in the cavity walls.He made two bold and controversial assumptions concerning the nature of the oscillators in the cavity walls.

26


1)The energy of an oscillator can have only certain discrete values E

En = nhf

where n is a positive integer called a quantum number,f is the frequency of oscillation, and h is a constant called Planck’s constant.• Energy of the oscillator is quantized. • Each discrete energy value corresponds to a

different quantum state, represented by the quantum number n.

27


Quantized Energy Levelsof a Harmonic Oscillator

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.

Figure shows allowed energy levels for an oscillator with frequency f, and the allowed transitions.

28

BLACKBODY RADIATION & PLANCK’S S HYPOTHESIS

Quantized Energy Levelsof a Harmonic Oscillator

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.

29

BLACKBODY RADIATION & PLANK’S HYPOTHESIS

Solution: The total energy of a simple harmonic oscillator having an amplitude ‘A’ is given by

E= ½ k A2=1/2( 25 N/m) (0.4m)2= 2.0 J

Hz56.0m

k

2

1f

30

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

En = nhf

n= En/hf = (2 J)/ (6.626 x 10-34) (0.56)

n = 5.4 x 1033

SJ: P-SE 40.2 The Quantized Oscillator contd.

Suppose in the previous problem, 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 in this one-quantum change.

Solution: E= En-En-1= hf= 3.7 x 10-34 J


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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 Sun’s surface emits as a black body, calculate its surface temperature. (b) Using the result, find max for the Sun.

32


Sun

Area=A=(4πR2) (Sphere)

nm504K10x75.5

10x898.2)b(

K10x75.5T

)K.m/W10x67.5()m10x96.6(4(1

W10x77.3

eA

PT

TeAP)a(

3

3

max

3

4/1

42828

264/1

4

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?

33


s/photons10x27.2)10x7.99(x)10x672.6(

10x150

hf

P

E

Pn

ondsecperemittedphotonsof.nonWhere

,nhfnEP

30

634

3

34

SJ: Section 40.1 P-6.

A sodium-vapor lamp has a power output of 10.0 W. Using 589.3 nm as the average wavelength of this source, calculate the number of photons emitted per second.

s/photons10x96.2)10x7.99(x)10x672.6(

10x150

hc

P

)c

(h

P

hf

P

E

Pn


,nhfnEP

19

634

3

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?

35


Solution: T=constant/λm= 2.898 x 10-3/560 nm

= 5.18 x 103 K

36

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.

Solution: Planck’s law gives intensity per wavelength.

Taking E to be photon energy and n to be the number of

photons emitted per second, we multiply by area and

wavelength range to have energy per time leaving the

hole.

Take: λ2= 501 nm, λ1= 500 nm, So λ=500.5 nmdλ= 501-500 = 1nmT= 7500 KkB= 1.38 x 10-23, h=6.627 x 10-34, c=3 x 108, d=0.050 x 10-3 m

37

s/10x3.1E

Pn


10x94.7)c2

(h)c

(hhfE,nhfnEP

15

19

21

1221

Tk)(

hc2

521

12

22

Tk

hc5

2

Tk

hc5

2

d&2

where

1e)2

(

)()4d

(hc2d

1e

1Ahc2P

A/Pd

1e

1hc2d)T,(I

B21B

B

Then P=10.32 x 10-4 Watts

Hole(circle

)

Area=A=(πd2/4)

38

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.

)waveradio(nm52.6)c(

)waveradio(cm68.9)b(

)colorblue,visible(nm484)Hz10x620(

10x3)a(

f

cUse)d(

eV10x91.1EHz10x46MHz46f)c(

eV10x28.1E,Hz10x10.3GHz10.3fSimilarly)b(

eV57.2)J10x6.1(

)Hz10x620(x)s.J10x627.6(hfE)a(

12

8

76

59

19

1234

39

SJ: Section 40.1 P-10. A simple pendulum has a length of 1.00 m and a mass of 1.00 kg. The amplitude of oscillations of the pendulum is 3.00 cm. Estimate the quantum number for the pendulum.

31

3

2

10x34.1hf

En,So

nhfE,Now

Hz498.0L

g

2

1f:noscillatioofFrequency

J10x41.4E

)9995.000.1)(ms8.9()kg0.1(E

)cosLL(mgmghE:energytotals'pendulumsimpleThe

radian03.0m1

m10x3

radius

arcTake

2

1. Introduction

1. What is Photoelectric Effect?

2. Apparatus for studying Photoelectric Effect

3. Experimental Observations

4. Classical Predictions

5. Clash between Classical predictions

6. & Observed Experimental results

7. Einstein’s model of the Photoelectric Effect

8. Explanation for the observed features of PE

1. Application

2. Conclusion

3. Summary40


MIT- MANIPAL

THE PHOTOELECTRIC EFFECT

Light of frequency (f)

(Photoelectrons)Emission of electrons

Photo electric current

Photoelectric effect

(Photosensitive metal)

The Photoelectric Effect

The emission of electrons from a metal plate when illuminated by light (electromagnetic radiation) of suitable wavelength (or frequency) is called photoelectric effect.

