QUANTUM MECHANICSFailure of classical mechanics:

Classical mechanics (Newtons laws and electromagnetic theory) fail to explain the phenomenon like radioactivity, photoemission, x-rays etc. New ideas were introduced to explain the parade axis. The new formalism, used to explain the microscopic phenomenon is known as quantum mechanics.

1. Black body radiation:A perfectly black body is one which absorbs all the radiations incident on it and emits the same when it is heated to various temperatures. The distribution of energy in the spectrum of a black body is shown in the figure below.The following conclusions are drawn from figure.

(i) At a given temperature the energy is uniformly distributed.(ii) For a given temperature the intensity of radiation increases with wavelength reaches a maximum, for a particular wavelength beyond which intensity decreases.(iii) As the temperature increases maximum shifts towards shorter wavelength region.(iv) The area under each curve represents the total energy emitted.The classical law proposed by Wien is max = constant or Emax = constant where max is the wavelength corresponding to Emax or Imax. T is absolute temperature.

This law succeeded in explaining the shorter wavelength region of the black body spectrum. The law proposed by Rayleigh and Jeans is where K is Boltzmanns constant, T is absolute temperature, is the wavelength of the emitted radiation. These laws succeeded in explaining the longer wavelength region of the black body spectrum. None of the classical laws succeeded in explaining the u-v region of the black body spectrum, which is popularly known as ultra violet catastrophe.

Max plank introduced a new hypothesis to explain the entire region of the black body spectrum. He assumed the particles emitted are considered as (simple) harmonic oscillators and the energy associated with each photon (quanta) is E = h, where h is planks constant, is the frequency of emitted radiation.

2. Stability of the atom and atomic spectra:According to Rutherford the entire mass and positive charge of the atom is concentrated in a small region called the nucleus. The electrons revolve round the nucleus in orbits. The electrostatic force of attraction between electron and the nucleus is compensated by centrifugal force.

where Ze = charge on the nucleus, r = radius of orbit, e = electric charge

v = velocity of electron, permitivity of free space.

The orbital motion of the electron cause continuous emission of radiation. The above facts support the stability of the atom. The emission of energy is possible only if an electron is an accelerated charge, according to classical physics. Then the electron cannot stay in a circular orbit but it follows a spiral orbit of decreasing radius and finally falling to the nucleus. It is contradiction to the observed stability of the atom. According to classical mechanic an exited atom continuously emit radiations of all wavelengths. This is contradictory to the observed discrete spectra of Hydrogen atom. Bohr introduced Quantum ideas to explain the stability of the atom and the discrete spectral lines. According to him stability can be achieved by (angular momentum) where mass of electron, velocity of electron, radius of the atom, integer (number of energy levels), planks constant.

The spectral lines of hydrogen atom can be explained by using the relation where Rydbergs constant, and various energy state.

3. Photoelectric effect:According to classical ideas, radiation consists of oscillating electric and magnetic fields. The intensity of radiation is being proportional to . The force exerted on the electron is i.e., where is the electronic charge. Therefore the kinetic energy of the emitted electron should depend on the intensity of radiation. This is contradiction to the experimental facts. Also classical physics fails to explain the existence of threshold frequency and to explain the instantaneous nature of photo emission. All the above facts were successfully explained by Einstein using the relation

max = hh0 where max = Kinetic energy of an emitted electron, planks constant, = frequency of incident radiation, 0 = threshold frequency.

4. Compton scattering:According to classical theory when a beam of incident on matter. The scattered should possess the same wavelength as the incident one. The scattering constant is found to be independent of the incident radiation. The distribution of scattered intensity should be symmetrical. The experimental observations revels the following:

(i) The scattering constant was found to depend on the wavelength of the incident .(ii) The scattered radiation was found to consists of two wavelength. One same as that of incident wavelength and the other a longer wavelength.(iii) The distribution of scattered intensity is not symmetrical.5. Specific heat of Solids:Specific heat of a solid is defined as

It is the amount of heat required to raise the temperature of unit mass of a substance through one degree centigrade.

The classical physics predicts that all solids have the same and temperature independent molar specific heats (= 3R).

