Lecture – 5

TITLE: Electromagnetic Radiation – Matter Interactions

Page-1

Objectives

In the previous lecture, we have learnt that the radiation or light is following

the Wave-Particle dual nature

It can be treated as electromagnetic wave or particle nature like photon

Similarly, the electrons can also be treated as particle and also as wave

corresponding to its momentum for describing the stable structure of atom.

When both of them interact with each other, we have to understand the

mechanism to follow for describing the experimental observations such as Compton

effect, absorption and emission of light by atoms.

In this lecture, the different kinds of treatment to understand the light-matter

interactions are described. We will start with the classical phenomenon and then proceed

to understand the quantum mechanical descriptions.

Page – 2

Classical treatment:

In classical theory as shown in Figure – 5.1, we model the atom as a heavy nucleus

with electron attached to it with spring the binding force between them.

The resonant frequency for this system is 0

If we treat the light as wave then the electric field can be represented as

0E E Sin t

Equation – 5.1

Figure – 5.1

The oscillating electric field will force the electron to oscillate. The displacement of

electron with respect to nucleus will produce an oscillating dipole.

It is known that an oscillating dipole emit electromagnetic radiation with the same

frequency of vibration. This emission of light is known as scattering of light by matter.

However, if the incident radiation frequency matches with the resonant frequency of

the system ( 0 ) then the resonance occurs. Energy transfer takes place.

The oscillating dipole ( ) ( )p t e x t

Equation – 5.2

Where e is the charge of the electron and x(t) is the time dependent displacement of

electron with respect to nucleus.

When this dipole oscillates, it emits radiation. As time passes by, due to the radiation loss

the emission dies or decays. This phenomenon can be modeled as a damping oscillator

whose solution will be x(t).

The differential equation of motion of the damping oscillator with damping constant

2

0( ) ( ) ( ) 0x t x t x t

Equation – 5.3

Where 2

0k

m , here k is the force constant of the spring.

+

t t

E E

Page – 3

With the initial values i.e. at t = 0

the displacement 0(0)x x and the velocity (0) 0x , the solution is

( )

20( ) [ ( ) ]

2t

x t x e Cos t Sin t

Equation – 5.4

The frequency 2

12 20( )

4

of the damped oscillation is slightly lower than the

frequency 0 of the undamped case.

Taking 0 , we get ( )

20 0( )

tx t x e Cos t

.

Equation – 5.5

Line profile of the emitted radiation

The damped oscillation ( )x t can be described as a superposition of monochromatic

oscillations with slightly different frequencies and amplitude ( )A

1( ) ( )

2

i tx t A e d

Equation – 5.6

The amplitude ( )A can be calculated by 1

( ) ( )2

i tA x t e dt

Equation – 5.7

The intensity

*

0 0 2 2

0

1( ) ( ) ( )

( ) ( / 2)I A A I

Equation – 5.8

Figure – 5.2

I(-0)

0

x0

t

e-(/2)/t

Page-4

Momentum of a Photon

Momentum vector magnitude E

pc

E is the energy.

E hp h

c c

parallel to the direction of propagation.

From relativity, we know that 2

0

2

2

2

1

E m c

p mV

mm

Vc

Ep mV V

c

Equation – 5.9

Minkowski’s four dimensional space & time ( E & three comp. of p)

If the system is confined to a single particle so, 2 2 2E p c is invariant under changes of

ref. frame. 2 2 2

2 22 2 2 2

4 2

22 42

2 42 2 40

022

2

1

1

1

1

E p c

E VE V c E

c c

Vm cc

m c V m ccV

c

Equation – 5.10

Photon is a quantity of energy. If we consider photon as a particle, then the energy 2E mc , with this infinite energy E .

So the only way of reconciling a speed V c with a finite energy is to assume rest mass

0 0m .

So, 2 2 2 0 ; EE p c pc

Rest energy 20 0m c for photon

So, 20. .K E E m c E so for photon all energy is K.E.

Page-5

Elastic Collisions of Photons – Compton Effect

The X-ray of wavelength 0 was incident on a target. With a crystal diffractometer, the

wavelengths of the scattered X-ray were measured by changing the angle as shown in

Figure – 5.3.