The emitted electrons are called as ‘photoelectrons’.

Photoelectric Effect Schematic

When light (an appropriate wavelength) strikes Emitter (E), photoelectrons are emitted.

Electrons collected at Collector (C) and passing through the ammeter (A) are a current in the circuit.

An evacuated glass/quartz tube

E & C: two electrodes in an evacuated quartz tube)E-Emitter: metal (photosensitive material) connected negative terminal of the battery.C- Collector plate C connected to the positive terminal of the battery.

EXPERIMENTAL OBSERVATIONS ON PHOTOELECTRIC EFFECT:

1. The number of electrons

emitted per second i.e.

photoelectric current is

directly proportional to the

intensity of incident light.

i

I

Photoelectric current (i) α Intensity of incident light (I)

Laws of Photoelectric Emission

2.For a given material, there exists a certain minimum frequency (fc) of

incident light so that photo-electrons can be ejected from metal surface. This minimum frequency is known as threshold /cutoff frequency ( fc).

Photoelectric

current

If f = fc

Light of frequency (f = fc)


3. If the frequency (f) is more than this frequency (fc) electrons escape from the metal

surface and move with certain amount of “kinetic energy” (K.E. =1/2 mv2).

photoelectric current

& electrons have KE

If f > fc

Light of frequency (f > fc)


KE of photoelectrons is directly proportional to frequency of incident light.

fv

fmv2

1

fKE

2

2

i.e. Maximum kinetic energy of the photoelectrons increases with increasing light frequency.

4. Electrons are emitted from the surface of the emitter almost instantaneously.

46



47

Laws of Photoelectric Emission5. The minimum value of the retarding potential

to prevent the electron emission is called the

stopping potential.

In such a case the

Kmax = e ΔVs

Photoelectric current versus applied potential difference for two light intensities.

Where Kmax- maximum KE of the electrons

e- electron charge (1.6 x 10-19 C)

ΔVs- stopping potential

Einstein’s Interpretation of em radiation

(A new theory of light)

1. Electromagnetic waves carry discrete energy packets (light quanta called photons now).

2. The energy E, per packet depends on frequency f. E = hf.

3. More intense light corresponds to more photons, not higher energy photons.

4. Each photon of energy E moves in vacuum at the speed of light c, where c = 3x 108 m/s.

5. Each photon carries a momentum p = E/c.48


Einstein’s model of the photoelectric effect

6. 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 -

f is called the work function of the metal. f It is the minimum energy with which an electron

is bound in the metal. 49


Einstein's photoelectric equation

All the observed features of photoelectric effect could be explained by Einstein’s photoelectric equation.

Observed features:1. Dependence of photoelectron kinetic energy on light intensity:Kmax is independent of the light intensity. Kmax+φ = hf: Equation shows that Kmax depends only on frequency of the incident light.Kmax increases with increase in frequency.

2. Time interval between incidence of light and electron of photoelectronsAlmost instantaneous emission of photoelectrons due to

one -to –one interaction between photons and electrons. 50


51

3. Dependence of ejection of electrons on light frequency. Failure to observe the photoelectric effect below a certain cutoff frequency follows from the fact that photons should have energy greater than the work function in order to eject an electron.

The cutoff frequency fc( the minimum frequency light needed to just to liberate electrons) is related to the work function (minimum energy needed to just to liberate electrons ) by fc = /h.If the incident frequency f is less than fc , no emission of photoelectrons. h- Planck’s constant.

4.Dependence of photoelectron kinetic energy on light frequency: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 52


A plot of Kmax of photoelectrons versus frequency of incident light in a typical photoelectric effect experiment.

Classical Predictions versus Experimental resultsPredictions made by a classical approach (wave nature of light) with experimental results (particle nature).1. Dependence of photoelectron kinetic energy on light

intensity:

(a) Classical predictions: 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.

(b) Experimental results: The maximum kinetic energy of photoelectrons is independent of light intensity. It depends directly on the frequency of the incident light.

53


54

Classical Predictions versus Experimental results

2. Time interval between incidence of light and ejection of photoelectrons:

(a)Classical predictions: If light is really a wave, there should be a measurable/ larger time interval between incidence of light and ejection of photoelectrons is required.

(b) Experimental results:Electrons are emitted from the surface of the metal almost instantaneously, even at very low light intensities.

3. Dependence of ejection of electrons on light frequency:(a) Classical predictions: Ejection of photoelectron should not depend on light frequency.

(b) Experimental results:No electrons are emitted if the incident light frequency falls below some cutoff frequency fc, whose value is characteristic of the material being illuminated.


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56

4.Dependence of photoelectron kinetic energy on light frequency:(a) Classical predictions: Photoelectron kinetic energy should not depend upon the frequency of the incident light.

(b) Experimental results:The maximum kinetic energy of the photoelectrons increases with increasing light frequency.


