Compton Effect

More experimental evidence of the quantized nature of radiation and the photon picture
was provided by the famous Compton effect observed when monochromatic X-rays of wavelnegth λ were scattered from a target material like graphite. At high angles of scattering an extra peak at wavelength λ' greater than the original
wavelength was observed in the intensity vs wavelength plot.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compdat.html#c1

This was termed as the Compton shift Δλ=λ-λ' was dependent on the scattering angle θ

The classical picture of radiation as an EMW predicted that the free electron in the material would be set into forced oscillations by the oscillating electric field of the EMW and would re radiate at the same wavelength and frequency as the incident radiation. There was no explanation for a peak in the Intensity for a shifted wavelength λ' from Classical Theory of EMW.

Compton viewed this effect as Relativistic Collison between an energetic photon
with E=hν and a stationary electron with rest mass m₀.

Relativistic collison theory is distinct from collisons in classical mechanics
due to the rest mass energy and energy and momentum are on the same footing.
(In fact the 3 mom and energy are the 4 components of the relativistic momentum 4-vector in 3 space+ 1 time dimensional Minkowski space)

The operative relation for energy is the famous equation of Einstein
E=[ m₀c²]γ where
1/γ=(1-v²/c²)^1/2
and m₀ is the rest mass of the particle.

The energy momentum formula is
E²=c²p²+ m₀²C⁴

For photons with rest mass ZERO E=cp=hν and hence p=h/λ where p is the photon momentum

Now we apply momentum and energy conservation along x and y directions and obtain the two key formula

E₀-E₁=K where E are the energy of the incident and scattered photons and K is the kinetic energy of the electron
or
c(p₀-p₁=K ......(1) as E=Cp for the photon and

p²=p²₀+ p²₁ -2 p₀ p₁Cos θ.....(2)

We also have the formula

K²/C² - 2Km₀=p²=p²......(3)

we substitute eqn. (2) in (3) and using (1) we have

1/p₁ - 1/p₀ = (1/m₀c) (1-Cos θ )........(4)

multiplying (4) by h and using p=h/λ we have

λ₁ - λ₀ = λ_comp (1-Cos θ)
=Δλ

where λ_comp= (h/m₀c) is called the Compton Wavelength and is measured for the electron to be 0.0243 Angstorms.

This analysis explains the shifted wavelength in the Compton effect. The other peak at the original incident wavelength may be understood as scattering from tightly bound electrons. If the incident photon energy is low this situation is like that of the incident photon colliding with the atom as a whole ( electron is tightly bound to the atom) so the mass in the Compton wavelength in the scattering formula is
the mass of the atom M which is far larger than that of the electron. Hence the Compton wavelength is vanishingly small and the Compton shift goes to ZERO.

Photoelectric Effect.

The Black Body radiation spectrum forced upon us the idea of quantized oscillators
radiating EMW and a departure from the concepts of Classical Physics. Another phenomena was soon discovered which seemed to defy a classical explanation in terms of EMW. This was the photoelectric effect which demonstrated that incident radiation on the (metal) surface are able to eject electrons. The electrons are called photo electrons and may constitute a current called a photo current

In 1886 Hertz did his famous experiment on this effect



http://physics.info/photoelectric/circuit.html

* Image from Flickr "Robbo Coach's" photostream

Several interesting features may be noted

1. The current I is NOT ZERO till a "reverse voltage is applied". This shows that
the electrons are ejected with some kinetic energy. The current saturates at some value of the forward voltage and this is dependent on the intensity of the incident radiation.

2. The cut off voltage is independent of Intensity of the radiation but depends on the frequency.

3. Millikan showed that the stopping potential varied linearly with the frequency and there was a "cut off frequency" ν₀ beyond which no electrons are emitted.

*Graph

An explanation of this phenomena based on the Classical Wave picture of EMW fails
due to the following reasons

1. Kinetic energy of photo electrons in particular the maximum kinetic energy K_max
=e V₀ where V₀ is the stopping potential should depend on the intensity as Intensity is proportional to the Electric Field amplitude squared in an EMW and this is the field that energizes the electron in the wave picture. But experiment shows otherwise.

2.The effect should occur for any frequency provide the incident radiation is intense enough to energize the electron. Experiment shows a cut off frequency specific to the material below which no photo electrons are ejected.

3. Effect should show a time lag as the wave energy is diffused over a spatial extent. Experiment shows that the effect is instantaneous.


These discrepancies inspired Einstein to apply the idea of Quantization due to Planck to EM radiation. Einstein postulated that EM radiation is composed of a large number of discrete packets of localized energy which behave like particles. The particles were called photons and their energy was quantized as E=hν where ν was the frequency of the radiation.

Einstein postulated that in the photoelectric effect 1 photon was absorbed by 1 electron to be energized and ejected provided the energy was adequate for the electron to overcome the surface atomic attractions.

So the kinetic energy of the ejected electrons would be K=hν-W where W is the energy for overcoming the surface barriers.