Figure – 5.3

Scattered radiation is composed by two lines:

(i) a component at the incident wavelength 0 , called the Thomson

component. Scattered radiation has the same frequency as the

incident radiation.

(ii) a component of different wavelength 0 .

Conclusion from experiment

(i) 0 is always positive.

(ii) is an increasing function of .

(iii) is independent of and the composition of the material used for

scattering.

X-Ray

source

Scatter

(electron)

Incident

beam

Scattered

beam

Crystal

(wavelength

selector)

Scattering

angle ()

Collimating

system

Detector

Page-6

Explanation

Elastic scattering between photons and free electrons (weakly bound to atoms can be

considered as free electrons) i.e. binding energy 0h .

Elastic Scattering K.E. is conserved

Figure – 5.4

We have taken into account the fact that p & E are related by relativistic invariance 2 2 2 2 4

0E p c m c .

Before Collision After Collision

Photon Energy 0

0

ch h

ch h

Momentum 0

0

hhc

along Oz hh

c

Electron

Energy 20m c

2

2 2 2 40

mc

p c m c

Momentum = Zero p making an angle with OZ

Photon

Momentum = 0hc

Energy 0E h

Electron

Z

X

O

Page-7

Conservation of energy and momentum

Energy 2 2 2 2 40 0 0h m c h p c m c

Equation – 5.11

Momentum

0 cos cos

0 sin sin

h halong OZ p

c c

halong OX p

c

0

0

cos cos

cos

sin sin

h hp

c c

h

c

hp

c

Equation – 5.12

Figure – 5.5

Photon

Momentum = 0hc

Energy 0E h

Electron

Z

Page-8

2 222 2 2

02 2

22 2 2 2 20 02

22 20 02

2 2 2 2 20 0

cos cos

cos 2 cos sin

2 cos

2 cos

h hp

c c

h

c

h

c

p c h

Equation – 5.13

From 1st equation

22 2 2 2 4

0 0 0p c h m c m c

Equation – 5.14

Equating Equation – 5.13 & Equation – 5.14,

22 2 2 2 2 4

0 0 0 0 0

22 2 40 0

2 cosh h m c m c

h m c

2 2 40 0 02h m c m c

2 20h 2 2h 2 2 2

0 02 cosh h 2 2h

2 20 0 0

2 20 0 0

0

2 2

2 2 1 cos

2

h h m c

h m c h

2h

2h

020

31002

0 0 0

20 0

00

0 0

0

1 cos

1 cos where 9.1 10

1 11 cos

1 cos

0 ; 0

21 cos ;

90 ;o

m c

hm Kg

m c

h

m c

h

m c

h h

m c m c

hCompton wavelength

m c

Page-9

So, we get

00

1 cosh

m c

Equation – 5.15

(i) is positive.

(ii) independent of 0 .

(iii) depends on only.

Hard X-ray 1

1

pm

h MeV

Energy of Compton Wavelength,

20

0

rest massof theelectronhc hc

m ch

m c

20 0 0 0h m c photons gives up very little energy.

20 0 0 0h m c photons gives up most of its energy.

Equation – 5.16

Page-10

Absorption of Photons (Inelastic Collision)

Let us define 12h is the energy difference between the two states of an atom with energy

2E and 1E ( 2 1E E ground state) 2 1 12E E h .

If there is a collision between an atom and a photon as shown in Figure – 5.6 and after the

collision they form one particle, then the initial momentum of the incident must be

preserved by the final single particle (momentum conservation). Or we can say that after

collision the atom must possess some kinetic energy that it did not have previously.

This kinetic energy can have been taken from the energy of the incident photon.

This is possible only when 12h h . We have to determine the relationship between

and 12 .