NOTE: In short, all experimental results

(photoelectric effect) contradict all four

classical predictions, supporting particle

nature of light when it interacts with matter.

Application of photoelectric effect

Photomultiplier tube

Explain the device, theory, and its working

57


Summary

Einstein successfully extended Plank’s quantum hypothesis to explain photoelectric effect.

In Einstein’s model, light is viewed as a stream of particles, or photons, each having energy E = hf , where h is Plank’s 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.


2010-11MIT- MANIPAL


59

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. FindA. The maximum kinetic energy of the ejected

photoelectrons andB. The cutoff wavelength for sodium.

Solution: (A) The energy (in eV) of each photon in the illuminating light beam is E=hf=h(c/λ)=[(6.627 x 10-34) (3 x 108)]/[(1.6 x10-

19) (300nm)E= 4.13 eV

MIT-MANIPAL 60

Kmax = hf - =4.13 eV-2,46 eVKmax= 1.67 eV

(B)

Work function= φ= h fc = h(c/λc)

λc = hc/φ= [(6.627 x 10-34) (3 x 108)]/[(2.46 x 1.6 x10-19V)

λc=(504 nm)

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?

61


(a) Cut off wavelength λc = hc/φ= )= [(6.627 x 10-34) (3 x 108)]/[(4.2 x 1.6 x10-

19V)

λc=(296 nm) Cut off frequency fc= c/λc

fc = 1.01 x 1015 Hz

MIT-MANIPAL 62

Einstein’s Photoelectric effect

E= φ + Kmax

h(c/λ)= φ+ e Vs

Vs=[h(c/λ)- φ]/e=

[Where φ= [4.20 eV)/1.6 x 10-19]

Stopping potential= Vs= 2.71 V

63

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?

34

19

c

19

834

max

10x627.6

)10x6.1(x38.1

hf)b(

eV38.1602.0982.1,So

eV982.110x6.1x)nm625(

)10x3)(10x627.6(hcE

KE)a(

SJ: Section 40.2 P-16. The stopping potential for

photoelectrons released from a metal 1 is 1.48 V larger

compared to that in metal 2. If the threshold frequency

for the metal 1 is 40.0 % smaller than for the metal 2,

determine the work function for each metal.

64


Given: ΔVs1=ΔVs2+1.48 V fc2= 60% fc1=0.6 fc1 implies φ2=0.4 φ2.....(1)(because hfc=φ)

From Einstein's photoelectric equation:

hf=φ1+e ΔVs1 and hf=φ2+e ΔVs2 gives φ1- φ2=1.48 eV …..

(2)

From Eqns (1) & (2):We have φ1=2.22 eV & φ2=3.70 eV


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 ( = 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 ( =

587.5 nm)?

65

V216.0Vs

eV216.0eV9.110x6.1x)10x5.587(

)10x3(x)10x627.6(EVse)b(

eV9.1

eV376.010x6.1x)10x1.546(

)10x3(x)10x627.6(Ve

hcVeE)a(

199

8

34

199

834

ss

66

SJ: Section 40.2 P-15.Lithium. Beryllium, and mercury have work functions of 2.30 eV, 3.90 eV and 4.50 eV, respectively. Light with a wavelength of 400 nm is incident on each of these metals. Determine (a) which metals exhibit the photoelectric effect and (b) the maximum kinetic energy for the photoelectrons in each case.

nm276:)eV5.4(mercuryFor

nm318:)eV9.3(berylliumforSimilarly

nm54010x6.1x)eV3.2(

)10x3(x)10x627.6(hc:LithiumFor)a(

c

c

19

834

c

Incident light: λ=400 nm. Only Lithium has λ < λc i.e f > fc

So, only Lithium exhibits photoelectric effect.

67

• Introduction

• What is Compton Effect

• Schematic diagram of Compton’s apparatus

• Experimental Observations

• Classical Predictions

• Explanation for Compton Effect

• Derivation of the Compton Shift Equation.

• Conclusion

• Summary

68

THE COMPTON EFFECT

69

SUMMARY OF PHOTON PROPERTIES

Relation between particle and wave properties of light

Energy, frequency, and wavelength, E = hf = hc / Also we have relation between momentum and wavelength of a photon as follows

For photon (light), m = 0, E= pc . Also c = f

λ

h=

λf

hf=

c

E=p

A.H.Compton gave direct confirmation of the existence of the particle nature of electromagnetic waves.

In 1923, he discovered that when a monochromatic beam of a high frequency radiation such in X-rays, gamma rays is scattered by a substance, the scattered radiation (scattering of X-rays from electrons) contains two components:

Compton Effect

(a)one having high wavelength λ’ (or low frequency or low energy) known as modified radiation.

(b)and the other having the same wavelength as the original λ0 one known as unmodified radiation.

Compton (1923) measured intensity of scattered X-rays as function of wavelength for different angles.

In such a scattering, a wavelength shift (λ’-λ0) in wavelength for the scattered X-rays takes place, which is known as Compton shift.

This phenomenon is known as Compton effect.

