11

Quantum Mechanics

• Reference: • Book 1(theory): Concepts of Modern

Physics “A. Beiser”• Book 2(theory): Solid State Physics “S.O.Pillai”

Lecture -II

In the year 1927, Davisson and Germer conducted their famous experiment which was the experimental verification of De-Broglie's hypothesis i.e. = h/p.

2o

o

o

1 2eVm v eV v

2 m

Using De-Broglie's equation

h=

m v

o

o

h 12.28= A

2eVm V

Using Bragg’s Law: The condition for diffraction of X ray is

where n = 1, 2, 3For Ni d=0.91

2dsin n

o

A

And θ=90-Φ/2

If

V=54 Volt0

0

A67.1

A65.1

Davisson and Germer -- VERY clean nickel crystal. Diffraction is electron scattering from Ni atoms.

e

e

e

e

e

e

e ee

e e

scatter off atoms

e det.

move detector around,see what angle electrons coming off

Ni

e

e

eeee

e

e

e det.

Ni

Observe pattern of scattering electrons off atomsLooks like …. Wave!

# e’s

scatt. angle 00

See peak!!

Number of electrons ( density of electrons / Intensity of electrons)

Vs Scattering angle

5

• Observations: – Intensity was stronger for certain

angles for specific accelerating voltages (i.e. for specific electron energies)

– Electrons were reflected in almost the same way that X-rays of comparable wavelength

– Current vs accelerating voltage has a maximum, i.e. the highest number of electrons is scattered in a specific direction

– This can’t be explained by particle-like nature of electrons electrons scattered on crystals behave as waves

o

111

o

X-ray

2dsin

d 0.91A; 65 50

1.65A

12.281.67A

V

6

This is in excellent agreement with wavelengths of X-rays diffracted from

Nickel!If electrons are “just” particles, we expect a smooth monotonic dependence of scattered intensity on angle and voltage because only elastic collisions are involved.

10

Single Electron Diffraction

Q: A proton and electron have equal kinetic energies. Compare their De-Broglie wavelengths.

Assignment:

Derive the expression for de-Broglie wavelength of electron in each case.(1)Non relativistic (m=m0):

(2)Relativistic Case: ( K.E (eV) is comparable to rest mass energy.

eVm

h

02

2/1

200

20

21

2)2(

cm

eV

eVm

h

cmKK

hc

Where K is Kinetic energy and m0 is rest mass of electron

11

12

G.P. Thompson exhibit wave nature of high energetic electrons. e- are produced from a heated filament F and accelerated through a high positive potential given to the anode A. The whole apparatus is kept highly evacuated. The e- beam passes through a fine hole in a metal block B and falls on a gold foil G of thickness 10-8 m. The e- passing through the foil are receivedon a photographic plate P. The foil consists of very large number of microscopic crystals oriented at random. Therefore some of them are always at the correct angles to give rise to diffraction according to Bragg’s law. Hence, on processing the plate, diffraction rings is obtained. The diameters of these rings are measured.Let AB be a beam which has undergone a Bragg’s diffraction by a small crystal in the gold foil G, and falls on a photographic plate at a point ‘E” at a distance r from the central Pont C . The incident and the first order diffracted rays make equal angle with Bragg plane. Let BC=L

13

From Bragg's Law, 2dsin =n

where d is the crystal lattice spacing. therefore for the first order,

d (as is small)2sin 2

but, from the figure, 2 =tan2 =

r L

d= (1)L r

Now, the De-Broglie wavelength associated with the moving particle is

h = (2)

mvWhere m is the

-

relativistic mass and v the velocity of the particle. In Thompson’s

experiment the particles e are accelerated through a p.d. of about 15 to 60 kV.

For such energetic e the relativistic mass formula

o

2

2

o

2

2

mis m=

v1

c

h h =

m vmv

v1

c

22 2 2 1/ 2

o o 2

The relativistic expression for kinetic energy is

v K=mc m c m c (1 ) 1

c

Therefore, if the e (charge e) is accelerated from the rest through a potential of Vvolts,

22 1/ 2

o 2

21/ 2

2 2o

1/2

v m c (1 ) 1 eV

c

v eV (1 ) 1 1 z (3)

c m c

v=c 1 (1 z)

2

1/ 22

2

2

1/ 2 1/ 22 2

o o

(4)

Multiplying eq (3) and (4)

v c 1 (1 z) (1 z)

v1

cz 2eV eV

=c[2z(1+ )] =c[ (1 )]2 m c 2m c

Substituting this value

1/ 22

oo 2

o

h eVin eq (2) = (1 )

2m c2eVm c

m c

This is the relativistic expression for the De-Broglie wavelength of an e- accelerated through a high p.d of V volts. It reduces to the expression if relativistic effect is ignored (i.e., v<<c).

Substituting the above value of in eq. (1)

Comparison of the value of d obtained in this way agreed to within 1% with values determined by using X-rays of known wavelengths. For example, in case of gold foil the value of λ obtained by the above formula was 4.08Ao while that obtained by X-rays method was 4.06Ao. Thus the De-Broglie conception of matter waves was verified.

1/22

oo

h eV(1 )

2m c2m eV

o

h

2m eV

1/22

oo

L h eVd (1 )

r 2m c2m eV

16

For first order diffraction : n=1

From figure, 2θ =R/D Hence d= D λ/R

d = λ /2sinθ =λ/2θ for small θ

The interference rings in Figure were produced by sending X-rays of wavelength =1Å through a poly-crystalline thin film of copper (d=2.55 Å) of thickness t=1m. To produce the same set of rings but with electrons m=10-

27 gm instead of X-rays, what kinetic energy (in eV) should the electrons in the incoming beam have?

12

2oo

D h eVd 1

R 2m c2m eV

The diffraction patterns simulated above compare the effects of x-rays passing through a thin foil with those of high energy electrons passing through the same medium. Notice how similar the patterns are to each other when the de Broglie wavelength of an electron beam equals the wavelength of the original x-rays.

17

18

A SUMMARY OF DUAL ITY OF NATUREWave particle duality of physical objects

LIGHT

Wave nature -EM wave Particle nature -photons

Optical microscope

Interference

Convert light to electric current

Photo-electric effectPARTICLES

Wave nature

Matter waves -electronmicroscope

Particle nature

Electric currentphoton-electron collisionsDiscrete (Quantum) states of confined

systems, such as atoms.