Figure – 5.6

Considering the properties of the relativistic particles:

Before Collision After Collision

Photon

Energy h

Momentum hc

Atom

Energy

2

1 1

1 rest mass

E m e

m

2 2 2 4

2W P c m e

Momentum Zero

2 1 122 2 2

where,

P

E E hm

c c

Photon

Momentum = hc

Energy E h

Atom

Atom - photon

system

E1

E2

Page-11

From the conservation of energy

2 2 2 4

1 2h E P c m e

Equation – 5.17

same as photonhMomentum pc

Equation – 5.18

2

1 12 2E h m c

From Equation – 5.17,

2 2 2 2 4

1 2

22 2 2 2 2

1 1 1 12

2 22 2 2 2

1 1 12 122

2 2

12 1 12 1212

1 1 12 1

1212 2

1

2

2

21 1

2 2 2

12

E h p c m c

E E h h p c E h

hc E E h h

c

h E h h

E h E h E

h

m c

Equation – 5.19

Page-12

So the incident photon frequency must be greater than the theoretical frequency of the

spectral lines 2 112

E E

h

.

12 10

2

1

1 1010

10 100

h to eV

m c to GeV

Comment:

(i) In the optical transitions 2

12 1h m c typically, 1012

2

1

~ 10h

m c

which means the shift

will be very small compared to other broadening 610 cannot be observed. Resonant

condition is applicable i.e. 2 1E E h .

(ii) When 12h is not very small compared to 2

1m c , the shift is appreciable.

Page-13

Emission of Photons:

Before Collision After Collision

Photon

Energy ~ h

Momentum ~ hc

along Oz

Atom

Energy 2E 2 2 2 4

1W P c m c

Momentum 2

22

0

Em

c

p along Oz

2 12121 2

E hEm

c c

2 2 2 2

2

2

2 2 2 2 2 2 4

2 2 1

2 2 2 4

1

22 2 4

2 2 1 2 12

2 2 2 2

2 2 2 12 2 12

2 2

2 2 12 12

2 2 2

2 12 12 1212

2 2 2

1212 2

2

0

2

2

2 2

2 2

2

2 2 2

12

hp p c h

c

E W h

E h W

E h h E p c m c

h m c

E h E m c E h

E h E E h E h

h E E h h

E h h h

hE E h E

h

m c

Equation – 5.20

Page-14

Application to rays

19 22

rays Very High Energy

10 to 10 Hz

Nuclear transitions also very high frequency 2

12 1h m c

Emission of ray photon is interpreted as the radiative transition between two energy

states of the nucleus (analogous to optical transition).

Frequency Displacement Line Width

Figure – 5.7

Resonance absorption and emission is not possible as in the case of optical transition. By

increasing temperature it is possible. So to get the information about the nucleus from

this is difficult.

What Mössbaur discovered that by cooling the source and the absorber the intensity

absorption increased.

It means that, below a certain temperature emitted or absorbing nucleus is embedded in a

crystalline lattice and it is the white crystal which recoils. 23~ 6 10Mass of Crystal Mass of Nucleus

The velocity of recoil is negligible.

No recoil energy loss extremely sharp lines (natural width) could be obtained.

2 2

0 3exp

2

where, . .

' .

.

Debye Waller Factor

E Tf

K

T Abs Temp

K Boltzmann s Const

Degree Temp of the Solid

Equation – 5.21

There are three main hyperfine interactions that can be observed by Mössbauer

spectroscopy. They are (i) Isomer Shift, (ii) Quadrupole Splitting and (iii) Nuclear

Zeeman Splitting.

For example :

The isomer shift of 57 3 1 14.42 2

Fe KeV in Ferricinium bromide is 82 10 eV .

8

0

3

0

0

0

810

3

0

2

3

2 10

14.4 10

2 103 10

14.4 10

6 100.04

14.4 10

h h eV

h eV

c

c v

v v

c v c

v c

cms

Mössbaur Spectrum of Ferricinium bromide

Figure – 5.8

Recap

In this lecture, we came to know that the classical physics has limitation to describe the

several experimental observations such as Compton effect and absorption and emission

of light by atoms.

We understood the different kinds of treatment for describing the light-matter

interactions: both the classical phenomenon and then the quantum mechanical

descriptions.

We also come to know the recoil energy required when both light and matter interact.

Based on this, we came to know the application of this recoil energy in Mossbaur

spectroscopy.