Scattered beam of X ray

What is Compton Effect ?

71

THE COMPTON EFFECT

Compton shift

Incident photon (X-ray photon)

Photons interacting with electrons:Photons interacting with electrons:

According to classical theory, electromagnetic waves of frequency f0 incident on electrons should have two effects.

(1) Radiation pressure should cause the electrons to accelerate in the direction of propagation of the waves.

Classical Predictions

73

(2) The oscillating electric field of the incident radiation should set the electrons into oscillation at the apparent frequency f’. This apparent frequency f’ is different from the

frequency f0 of the incident radiation

because of the Doppler effect.

Each electron first absorbs em radiation as a moving particle and then re-radiates em waves as a moving particle, thereby exhibiting two Doppler shifts in the frequency of radiation.

Classical Predictions

MIT-MANIPAL 74

Thus, Classical theory failed to explain the scattering of

x rays from electrons. On the basis of quantum theory,

Compton explained satisfactorily the modified

radiation.

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.

λ0- wavelength of the incident radiationλ’- wavelength of the scattered radiation.From the experimental observation it was found that

1. The difference (λ’-λ0), which indicates the enhancement in the wavelength, is called Compton shift. Compton shift (Δλ) is independent of the wavelength of the incident radiation.

Experimental Observations and Results

77

2. Δλ is independent of the nature of the scatterer as it is same for all scatterer. Compton made measurement by replacing graphite with other materials, and found that λ’ is independent of the target material.

3. Δλ depends only on the scattering angle θ as the scattering angle is increased Δλ also increases. Scattered x-ray intensity versus wavelength for Compton scattering at = 0°, 45°, 90°, and 135°.

Δλ is maximum for θ=1800 & Δλ =0 for θ=00


4. The incident X-ray photon with energy (E0=hc/λ0), when collides with the electron, it gives some of its energy to (a)the electron which recoils with a velocity (v)

in the direction making an angle (φ) with the direction of the incident photon.

and remaining energy to(b) the photon with reduced energy (E’=hc/λ’) scattered in the direction θ with the original direction.


.lyrespectivephotonsscattered&incidenttheofswavelength'&

lightofspeedtheisc

,ttanconss'Plancktheish,Where

K'

hchc

electronofKE'EE

Where

0

e

0

0

5. From the law of conservation of energy and conservation of momentum, Compton derived an equation for the change in wavelength Δλ (known as Compton shift) of the X-rays, which is given follows.

nm00243.0cm

h

.electrontheofwavelengthComptonthecm

hfactorThe

,equationshiftComptontheasknownisressionexpThis

kg10x11.9electrontheofmassmWhere

)cos1(cm

h)(shiftCompton

e

e

31

e

e

0

1


Prediction were in excellent agreement with the experimental results.

80

The total energy ‘E’ relativistic particle is given by

THE COMPTON EFFECTGeneral Information

A relativistic particle is a particle which moves with a relativistic speed; that is, a speed comparable to the speed of light (v˜c)

2 2 2 2 4E p c m c

Here p and m are the momentum and mass of the particle: c is the speed of light.

But, photon mass m=0, So, the relativistic energy of a photon: E2=p2c2

http://en.wikipedia.org/wiki/Relativistic_speed

http://en.wikipedia.org/wiki/Speed

http://en.wikipedia.org/wiki/Speed_of_light

81

v = speed of the electron & c = speed of light in vacuum, m is the mass of the electron.

When a subatomic particle like electron travels with a speed comparable with the speed of light (v~c):

Its relativistic energy :

Ee2=pe

2c2+m2c4

Where pe= mv (Relativistic momentum of electron)

Where

2

2

cv

- 1

1 factorLorentz γ

82

NOTE: A non relativistic particle

(Newtonian Mechanics) is a particle which

moves with a speed very small compared

to the speed of light (v<<c).

For non relativistic particle: =1.

So, p=mv

Newtonian definition of momentum is

valid at low speeds.

http://en.wikipedia.org/wiki/Speed



Derivation of the Compton Shift Equation

Figure shows an X-ray incident photon is treated as a particle having energy E = hf0 = hc/0 and zero rest energy.

They collide elastically with free electrons initially at rest as shown in figure.

83

THE COMPTON EFFECT

incident photon

scattered photon

0- wavelength of the

incident X-ray photon.’- wavelength of the

scattered X-ray photonm- rest mass of the electronh- Planck’s constant & c-speed of photon

84

The incident X-ray photon with energy (E0=hc/λ0), when collides with the electron, it gives some of its energy to 1. the electron which recoils with a relativistic

velocity (v) in the direction making an angle (φ) with the direction of the incident photon.

The relativistic energy of the scattered electron of mass ‘m’:

Ee2=pe

2c2+m2c4…………..(1)

Where pe= mv (relativistic momentum of

electron)

and (b) remaining energy to the photon with reduced energy (E’=hc/λ’) scattered in the direction θ with the original direction.