Dulong and Pettits law showed that at room temperature, Atomic weight Specific heat = 6.4 (Atomic specific heat). But in the case of diamond and silicon it is not true (For these two specific heat increases with temperature). The experiment showed specific heat decreases slowly with fall of temperature and finally tends to zero at absolute zero at 0 K. This is contradictory to Dulong and Pettits law.

The experimental observations were successfully explained quantum mechanically by Einstein.

deBroglie waves or matter waves:deBroglie suggested that particles like electrons, protons, neutrons etc exhibit dual nature i.e., material particle can behave both as wave as well as particle. He proposed the following assumptions.

(i) The Universe is made up of particles and radiations. These entities must be symmetrical.

(ii) Nature loves symmetry.

According to deBroglie, moving particle can be associated with a wave. The waves associated with (material) moving particle are known as deBroglie waves or matter waves.

Expression for deBroglie wave:

According to Quantum theory of radiation, the energy associated with a photon is given by

E = h where planks constant, = frequency of incident radiation

We know that =

wavelength, velocity of light.

The equation becomes,

.....................(1)

According to Einstein, mass and energy are inter-convertibles.

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

where mass of the particle.

From equations (1) and (2)

where the momentum of photon.

If the material particle of mass moving with a velocity then the wavelength associated with the particle is given by

...................(3)

Different forms of deBroglie equations for wavelength:Consider a beam of electrons travelling through a potential difference , then the electrons acquire kinetic energy

where is the electronic charge.

Equation (3) becomes

According to kinetic theory of gases,

where Boltzmanns constant, absolute temperature

Equation (3) becomes

deBroglie wavelength for a particle moving a wave with very high speed or velocity. The mass of moving particle is given by

where mass of moving particle, rest mass of particle, velocity with which particle moving.

Equation (3) becomes,

Note: Wavelength associated with an electron of mass Kg

GP Thomsons experiment:

Construction and working:

High energy electrons produced by a cathode are passed through two perforated discs (D1, D2) to get a narrow beam and then passed through an aluminium cylinder which serves as an anode. While passing through the aluminium cylinder electrons get accelerated to very high velocities. This high energy electron beam is made incident on a gold foil, which act as a diffraction grating. The diffraction pattern is obtained on the screen.

Theory:

High energy electrons on passing through a thin poly crystalline gold foil undergo Braggs reflection and form a series of rings on the photographic plate (screen).

According to Braggs law

where order of the diffraction pattern, spacing between the layers where the atoms are accommodated. The incident beam AB passes through the film at B. BP is the reflected beam. Let OP= R & BO = .From figure , PBO= 2, where = glancing angle

P

R

A B O

For small angle,

..............(1)

From le BOP

Equation (1) becomes,

For the order

The wavelength calculated from this relation is in good agreement with wavelength calculated using the relation

DavissonGermer experiment:

Construction:

DavissonGermer experiment consists of an electron gun in which thermal electrons are produced by heating the filament using low tension battery. Two aluminium diaphragms D, D are used to collimate the electron beam, which is then accelerated by an anode (A). Electrons ejected from the electron gun are made to be incident on a single crystal Nickel (Ni) which can be rotated horizontally by means of a handle H. The scattered electrons are received by the ionisation chamber, which is connected to a sensitive galvanometer. The ionisation chamber is moved along a circular scale with the crystal N at its centre to measure the current.

Working: This experiment can be performed by

a. Normal incidence method

b. Oblique incidence method

a. Normal incidence method:

A known potential difference is applied between the filament F and anode A. The ionisation chamber is set at different angles and for each scattering angle the current is noted. The current is proportional to the number of electrons entering the chamber per second. For each scattering angle the ionisation current was noted. As the variation involves angular changes, a plot using polar co-ordinates is well suited for studying dependence of current on scattering atom.

Initially the accelerating potential was kept at 40V. The variation was found to be

A smooth curve (without any hump) as shown in the fig(i). When the experiment was repeated with accelerating potentials at 44V, 54V and 68V the graph were shown in fig(ii), fig(iii) and fig(iv). As the accelerating potential increases a distinct hump was formed and it was more pronounced for = 500 and V= 54V. Above 54V the hump was declined and faded away.