The maximum kinetic energy K_max=hν -W₀ where the second term was called the work function of the material and referred to the minimum energy for electrons to overcome the surface atomic attractions.

The photon picture was able to explain the experimental results

1. Intensity of radiation was proportional to the number of photons or their density. However each electron could absorb only 1 single photon. Hence the maximum kinetic energy was independent of the intensity.

2. At threshold when an electron is Just ejected with zero kinetic energy ( corresponding to the stopping potential K_max=0 and hν₀=W₀ where ν₀ is the cut off frequency
and no emission was possible below this.

3. As the energy of radiation was highly localized in the photon picture the absorption is instantaneous and hence there was no time lag as observed experimentally but was predicted by the classical wave picture.


From the equation K_max=eV₀=hν -W₀
its is seen that the stopping potential V₀ varies linearly with the frequency and from the slope of the graph of the stopping potential vs frequency
we can fix the Planck's constant h knowing e and m for the electron, which comes out to be close to the measured value.

Black Body Radiation III

The Planck Distribution
------------------------

We saw that the Rayleigh Jeans law fails to explain the black body spectrum at higher frequencies. Planck in trying to explain this discrepancy noticed that the average total energy for the modes was kT at low frequencies but must fall to zero at high frequencies to describe the experimental curve. This required a high frequency cut off for the oscillators in violation of the equipartition theorem which assumes that the oscillators may have any energy from 0 to infinity with equal a priori probability resulting in an average total energy per mode independent of frequency.

Planck departed from this hypothesis and postulated that the oscillators have
discrete energies which are equally spaced that is E=0, ΔE. 2 ΔE..... and ΔE=hν was proportional to the frequency and the proportionality constant h was refered to as the Planck's constant.

As ΔE=hν the energy of the oscillators E=nhν with n=0,1,2,3.....
An average energy computed from this discrete values using the Boltzmann
distribution as in the Classical Equipartition Theorem ( using summations over discrete values instead of integrals) yielded the Plancks distribution formula for the average energy

E_av=[hν/ e^(hν/kT) -1]

It is obvious from the formula that the desired high frequency cut off for the oscillators is obtained as E goes to zero with large values of frequencies and E
goes to (kT) ( expanding the exponential to first order) for small values of frequency.

Multiplying this by the Jeans number of modes in an infinitessimal frequency interval we obtain the energy density of the cavity radiation to be

ρ(ν)dν=(8πν²/c³ [hν/ e^(hν/kT) -1] dν

This is the Planck Black body Spectrum and completely matches the experimental black body curve ( upto a proportionality constant).

http://www.youtube.com/watch?v=cW4vmr3hb2o&feature=related 

Black Body Radiation II

In the last 2 sessions we have established that the Cavity Radiation spectrum is identical to the black body radiation spectrum and this is expressed as

ρ(ν)dν=constant. R(ν)dν at some temperature T deg K at thermal equilibrium.

The cavity radiation is due to the energized atomic oscillators and at ε is a standing wave configuration ( assuming no dissipation at walls)

To calculate the energy density of the cavity radiation in an infinitesimal frequency interval dν we need

(i) The number of modes in that interval

(ii) The average total energy in each mode

so ρ_T(ν)= # of modes X average energy/volume

To count the modes for simplicity we assume a cubic cavity and we will ses that the final result is independent of the geometry of the cavity.

As a warm up exercise we first compute the modes for 1 dimension ( line cavity of length L). We solve the 1 d wave equation with the boundary condition of nodes at the walls at x=0 and x=L. The boundary conditions restricts the modes to be such that the wave vector k=(nπ/L).

from standard relations between frequency and wave vectors its easy to see that the number of modes in the frequency interval dν is dN=(2L/C) X dν

This is easily generalized to 3d with a cube of side L and
assuming independent propagation in (x, y,z) directions due to parallel walls and

n²=n²_x + n²_y
+ n²_z

Using the same construction as 1d we mark the n's on a 3 d lattice and we see that
the number of modes = volume in lattice space. To find the number of modes in the frequency interval dν it is required to find the volume between two spheres of radius N and N + dN in the first octant as all
n > 0.

N= (1/8)(4π/3)n³=(π/6)(2L/c)³ ν³

so that dN=N(ν)dν=(4π/c)³ L³ν²dν X 2

for the 2 states of polarization of EMW

so the density is dN/L³ = (8π/c³)ν² dν

This is to be multiplied by the average total energy of each mode which may have any energy continuously between 0 and infinity.

Since the standing waves are in Ε with each other the equipartition theorem may be applied to find the average total energy to be E=kT.
(potential and kinetic)

So ρ(ν)dν=kT (8π/c³)ν² dν

This is the famous Rayleigh Jeans law and shows a quadratic dependence on the frequency. It is quite obvious that the law predicts an indefinite increase in the
energy density with frequency. It describes the black body spectrum at low frequencies but breaks down at high frequencies ( ultraviolet direction). Hence this came to be known as the "Ultraviolet Catastrophe"

It actually marks the failure of classical physics comprising of Mechanics and
EM.