85

In the scattering process, the total energy and total linear momentum of the system must be conserved.

y mv sinφ

Incident photon

Energy=E0=hc/λ0

Momentum=p0=h/λ0

Scattered electronwith relativistic momentum (pe= mv) φ

Scattered photon

(h/λ’ cosθ+ mv cosφ)θ

(Momentum=p’=h/λ’)

x

h/λ’ sinθ

EnergyE’=hc/λ’

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

:component x v φ m + θ cosλ'

h=

0λ

hcos

:component y v0 φ m - θ λ'

h= sinsin

Where h/0 = p0 is the momentum of the incident photon h/’ = p’ is the momentum of the scattered photon

86

THE COMPTON EFFECT

2

2

cv

- 1

1 factorLorentz γ

Rewriting the above equations as

:component x ' φ p θ p-p cose

cos0

:component y ' φ e θ p sinsin p

Squaring and adding the above equations give

p= p'+ θ cosp'p2- p 22

0

2

0 e


2010-11MIT- MANIPAL

THE COMPTON EFFECT

)A......(..........p'

hcos

'h

2h 2

e2

2

0

2

2

0

2

88

Applying the law of conservation of energy to the process gives

ENERGY BEFORE COLLISION:

(Energy of the incident photon+ rest

mass energy of the electron) = E0 + mc2

ENERGY AFTER COLLISION

(Energy of the scattered photon +

energy of the recoiled electron)= E’+Ee

ENERGY BEFORE COLLISION=ENERGY AFTER

COLLISION

E0 + mc2= E’+Ee

MIT-MANIPAL 89

Squaring both sides by taking [(a=E0 & b=(E’-mc2)]

[E0-(E’-mc2)]2=Ee2

E02+(E’-mc2)2-2E0(E’-mc2)=Ee

2

E02 + E’2 + m2c4 - 2E’mc2 - 2E0E’ + 2E0mc2 = pe

2 c2+m2c4

4222

e

2

00

22

242

2

2cmcpmc

hc2'

ch2mc

'hc

2c

m+λ'ch

λch

2

2

0

22

Rewrite as: E0- (E’-mc2)=Ee

MIT-MANIPAL 90

2

22

e

2

00

22

2

hcbysidesbothDivide

cpmchc2

'

ch2m

'

hc2

2

2

0

2

λ'h

λh

2

e

00

2p

mc2'

h2

'mc

2

2

0λ'h

λh

cos

'h2

'hhmc2

'h

2'

mc2

pfor)A(EqnSubstitute

0

22

000

2

2

e

2

0λ'h

λh

91

getwe'byMultiply

cos'

h2mc2'

h2

'mc

2

0

000

)cos1(mc

h)'(Gives

)cos1(h2mc)'(2

cosh2mc'2h2mc2

0

0

0

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.

)θ cos - 1(cm

h λ - 'λ

e0

92

THE COMPTON EFFECT

SJ: P-SE 40.4 Compton scattering at 45°X-rays of wavelength 0 = 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. (a) Calculate their wavelength. (b) 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?


2010-11MIT- MANIPAL

THE COMPTON EFFECT

) cos - 1(cm

h - '

:bygivenisraysXscattedtheofwavelengthinshiftThe)a(

e

0

MIT-MANIPAL 94

nm200710.0'

)45cos1())(3x10(9.1x10

10x627.610x2.0

) cos - 1(cm

h '

:bygivenisraysXscattedtheofwavelengthThe

831-

349

e

0

(b) When θ increases larger than 450, cos θ value decreases and (1-cosθ) increases.

As the Compton shift is directly proportional to the term (1-cosθ), the shift also increases.

MIT-MANIPAL 95

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

s/m.kg10x47.9nm700

10x627.6hp

eV78.1J10x84.210x700

)10x3(x)10x627.6(hcE

28

34

19

9

834

96

SJ: Section 40.3 P-22 X rays having energy of 300 keV undergo Compton scattering from a target. The scattered rays are detected at 37.00 relative to the incident rays. Find (a) the Compton shift at this angle(b) The energy of the scattered x-rays and © the energy of the recoiling electron.

m10x88.4) cos - 1(cm

h - '

:bygivenisraysXscattedtheofwavelengthinshiftThe)a(13

e

0

MIT-MANIPAL 97

keV32keV268keV300'EEK)c(

keV268'

hc'E

m10x63.410x88.410x14.4'

m10x14.4)

10x6.110x300

(

)10x3)(10x627.6(

E

hchcE:findTo

''

hc'E)b(

0

e

1213

12

3

834

0

0

0

00

0

12

19

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?

98

THE COMPTON EFFECT

0

e0

e

0

e

000

70gives

h

cm1cos)cos1(

cm

h

) cos - 1(cm

h-2 - '

:bygivenisraysXscatteredtheofwavelengthinshiftThe

MIT-MANIPAL 99

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.