Theory:

If e is the change on the electron then the kinetic energy is given by

Where v is the velocity, V is volts (potentials)

According to de-Broglie

If electron behaves like a wave

where d is the lattice space for (1,1,1) plane for Nickel, d = 2.15

For normal incidence, we have , n = 1

This proves wave particle duality of an electron.

b. Oblique incidence method: The accelerating potential kept at 60V, the galvanometer deflections are noted at different glancing angle. A graph of galvanometer deflection versus glancing angle is plotted as shown above. As seen from the graph the maxima are observed at and .For oblique incidence

The theoretical value of wavelength () is given by

Experimental and theoretical values of wavelength are in good agreement with each other.

Properties of deBroglie wave:1. To arrive at Bohrs postulate from deBroglie wave.

DeBroglie concept of matter waves provide a theoretical explanation to Bohrs postulate. According to deBroglie the electron is not a particle (moving mass). But it moves as a wave as shown in the above figure.

If the wave train is in phase, the circumference of the circle (orbit) is an integral multiple of wavelength.

....................(1)

According to deBroglie,

From equation (1)

(The angular momentum of an orbital electron is an integral multiple of )

2. DeBroglie wavelength of a particle at relativistic velocity.

According to Einsteins theory of relativity

where

rest mass of the particle

mass of the particle moving with the velocity .

particle velocity

velocity of light.

According to DeBroglie

3. Phase velocity of a DeBroglie wave.

According to DeBroglie,

If be the frequency of a matter wave then

The phase velocity of a DeBroglie wave is given by

4. Group velocity.

A particle of mass moving with a velocity v is thought of to be a group of waves moving in a single direction. The velocity with which the energy transported is called Group velocity.

A wave pocket consists of a group of waves the amplitude and phases are constructively interfere over a very small region of space in which particle can be located. Just outside this pocket they interfere destructively and the amplitude tends to zero.

Consider a free particle of mass m having phase velocity and group velocity . If be the deBroglie wavelength then ...............(1)

The phase velocity is given by

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

The energy of a particle is given by

.....................(3)

Substitute equation (1) and (3) in equation (2)

HEISENBERGS UNCERTAINTY PRINCIPLE:It is impossible to determine simultaneously both position and momentum of a particle accurately.

If x and p are the uncertainties in position and momentum of a particle respectively then

or

Heisenbergs uncertainty principle can also be stated as In any simultaneous determination of energy and time in a physical process, the product of corresponding uncertainties is equal to or greater than .

If E and t are uncertainties in energy and time of a particle respectively then

GammaRay microscopic experimentConsider a photon from a source S incident on electrons. Some of these electrons bounce into microscope. This will enable the observer to see the flash of light and thus to determine position and momentum of the electron simultaneously. But the accuracy in determining the position of electron by a microscope is limited by limit of resolution of the microscope. If x is the least distance between two points in the field of view which can be distinguished as separate. Then

where n is the refractive index of the medium between the object and objective.

when (i.e., in air medium)

................(1)

This means that if the position of the electron changes by x, the microscope will not be able to detect it. x is made as small as possible by illuminating the electrons by radiations of very short wavelength such as rays.

When photon interacts with electron, there is a recoil of the electron and hence its momentum changes. Initial momentum of the photon is given by where is the frequency of radiation.

The photon undergoes scattering in the direction of OA and makes an angle with the incident direction. In turn the electron recoils with a velocity v in a direction OB making an angle with the incident direction.

According to the law of conservation of momentum

( cos )The scattering angle can be vary from to .

[ cos ]

[ cos ]

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

Multiplying equation (1) and (2), we get

Illustartion: . Why electrons does not exist inside the nucleus?

We know that the radius of nucleus is of the order of m. This means that the uncertainty for an electron to be confined inside the nucleus cannot exceed the diameter of the nucleus. Therefore m.

According to Heisenbergs principle,

But the maximum energy of an electron in atom is of the order of 4MeV. Hence electrons cannot exist inside the nucleus.

PROBLEMS:1. Calculate the deBroglie wavelength associated with an electron accelerated by a potential of 60 V.Solution:

2. Calculate the deBroglie wavelength associated with a marble of mass 10gms moving with a velocity 1000 cm/s.

Solution:

3.Calculate the wavelength associated with an electron of energy 200 eV.

Solution:

where E is the kinetic energy

4.A proton of mass Kg has a wavelength of 0.5Ao. Calculate the energy in eV.

Solution:

5.Calculate the wavelength associated with a thermal neutron at 300K. Given that mass of a neutron is Kg and Boltzmanns constant is J/K.

