Unit 1 Modern Physics Compiled by Dr Santhosh D Shenoy

7/24/2019 Unit 1 Modern Physics Compiled by Dr Santhosh D Shenoy

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UNIT-1

MODERN PHYSICS

Introduction to blackbody radiation spectrum:A body which absorbs all radiation that is incident on it is called a perfect blackbody. When radiation

allowed to fall on such a body, it is neither reflected nor transmitted. Such a black body after absorbing the

incident radiation gets heated and starts emitting radiation of all possible wavelengths. In practice, perfectblackbody does not exist and we can have objects that are only close to a blackbody. A blackbody, on heatingcan emit all radiations it has absorbed and is called blackbody radiation. Figure shows the energy distributioncurves in which energy density E (energy emitted by the blackbody per unit area of the surface) is plotted as afunction of wavelength at different temperatures of the blackbody.

The important points that can be noted down from these curves are1. All curves shows a peak suggesting that the emitted intensity is maximum at a particular wavelength.2. An increase in temperature results in an increase in the energy emitted.3 .As the temperature increases, the peak shifts to lower wavelength.

Laws of blackbody radiation:1. Stefan-Boltzmann law

It states that the total energy density E0of radiation emitted from a blackbody is directly proportional tothe fourth power of its absolute temperature T.

i.e., E0T4

Or E0= T4

where is a constant called Stefans constantwith numerical value equal to 5.6710-8Wm-2K-4. This lawagrees well with the experimental results.

2. Wiens displacement law

It states that the wavelength m corresponding to the maximum emissive energy decreases withincreasing temperature.

i.e.,m

T

Or mT = b,where b is called the Wiens constant and is equal to 2.910-3mK.

3. Wiens law

Using laws of thermodynamics and classical concepts, Wien developed an expression for energy density


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as,

() =

where C1and C2 are constants.

This law holds good for smaller values of wavelength.

4. Rayleigh-Jeans law

Rayleigh derived an expression for the energy density of radiation based on classical theory which isgiven by,

() = where k is called Boltzmanns constant and its value is 1.381 x 10-23JK-1

This law holds good only for large values of wavelength.

As per the Rayleigh-Jeans law, the radiant energy increases with decreasing wavelength and ablackbody must radiate all the energy at very short wavelength. But in actual practice, it doesnt happen so. Thefailure of Rayleigh-Jeans law to explain the aspect of very little emission of radiation beyond the violet regiontowards the lower wavelength side of the spectrum is referred as ultraviolet catastrophe.

Both Wiens law and Rayleigh-Jeans law indicates failure of classical theory in explaining blackbody radiation.


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Plancks radiation law:Max Planck proposed a law based on quantum theory. According to this, atoms or molecules absorb or

emit radiation in quanta or small energy packets called photons. If is the frequency of photons, then itsenergy can be explained as E=hwhere h=6.63x10-34Js is called Plancks constant. Applying quantum theoryPlanck obtained an expression for energy density of blackbody radiation as,

() =

----------------- (1)This law agrees well with the experimental observation of blackbody radiation and is valid for all wavelengths.

1. For shorter wavelengths:

i.e., when is small, is very largeOr i.e., >>1 hence ( )

Substituting this in Plancks radiation law i.e., in eq(1) then,

(

)

=

i.e., () = Where C1=8hc and C2=

hc

k

Hence at smaller wavelengths, Plancks radiation law reduces to Wiens law.

2. For longer wavelengths:

i.e., when is large, is very smallExpanding the power series , we have

= + ! +()! +

Sinceis very small, its higher power terms can be neglected

Then the above expression becomes

= + hckT

or = hckT

Substituting this in Plancks radiation law i.e., in eq(1),

(

)

=

i.e.,() = Hence at longer wavelengths, Plancks radiation law reduces to Rayleigh-Jeans law.