Since

φ e

+ θ cosp=p cosp'0

:component x

:momentumofonConservati)a(

:Given

)1......(p'

h(

0

h

)

e + = cos

)2...('

hpp'gives sinpsin'p

Since

:component y

ee

'

φ θ p

esinsin p

100

0

0

0

43forsolve

Ehc

substitute0

hc

0E

MIT-MANIPALBE-PHYSICS-INTRODUCTION TO

QUANTUM PHYSICS-2010-11101

keV278'EEK

s/m.kg10x22.3'ppe:3eqnfrom)c(

s/m.kg10x21.3c'E

'pThen

0

22

22

keV602)cos2(

0

hc

'

hc'E:eqnuse)b(

MIT-MANIPAL 102

SJ: Section 40.3 P-24. A 0.110 nm photon collides with a stationary electron. After the collision, the electron moves forward and the photon recoils backward. Find the momentum and the kinetic energy of the electron.Given: λ0=0.110 nm

keV8.10'

hc'Egivesnm115.0'

m10x86.4

180where)cos1(mch

0

12

0

keV3.11

0

hc

0E:eqnUse

103

Momentum

p0

pe

p’

No need to resolve into components:

p0= pe + (-p’)

=117.86 x10-25 kg.m/s

eV478'EEKe0

THE COMPTON EFFECT

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

104

Photons and Electromagnetic Waves

Evidence for wave-nature of light• Diffraction • Interference

Evidence for particle-nature of light• Photoelectric effect• Compton effect

105


MIT- MANIPAL

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. 106


MIT- MANIPAL


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.

De Broglie

107


h

p

de Broglie wavelength

346.63 10 Jsh

Planck’s constant

He found that the wavelength of a photon can be expressed as

de Broglie Hypothesis

momentum

108

The wave Properties of Particles

In 1924, Louis de Broglie postulated that like photons, perhaps all forms of matter (electrons/protons) have wave and particle properties.

Thus, the waves associated with material particles (like electrons , protons) are called matter waves or De-Broglie waves.De Broglie suggested that material particles of mass ‘m’ momentum p (=mv) have a characteristic wavelength (λ) that is given by the same expression:

mv

h

p

h

Momentum

ttanconss'PlanckWavelength

109

Furthermore, in analogy with photons, particles obey the Einstein relation E=hf, where E is the total energy of the particle.Then the frequency of the particles is

hE=f

frequency of the particle

Energy of the particle

Planck’s constant

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

Ve v m 22

1 Where m = mass, v = velocity

p = m v = Ve m 2

110

SJ: P-SE 40.5 -2Calculate the de Broglive wavelength for an electron (me= 9.11 x 10-31 kg) moving at 1.00 x 107 m/s .

m11

10x28.7

)710x1)(kg3110x11.9(

s.J3410x63.6

v m

h

p

h

SJ: P-SE 40.5-33Calculate the de Broglive wavelength for an proton (mp= 1.67 x 10-27 kg) moving at 1.00 x 106 m/s .(Answer: 3.97 x 10-13 m)

111

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

m34

10x32.3

)s/m40)(kg310x50(

s.J3410x63.6

v m

h

p

h

MIT-MANIPAL 112

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.When a charged particle is accelerated from rest through a potential difference ΔV, its gain in kinetic energy ½ mv2 must equal the loss in potential energy q ΔV of the charge-field system:

Vmq2

h

p

h

Vmq2p

momentummvp,Vq

v2

1

2m

p

Vq m 2

2

MIT-MANIPAL 113

Assignment:SJ: Section 40.5 P-34

Calculate the de Broglie wavelength for an electron that has kinetic energy (a) 50.0 eV and (b) 50 keV

Answers: (a) λ=0.174 nm (b) λ=5.49 x 10-12 m

MIT-MANIPAL 114

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 energy.

Davisson -Germer experiment

&

Electron Diffraction pattern(Go through the details of the experiments)

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

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.

115

MIT- MANIPAL


Davisson and Germer experiment This is an experiment on diffraction of accelerated electrons by crystals to establish the wave nature of electrons.

The regular spacing of the atoms in a crystal act as a grating for an electron beam producing a diffraction pattern by electron scattering.

The result of this experiment resembles the result of the x ray diffraction giving an experimental evidence for de Broglie hypothesis.

Experimental set up

Detector

Nickel Crystal

Electron Gun

117

A beam of electrons from a heated filament

accelerated to a potential V is collimated and

allowed to strike a single crystal of nickel.

Electrons are scattered in all directions by the

atoms in the crystal. The intensity of the

scattered electron beam is measured by a

detector which can be moved to any angle

relative to the incident beam.

The most intense reflection of electron beam at

an angle = 50, for an accelerating potential 54

V is observed.

An electron of mass m accelerated to a potential V has kinetic energy where , v = velocity.

Momentum,

de Broglie wavelength

Substituting V=54V, we get the experimental value for the de Broglie wavelength as = 0.167nm.

Ve v m 2

2

1

2 m e Vp mv

V e m 2

h

v m

h

p

h

119


Bragg’s Equation

The theoretical value for the de Broglie wavelength can be calculated from the Bragg equation

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.

nsina

a-lattice spacing

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?