Solution:

6. Calculate the deBroglie wavelength of a proton moving with of velocity of light. Given mass Kg.

Solution:

7. Calculate the deBroglie wavelength of 143 V electron. Also calculate Braggs angle for the first order reflection from (1,1,1) plane of a Nickel for which .

Solution:

Where

8. Calculate the deBroglie wavelength associated with a neutron at

Solution:

9. Calculate the wavelength of an electron having kinetic energy 1.5 MeV. Given that mass of electron is Kg.

Solution:

10. 10 KV electrons are passed through a thin film of a metal for which atomic spacing is m. Calculate the deBroglie wavelength and what is the angle of deviation from the first order diffraction pattern?

Solution:

11. Electrons initially accelerated through a potential difference of 5 KV. Compute (a) momentum (b) deBroglie wavelength.

Solution:(a)

(b)

12.Find the speed and momentum of a proton whose total energy is eV.

Solution:

13.A microscope using photons is employed to locate an electron in an atom to within a distance of 0.2Ao. What is the uncertainty in momentum of the electron located in this way?Solution:

14. The speed of an electron is found to be to an accuracy of 0.003%. Find the uncertainty in determining position of electron. Given that , .Solution:

15. The deBroglie wavelength of a proton is 0.589Ao. Find the energy in eV.Solution:

16. Calculate the deBroglie wavelength and momentum of thermal neutron at 27oC.Solution:

Momentum:

17. Calculate the deBroglie wavelength associated with 54V electrons and calculate the speed of matter waves of these electrons.Solution:

18. An electron has a speed of . The accuracy measurement is 0.004%. Calculate the uncertainty in locating its position.Solution:

EXERCISE

1. Explain the failure of classical theory in explanation of i) Photoelectric effect ii) Black body radiation iii) Atomic spectra iv)Compton effect. v) stability of atom2. With relevant theory explain G. P. Thomsons experiment on electron diffraction.

3. With relevant theory explain Davisson and Germers experiment to substantiate concept of matter waves.4. Explain, how the concept of group velocity gives the quantum picture of a material particle. Establish the relation between phase velocity and group velocity for a non-relativistic free particle.5. Explain the gamma ray microscope to bring out the concept of uncertainty principle6. a. State Heisenbergs Uncertainty principle? Describe ray microscopic experiment to support Heisenbergs uncertainty principle?

b. Why electrons does not exist inside the nucleus?

CONCEPT QUESTIONS

1. Why G-P Thomsons method is called absolute method?

a. Wavelength is determined by dynamical value so G-P Thomsons method is called absolute method i.e., is calculated dynamically.

2. One cannot easily identify wavelength of marble of mass of 10gms?

a. In marble wavelength is very small. So we cannot identify the wavelength of marble of mass 10gms.

3. For what frequency will the quantum effects be noticeable at room temperature?

Ans: Quantum effects will be noticeable when =1. At room temperature T=300K,

hence = = =6.25x1012Hz. Oscillations of frequency ~ 1012Hz or more will show quantum effects.

4. Is atomic heat for all metals a constant?

Ans: At ordinary temperature, the atomic heat of most metals is a constant. But as temperature decreases, atomic heat decreases and is zero for 0K.

5. Is electromagnetic radiation a wave or a particle?

Ans: Electromagnetic radiation manifests either as waves or as photons(particle) and hence is said to have wave particle duality.

6. Do x-rays have energy and momentum.

Ans,. X-ray like all electromagnetic radiation has energy E= h and momentum

7. What is Compton wavelength?

Ans. Compton wavelength is the wavelength shift when the scattering of radiation with matter takes place at 900.

8. A charged electron moves in a circular orbit, yet the atom is stable.

Ans. According to Bohrs theory, the electron moving in a stationary orbit does not radiate and hence does not loose energy. Therefore the atom is stable.

9. Is the hydrogen spectrum discrete or continuous? Explain.

Ans. Hydrogen spectra is discrete according to Bohr, when electron radiates it jumps from one stationary state of higher energy to that of lower energy giving rise to a discrete spectra.