Photoelectric effect:


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Emission of electrons from a metal surface when light of suitable energy falls on it is calledphotoelectric effect. The experimental setup for observing photoelectric effect consists of a pair of metal plateelectrodes in an evacuated discharge tube connected to a voltage source as shown:

When light of suitable energy is incident on the cathode, electrons are emitted and a current flows across thedischarge tube. Some special features of photoelectric emission are:1. It is an instantaneous process-there is no time interval between the incidence of light and the emission of

photoelectrons.2. There is a minimum frequency for the incident light called threshold frequency, below which no photoelectric

emission occurs. This depends upon the nature of the material of the emitter surface. The energycorresponding to the threshold frequency, called the work function is the minimum energy required to releasean electron from the emitter surface.

3. For a given frequency of the incident light, photocurrent is directly proportional to the intensity of theincident light.

4. The photoelectron emission can be stopped by applying a reverse voltage to the phototube. ie., by making theemitter electrode positive and the collector negative. The negative collector potential required to stop thephotoelectron emission is called the stopping potential.

Einsteins Theory:Photoelectric effect can be explained on the basis of quantum theory of light. When the energy equal to

work function of the metal is incident on the metal surface, the incident photon liberates electrons from theirbound state. When the incident photon carries energy in excess of the work function, the extra energy appears as

the kinetic energy of the emitted electron. When the intensity of light increases, the number of photoelectronsemitted increases but their kinetic energy remain unaltered. When a photon of frequency is incident on ametal surface of work function , then,

h= + (mv

2)max

where (12mv

2)max is the kinetic energy of the emitted photoelectrons.

This is known as Einsteins photoelectric equation.Since = h0, it can also be written as,


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(12mv

2)max = h- = h(-0)

If V0is the stopping potential corresponding to the incident photon frequency , then,

(12mv

2)max = h-

(12mv

2)max = h(-0) or (mv

2)max = eV0

Compton Effect:

When X-rays are scattered by a solid medium, in addition to the scattered X-rays of same frequency,there exist some scattered X-rays of a slightly lower frequency (higher wavelength). Compton observed thisphenomenon and is called Compton Effect.Compton Effect can be explained on the basis of the quantum theory and laws of conservation of energy andmomentum. Consider an x-ray photon of energy hincident on an electron at rest.

After the interaction, the X-ray photon gets scattered at an angle with its energy changed to h andthe electron which was initially at rest recoils at an angle . It can be shown that the increase in wavelength isgiven by

=

h

m

0

C

1- cos )

where m0is the rest mass of the electron.When = 900, =

h

m0C= 0.0242. This constant value is called Compton wavelength.

Wave particle dualism:The Photoelectric Effect and Compton Effect conclusively established the particle behavior of light. The

phenomena of interference, diffraction and polarization give exclusive evidence for the wave behavior of lightHence we have to conclude that light behaves as an advancing wave in some phenomena and it behaves as aflux of particles in some other phenomena. Therefore we say that light exhibits wave-particle duality.

De-Broglies hypothesis:

De-Broglie extended the wave particle dualism of light to the material particles. This is known as de-Broglie hypothesis. According to this hypothesis, material particles in motion possess a wave character. Thewaves associated with material particles are called matter waves or de-Broglie waves.According to Plancks theory of radiation,

E = h --------- (1)where is the frequency associated with the radiation.

According to Einsteins mass-energy relation,E = mc2 --------- (2)

where m is the mass of the photon and c is the velocity of light


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Combining (1) and (2),

i.e., h= mc2 =>hc = mc

2 (since =)h = mc

Therefore momentum associatedwith the particle is given by p = mc,

Or, = where is called de-Broglie wavelength.

De-Broglie wavelength associated with the accelerated electron:A beam of high energy electrons can be obtained by accelerating them in an electric field. Consider an

electron starting from rest when accelerated with a potential difference V, the kinetic energy (E) acquired by theelectron is given by,

E=

1

2

mv

2

and alsoE= eV

Thus,1

2mv2= eV

Orm2v2

2m= eV

i.e.,p

2

2m

=eV =E

where v is the velocity of the electron, m its mass and p the momentum.Now the momentum may be expressed as,p =

2mE =

2meV

Hence the de-Broglie wavelength =h

p

=

h

2mE

=

h

2meV

To test de-Broglie hypothesis, Heisenberg and Schrdinger formulated theories whereas G.P.Thomson,Davisson and Germer conducted experiments.