120


MIT- MANIPAL



• What is a Quantum Particle?

• How to represent a quantum particle?

• Wave packet• Phase velocity• Group velocity

• Double Slit Experiment

• Conclusion• Summary

122


MIT- MANIPAL

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.

123


MIT- MANIPAL


How to represent a quantum particle?

To represent a quantum particle, 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. “Localized” - definite position, momentum, confined in space

But an ideal wave is infinitely long (unlocalized) as shown in figure below. 12

4


Wave packet

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.

125

THE QUANTUM PARTICLEWave packet

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.

126


The small region of constructive interference is called wave packet.A wave packet is localized – a good representation for a particle!

Wave packet

This is a localized region of space that is different from all other regions.

This location of the wave packet can be considered as a particle and it corresponds to the position of the particle.

This localized region (wave packet) can represent the particle and let us show that the wave packet has another characteristic of 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

ω= ( )t -xk cos A y2 22

ω=and

128


MIT- MANIPAL


w- angular frequency (ω) which is given by: ω = 2πf

k- the wavenumber or propagation constant given by

k = 2π/ λ



The resultant wave is, y = y1 + y2 t

2ω+ ω

-x 2

k + k cost

2Δω

-x 2

Δk cos 2A =y )()]([ 2121

Amplitude varies with t and xWhere 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.

Wave velocity (or phase velocity): vp

Individual wave inside the wave packet

travels with different velocity called wave

velocity or Phase velocity (vp).

The rate of advance of a crest on a

single wave, which is a point of fixed

phase is called Wave velocity Phase

velocity (vp).

The red dot moves with the phase velocity.

http://en.wikipedia.org/wiki/Phase_velocity

http://en.wikipedia.org/wiki/Phase_velocity

http://en.wikipedia.org/wiki/File:Wave_phase.gif

131

Expression for vp: The individual wave

(plane progressive wave) travelling in

the positive direction of x-axis is given

by

y = A cos (kx-wt)

w- angular frequency =2πf,

k- the wave number: k = 2π/ λ

(kx-wt)- is the phase of wave motion and is equal to a constant .

i.e. (kx-wt)= a constant

Differentiating w.r.t .t. we get

k dx/dt -w=0

Wave velocity /phase velocity: vp = dx/dt, is given by

vp = ω / k

Group velocity (vg)Group velocity is the velocity with which

the envelope enclosing a wave group- called

wave packet formed due to superimposition

of two or more travelling waves of slightly

different wavelengths, is transported.

It is the velocity with which the energy

transmission occurs in a wave.

134

dkdw

kw

)2/k()2/w(

xiablevarspaceof.coefftiablevartimeof.coeff

vg

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

This wavepacket (envelope) can travel through space with a different speed than the individual waves.

For the superposition of two waves, the factor

t2

wx

2

kcosA2

is of the form of a wave, so it moves with a speed given by:

NOTE:A 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

(w=2πf (angular frequency) and k=2π/λ (wave number)

The group speed, the speed with which the envelope (energy) moves. This is given by

For a superposition of a very large number of waves to form a wave packet, this ratio becomes a derivative.In general these two speeds are not the same.

135


vp = ω / k =fλ

Relation between group speed )(vg) and phase speed (vp)

p

pp

dk

dk

dk

)kv(d =

dkdω

=υButg

p==.,e.i

==,havewe

phase

phase

υυω

λω

υ

k k

f k

g = p – dd p

λ

υ

Substituting for k in terms of , we get

136

MIT- MANIPAL


Wherew=2πf-angular frequency

K=2π/λ- wave number

f- wavelengthλ -wavelength

Relation between group speed(vg) and particle speed (u)The group speed, the speed with which the envelope

(energy) moves. This is given by )1....(..........

dkdw

vg

momentump

wavetheofenegryEwherepE

kw

hp 2π

=ph 2π

=λ

2π=k :numberWave

hE

2f 2 :frequencyAngular:knowWe

138

)2........(dpdE

dkdω

= v:(1)eqnThereforeg

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

mu)p(

....(3)u =mp

=dpdE

or

m2 dp p 2

= dE and m2

p mu

21

E 2

2

velocityparticle theu,dpdE

dkdω

:velocityGroup

(3)&(2) From i.e.,

g

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.

139


MIT- MANIPAL



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.

140

2vv:)2(&)1(From

)2......(2u

muh

x)2h

mu( f =or v

muh

= and hf mu21

E1)u........(vGiven

g

phase

2

phase

g

2

The double-Slit Revisited This experiment

crystallize our ideas about the electron’s wave-particle duality.

A parallel beam of mono-energetic electrons is incident on a double slit as shown in figure.

We assume the slit widths are small compared to the electron wavelength, so that no need to worry about diffraction maxima and minima.

Electrons detector

An electron detector is positioned far from

the slits at a distance much greater than

d (the separation distance of the slits.

D>>d.