10. The classical concepts are not valid in the region of atomic dimension. Explain.

Ans. For atomic dimension, the plancks constant h= 6.625 x 10-34 Js of significance and hence leads to quantum effects. Classical concepts are valid only when

11. Even when monochromatic X-rays are used, the Compton spectrum contains more than one line. Why?

Ans. The Compton spectrum contains one incident light of wavelength and another scattered light of wavelength due to recoil of the electron. Hence when monochromatic X-rays are used, the Compton spectrum contains more than one line due to inelastic scattering.

12. Waves are associated with sub atomic particles not with macroscopic particles.

Ans. Sub atomic particles have wavelength in while the wavelength of one gram mass with the same energy has a very small value of wavelength of nearly 10-22 m and hence is not measurable.

13. If an electron and a proton have the same kinetic energy, which one will move farthest?

Ans.

Since > ; > so the electron will move faster.

14. An electron and newtron have the same kinetic energy. Compare the momentum.

Ans.

Sine mass , the momentum of the neutron is greater than that of ther electron.

15. An electron and a neutron have the same de Broglie wavelength. Which one will move faster and why?

Since the mass of the proton is greater than that of the electron, the velocity of the electron is greater than that of the proton.

16. Do the de Broglie waves produce dispersion in vacuum.

Ans. De Broglie waves produce dispersion in vacuum because the phase velocity of the wave depends on the wavelength.

17. Explain why electron cannot exist in the nucleus.

Ans. According to uncertainty principle, if the electron is confined in the nucleus of diameter 10-14m the corresponding momentum of the electron would be of the order of 10-20m kg ms-1. Then the Kinetic energy of the electron would be of the order of 100 Me V. The maximum energy that the emitted particle has is only 4MeV, hence an electron cannot exist in the nucleus.

18. Can we observe the de Broglie wavelength with a speeding tennis ball. Explain.

Ans. No. A speeding tennis ball of mass ~. 1 Kg can have velocity ~106ms-1. Hence the corresponding wavelength is ~

=

The wavelength is too small to be observed or measured.

19. Does the concept of Bohr orbit violate the uncertainty principle. Explain.

Ans. No. Infact the concept of Bohr orbit reinforces the uncertainty principle. Uncertainty principle shows that if the minimum energy is 13.6 eV it must have a dimension of an angstrom.

20. If the Plnak, s constant were to be much (i) smaller (ii) bigger than the actual value explain how the quantum phenomenon would be.

Ans. (a) If the Plank constant was smaller, then even macroscopic and particles even smaller would have quantum effects.

(b) If the Plank constant was larger, even macroscopic objects would show quantum effect like discrete energy levels and wave nature.

21) The concept of trajectory has no meaning in quantum mechanics. Explain.

Ans. Accordiong to uncertainty principle , the position and momentum of a particle cannot be simultaneously determined precisely hence trajectory in a phase space has no meaning.

21. Is de Broglie hypothesis of matter waves applicable to an electron in an atom. Justify.

Ans. Yes, de Broglie wavelength is associated with the motion of an electron in an atom but these waves give rise to standing waves hence the electrons are in stationary orbits.

22. How can the wavelength of an electron be =h/p when the momentum p represents that the electron is a particle.

Ans. Due to wave particle duality, it is possible for the electron to have momentum p corresponding to a particle and a wavelength associated with its wave nature.

23. Can matter wave travel faster than light? Justify your answer.

Ans. No. Even though the phase velocity of the matter wave is greater than the velocity of light in vacuum, the group velocity is lesser.

1646K

Wiens region

Rayleigh-Jeans region

uv region

1095K

904K

I

max

max

max

Cathode

C

A

Aluminium cylinder

Electron beam

D1

D2

Screen

Gold foil

Diffraction pattern

Photo graphic plate

N

H

E

Scattered electron beam

Incident electron beam

D1 D2

A

F

L.T

H.T

S

G

R

L.T = Low tension battery

H.T = High tension battery

N = Nickel crystal

E = Electron gun

A = Aluminium cylinder

S = Circular scale

R = Ionization chamber

F = Filament

D1, D2 = Aluminium diaphragms

G = Galvanometer

Direction of incident beam

V = 40V

V = 44V

V = 54V

= 50o

V = 68V

Fig(i)

Fig(ii)

Fig(iii)

Fig(iv)

31o

62o

Glancing angle

Deflection

A B

S O

Electron

Microscope

Objective

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