Davisson-Germer experiment:The electron diffraction experimental setup used by Davisson and Germer to verify de-Broglies

hypothesis is as shown:

The filament F is heated to produce electrons via thermionic emission. These electrons are passedthrough a narrow aperture forming a fine beam of accelerated electrons. The electron beam was then made toincident on a single crystalline sample of Nickel. The electrons scattered at different angles were counted usinga detector. The experiment was repeated by recording the scattered electron intensities at various positions ofthe detector.


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A sharp maximum occurred in electron density at an angle () of 500with the incident beam for anaccelerating potential of 54 V. The angle of incidence corresponding to this is 250and from figure, glancingangle (angle of diffraction) is 650. From X-ray diffraction experiment, spacing of the planes responsible fordiffraction was found to be 0.091 nm.Assuming first order diffraction, Braggs law can be written as,

= 2d sin= 20.09110-9

sin65 = 0.165 nm.

By de-Broglie relation,

= = . (..)= 0.167nmThus Davisson and Germer experiment directly verifies the de-Broglie hypothesis.

Characteristic properties of matter waves:1. Matter waves are associated with moving particle.2. Wavelength of matter waves is inversely proportional to the velocity with which the particle is moving

(=h

mv). Hence a particle at rest has an infinite wavelength.

3. Wavelength of matter waves is inversely proportional to the mass of the particle. Hence wavelike behaviorof heavier bodies is not very evident whereas wave nature of subatomic particles could be observed

experimentally.4. Wave function is used to define a matter wave which is related to the probability of finding a particle at anyplace at any instant.

5. Matter waves are represented by a wave packet made up of a group of waves of slightly differingwavelengths. Hence we talk of group velocity of matter waves rather than the phase velocity (velocity of asingle wave). The group velocity can be shown to be equal to the particle velocity.

Phase velocity:General expression for a wave is Y = A cos (t-kx)where Y = Displacement at any instant t, A = Amplitude of vibration, = 2is the angular frequency and

k =2

is the wave vector or wave number.

Phase velocity or wave velocity of a wave is the velocity of the wave when phase is constant.i.e., t-kx = constantor, kx = t + constant

or, x =tk

+ constant

Hence Phase velocityvp= =

Group velocity:The de-Broglie waves are represented by a wave packet and hence we have group velocity associated

with them. Group velocity is the velocity with which the wave packet travels.Consider two waves having same amplitude but having slightly different frequency and wave numberrepresented by the equationsY1= A cos (t-kx)Y2 = A cos [(+) t (k+ k) x]The resultant displacement due to the superposition of the above two waves is,

Y = Y1+ Y2

= A cos (t-kx) + A cos [(+) t (k+ k)x]


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Since, cos A + cos B = 2cos (2

BA +) cos (

2

BA),

Y = 2A cos {(2

2 +) t (

2

2 kk +) x} cos {(

2

) t (

2

k) x}

As the difference in frequency of the two waves is very small, we can assume,2+ 2and 2k+k 2k

Y = 2Acos {(2

) t (2k ) x} cos (t-kx)

The velocity of the resultant wave (group velocity) is given by the speed with which a reference point, say themaximum amplitude point, moves. Taking the amplitude of the resultant wave as constant, we have,

2Acos {(2

) t (

2

k) x} = constant

or, (2

) t (

2

k) x = constant

or, x = + constant

Group velocity vg= = When and k are very small,

vg=

Relation between group velocity and phase velocity:Phase velocity, vp=

where is the angular frequency of the wave

= vp k ----- (1)

Group velocity, vg=

vg=dvk

dk (From eq (1))

= vpdkdk+ k dv

dk = vp+ (

2 )

dvd( )

where the propagation constant (or wave number) k =2

= vp+ (1)

dvd()

= vp+ (1)

dvd()

dd

= vp+ (1)

dvd

dd()

Butd()

d=1

Therefore vg= vp+ 1 dvd (1)