A typical wave interference pattern for the

counts per minute was detected by the

detector which is evidence for wave

nature of matter particles like electrons.

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

143

THE DOUBLE–SLIT EXPERIMENT REVISITED

Maximum intensity occurs at: d sinθ=mλ

m=0,1,2………………………. The minimum intensity occurs when the path difference between at the points A and B is half a wavelength, or when d sinθ=(m+1/2)λ m=0, 1, 2,…………………

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 is 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.

144


The result with both slits open (interference pattern) is shown in brown.

145


Results of the two-slit electron diffraction experiment with each slit closed half the time (blue).

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.

146

MIT- MANIPAL

d=0.060 m y=0.40 mm

D=20 cm

So, θ=tanθ-1(y/D)=tan-1(0.40 mm/20) cm θ=0.1145 rad

MIT-MANIPAL 147

Condition for maximum: d sinθ=mλ m=1 d sinθ=λ=

V105em2h

Vgives

Vem2hVe

m2p

mv21

K

hpgives

mvh

ph

2

2

2

2

2

2

148

SJ: Section 40.7 P-45

Neutrons traveling at 0.400 m/s are directed through a pair of slits having a 1.00 mm separation. An array of detectors is placed 10.0 m from the slits.(a)What de Broglie wavelength of the neutrons?(b)How far off axis is the first zero-intensity point on

the director array?

tanθ=y/D gives y=D tanθ=4.96 mm

Quantum theory predicts that, it is fundamentally impossible to make simultaneous measurements of a particle’s position & momentum with infinite accuracy. This is known as Heisenberg uncertainty principle. The uncertainties arise from the quantum structure of matter.

149

THE UNCERTAINTY PRINCIPLE

MIT-MANIPAL 150

For a particle represented by a single wavelength wave existing throughout space, 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.



Based on this argument Werner Heisenberg, a German physicist, enunciated the principle of uncertainty which says that “It is fundamentally impossible to make both simultaneous measurements of a particle’s position and momentum with infinite accuracy. The more precisely known the value of one, the less precise is the other.

The Heisenberg uncertainty principle: Statement: If a measurement of the position of a particle is made with uncertainty Δx and a simultaneous measurement of its x component of momentum is made with uncertainty Δpx, the product of the two uncertainties can never be smaller than

2/

2px x

Where

s.J10x05.1142.3x2

s.J10x61.6

2

h 3434

This means = 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 is large. If x is made zero, , & thereby p will become .

153


MIT- MANIPAL


4h

)p()x(2

)p)(x(

4h

)t()E(2

)t)(E(

Another case of uncertainty principle relates energy and time:

Where, E is the uncertainty in the energy emitted in the form of an electromagnetic wave in a de-excitation process and ∆t is the uncertainty in the time measurement.

This form of uncertainty principle suggests that energy conservation can appear to be violated by an amount ΔE as long as it is only for a short time interval Δt .

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.Solution:Assuming that the electron is moving along the x axis, the x component of the momentum of the electron is: px=mvx= (9.11 x10-31 kg) x (5.00 x 105 m/s) = 4.56 x 10-27 kg.m/s

The uncertainty in px is 0.00300%=0.0000300Δpx= (0.0000300) (4.56 x 10-25 kg.m/s)

= 1.37 x 10-31 kg.m/s

156

Minimum uncertainty in position:

mm383.0

)s/m.kg10x37.1(2

s.J10x05.1

p2x

31

34

x

157

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?

Hz10x0.8f

)s10x0.1(4

1

t4

1

th2h

Ef

2/t)fh(2

)t)(E(

6

8

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.

Δx= 2x 10-15 m


158

s/m.kg10x6.2x2

p 20

For an electron non relativistic approximation p=mev gives v=2.8 x 1010 m/s

Whole ‘v’ can not be greater than c

MIT-MANIPAL 159

Another solution from relativistic approximation for electron:

From non relativistic approximation for a proton:

v=p/m gives v=1.55 x 107 m/s less than one-tenth

the speed of light.

160

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 10-15 m.

Minimum uncertainty in the position =Δx= 1.0 x 10-15 m

MeV21.5J10x33.8K

m)x(32h

)x4(m2m2p

m2p

K

x2p

13

22

2

2

222

MIT-MANIPAL 161

SJ: Section 40.8 P-49An electron (me=9.11 x 10-31 kg) and a bullet (m=0.0200 kg) each have a velocity of magnitude of 500 m/s, accurate to within 0.0100%. Within what limits could we determine the position of the objects along the direction of the velocity?

162

INTRODUCTION TO QUANTUM PHYSICS QUESTIONS1. Explain (a) Stefan’s law (b) Wien’s 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 Planck’s radiation law.

[2]4. Write the assumptions made in Planck’s 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 Einstein’s photoelectric equation.[2]



INTRODUCTION TO QUANTUM PHYSICS QUESTIONS10.Which are the features of photoelectric effect-

experiment explained by Einstein’s 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]



INTRODUCTION TO QUANTUM PHYSICS QUESTIONS20.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]

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