Thus vg= vp- Relation between group velocity and particle velocity:

Energy of a photon E = h or =E

h ------ (1)


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We know the angular frequency of the wave = 2 or =(2E)h

d =(2h )dE ------(2)

Further, k =2 =

2ph since =

hp

dk = (2h )dp ------(3)

Substituting the value of dand dk from equations (2) and (3) in the expression for group velocity,

vg = ddk

= dEdp

------ (4)

If a particle of mass m is moving with a velocity vparticle, its energy is given by,

E =12mv

2particle=

2-------(5)

Substituting this in equation (4),

vg =dE

dp =

ddp

=pm

=mVparticle

m = vparticle

Hence vg = vparticle

Relation between velocity of light, group velocity and phase velocity:

Phase velocity, vp= where k is the propagation constant or wave number

We know the angular frequency of the wave = 2 or =(2E)h

Further, Wave number k =2 =

2ph since =

hpwhere p is the momentum of the wave

Thus vp=

=()

=Ep =

mcmv (Since E = mc

2)

=c

vvphase vparticle= c

2

But vg = vparticle

vphase vg= c2

Expression for de-Broglie wavelength using group velocity:

We know, the group velocity vg =d

dk, where angular frequency = 2and wave number k = 2

d = 2dand dk = 2d(

1)

Thus vg =ddk

=2d

2d(1

)=

d

d(1

)

Or, d(1

) =d

vg=

d

vparticleg -------- (1) [since vg = vparticle]

Total energy of a particle moving under an applied potential V is given by,

E =12mv

2+ V


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de-Broglie related E = hto above expression

Then, h=12mv

2+ V

Assuming V as a constant potential and differentiating the above equation,h d= m vparticle dv

or, d= (mv

h ) dvSubstituting this in equation (1),

d(1) = mdvh Integrating,1 =

mvh + constant

i.e.,1 =

ph + constant

Assuming constant of integration to be zero,1 =

ph + constant

or, = , the de-Broglie wavelength.

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VTU Model Question Paper

1 .a) 1) If the momentum of a particle is increased to four times, the de-Broglie wavelength isi) become twice ii) become for times iii) become one-fourth iv) become half

2) Blackbody radiation spectrum, maximum intensity is shifting towardsi) shorter wavelength ii) longer wavelength iii) no change iv) none of these

3) Group velocity of wave is equal toi) V phase ii) V particle iii) Velocity of light iv) none of these

4) de-Broglie wavelength of an electron accelerated by a potential of 60 V isi) 1.85 ii) 1.58 iii) 1.589 iv) 1.57 (4 marks)

b) Describe Davisson-Germer experiment to prove the dual nature of matter waves. (8marks)c) Explain phase velocity and Group velocity. Derive de-Broglie wavelength using Group velocity. (8 marks)

Dec 08/ Jan 091 a) 1) The de-Broglie wavelength associated with an electron of mass m and accelerated by a potential V is

i)h

2mVe ii)2mVe iii)

hVem iv)

h2Vem

2) Davisson and Germer were the first to demonstrate:i) The straight line propagation of light ii) The diffraction of photonsiii) The effective mass of electron iv) None of these

3) Electrons behaves as waves because they can be:i) Deflected by an electric field ii) Diffracted by a crystaliii) Deflected by magnetic field iv) They ionize a gas

4) In Davisson-Germer experiment, the hump is most prominent when the electron is accelerated by

i) 34 volts ii) 54 volts iii) 60 volts iv) 80 volts (04 Marks)b) Define phase velocity and group velocity. Show that group velocity is same as particle velocity. (08 Marks)c) Derive de-Broglie wavelength using Group velocity. (04 Marks)d) Compare the energy of a photon with that of a neutron when both are associated with wavelength of 1 .

Given that mass of neutron is 1.678 10 -27kg. (04 Marks)

June-July 2009

1) a) 1) An electron and a proton are accelerated through same potential. The ratio of de-Broglie wavelength


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e/p is

i) 1 ii)mm iii)

mm iv)

mm

2) Wavefunction associated with a material particle isi) Single valued ii) Finite iii) Continuous iv) all the above

3) In a blackbody radiation spectrum, the maximum energy peaks shift towards the shorter wavelength sidewith the increase in temperature. This confirms

i) Stefans law ii) Weins law iii) Rayleigh-Jeans law iv) Plancks law4) The group velocity of the particle is 3106m/s, whose phase velocity is

i) 6.06106m/s ii) 31010m/s iii) 3 nm/s iv) 1.51010m/s (04 Marks)b) Describe Davisson and Germer experiment for confirmation of deBroglie hypothesis. (08 Marks)c) Explain phase and group velocity. Calculate the de-Broglie wavelength of a bullet of mass 5 gm moving

with a velocity 20 km/hr. (08 Marks)

Dec.09/Jan.10

1 a)1) Wiens law is deduced from Plancks radiation formula under the condition ofi) Very small wavelength and temperature ii) Large wavelength and temperatureiii) Small wavelength and high temperature iv) Large wavelength and small temperature

2) The Compton wavelength is given byi) h/m0C

2 ii) h2/m0C2 iii) h/m0C iv) h

2/2m0C3) Which of the following relations can be used to determine de-Broglie wavelength associated with a

particle?

i)h

2mE ii)hVm iii)

h

2mVe iv) All of these4) If the group velocity of a particle is 3106m/s, its phase velocity is

i) 100 m/s ii) 3106m/s iii) 3108m/s iv) 31010 m/s (04 Marks)b) What is Plancks radiation law? Show how Wiens law and Rayleigh-Jeans law can be derived from it.

(06 Marks)c) Define group velocity. Derive relation between group velocity and phase velocity. (06 Marks)

d) A fast moving neutron is found to be have a associated de-Broglie wavelength 2. Find its kinetic energyand group velocity of the de-Broglie waves. (04 Marks)

May/June 2010

1 a) 1) In a blackbody radiation spectrum, the Wiens distribution law is applicable only fori) Longer wavelength ii) Shorter wavelength iii) Entire wavelength iv) None of these

2) The de-Broglie wavelength associated with an electron of mass m and accelerated by a potential V is

i)h

2mVe ii)2mVe iii)

hVem iv)

h2Vem

3) Electrons behaves as a wave because they can bei) Diffracted by a crystal ii) Deflected by magnetic field

iii) Deflected by electric field iv) Ionise a gas4) If the group velocity of de-Broglie wave is 4108m/sec, its phase velocity is

i) 12108m/sec ii) 2.25108m/sec iii) 5.33108m/sec iv) 1.33108m/sec (04 Marks)b) Explain duality of matter waves. (04 Marks)c) Define phase velocity and group velocity. Show that group velocity is equal to particle velocity.(08 Marks)d)Calculate the momentum of the particle and de-Broglie wavelength associated with an electron with a

kinetic energy of 1.5 keV. (04 Marks)January 2011


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1 a) 1) Green light incident on a surface releases photoelectrons from the surface. If now blue light is incidenton the same surface, the velocity of electronsi) increases ii) decreases iii) remains same iv) becomes zero

2) Rayleigh-Jeans theory of radiations agree with experimental results fori) all wavelengths ii) shorter wavelengths onlyiii) longer wavelengths only iv) middle order wavelengths only

3) The de-Broglie wavelength of an electron accelerated to a potential difference of 100 volts is

i) 1.2 ii) 10 iii) 100 iv) 124) The wave nature associated with electrons in motion was verified by

i) photoelectric effect ii) Compton effect iii) diffraction by crystals iv) Raman effect (04Marks)b) State and explain de-Broglies hypothesis. (04 Marks)c)Define phase velocity and group velocity. Obtain the relation between group velocity and particle velocity.

Obtain the expression for de-Broglie wavelength using group velocity. (08 Marks)d) Find the kinetic energy and group velocity of an electron with de-Broglie wavelength of 0.2 nm. (04 marks)

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Unit 1 Modern Physics Compiled by Dr Santhosh D Shenoy

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