Compton Effect (Quantum nature of light)

Compton effect was discovered by A.H. Compton.
When a beam of monochromatic X-rays of wavelength λ is scattered by a light element (carbon), then the scattered X-rays have maximum intensities at two wavelengths, one of them at the same wavelength λ and the other at slightly longer wavelength λ՛. This effect is known as Compton effect.
The difference in wavelength (λ – λ՛) is known as Compton shift.

Theory

The energy of incident photon, E₁ = hν
Momentum of incident photon, p₁ = hν/c
This photon strikes with an electron at rest.

Before collision

Momentum of electron, p₂ = 0
Energy of electron (rest mass energy), E₂ = m₀c²

After collision

The photon scattered from its original path by giving some of its kinetic energy to the electron and the electron recoils.
Energy of scattered photon, E₁՛ = hν՛
Momentum of scattered photon, p₁՛ = hν՛/c
The scattered photon move in a direction making an angle θ with the X-axis.
The energy of electron, E₂՛ = mc²
Momentum of electron, p₂՛ = mv
This electron moves in a direction making an angle ϕ with the X-axis.

From the law of conservation of energy

$E_{1}+E_{2}=E_{1}'+E_{2}'$

$\therefore \;\;\;\;\;h\nu +m_{0}c^{2}=h\nu '+mc^{2}$

$or \;\;\;\;\;mc^{2}=h(\nu-\nu') + m_{0}c^{2}$

On squaring either side, we get

$m^{2}c^{4}=h^{2}(\nu^{2}-2\nu \nu '+\nu'^{2}) + m_{0}^{2}c^{4}+2h(\nu -\nu ')m_{0}c^{2}$ …(1)

From the law of conservation of momentum

Momentum along X-axis

$p_{1x}+p_{2x}=p_{1x}'+p_{2x}'$

$\therefore \;\;\;\;\;\frac{h\nu }{c}=\frac{h\nu' }{c}cos\theta +mvcos\phi$

$or\;\;\;\;\;mvc\;cos\phi=h(\nu -\nu '\;cos\theta)$ …(2)

Momentum along Y-axis

$p_{1y}+p_{2y}=p_{1y}'+p_{2y}'$

$\therefore \;\;\;\;\;0=\frac{h\nu' }{c}sin\theta -mvsin\phi$

$or\;\;\;\;\;mvc\;sin\phi=h \nu '\;sin\theta$ …(3)

On squaring and adding equations (2) and (3), we get

$m^{2}v^{2}c^{2}=h^{2}(\nu -\nu '\;cos\theta)^{2}+h^{2}\nu '^{2}sin^{2}\theta$

$or\;\;\;\;\;m^{2}v^{2}c^{2}=h^{2}(\nu^{2} -2\nu\nu '\;cos\theta+\nu '^{2})$ …(4)

On subtracting equation (4) from (1), we get

$m^{2}c^{2}(c^{2}-v^{2})=-2h^{2}\nu\nu '(1-cos\theta)+2h(\nu -\nu ')m_{0}c^{2}+m_{0}^{2}c^{4}$ …(5)

From relativistic formula

$m=\frac{m_{0}}{\sqrt{1-v^{2}/c^{2}}}$

$or\;\;\;\;\;m^{2}(c^{2}-v^{2})=m_{0}^{2}c^{2}$

$or\;\;\;\;\;m^{2}c^{2}(c^{2}-v^{2})=m_{0}^{2}c^{4}$ …(6)

On equating equations (5) and (6), we get

$2h^{2}\nu \nu '(1-cos\theta )=2h(\nu -\nu ')m_{0}c^{2}$

$or\;\;\;\;\;\frac{\nu -\nu '}{\nu \nu '}=\frac{h}{m_{0}c^{2}}(1-cos\theta )$

$or\;\;\;\;\;\frac{1}{ \nu '}-\frac{1}{ \nu}=\frac{h}{m_{0}c^{2}}(1-cos\theta )$

$or\;\;\;\;\;\frac{c}{ \nu '}-\frac{c}{ \nu}=\frac{h}{m_{0}c}(1-cos\theta )$

$or\;\;\;\;\;\lambda '-\lambda =\frac{h}{m_{0}c}(1-cos\theta )$

$or\;\;\;\;\;\Delta \lambda =\lambda _{c}(1-cos\theta )$

This is the change in wavelength of X-rays and is known as Compton shift in the wavelength of X-rays.
Here λ_c is Compton wavelength and its value is given by
λ_c = h / m₀c = 0.0242Å

Conclusion

Compton shift is independent of the wavelength of the incident radiation and nature of the scattering substance.
It depends only on the scattering angle θ.

Special cases

If θ = 0º , then Δλ = 0
If θ = 90º, then Δλ = λ_c = 0.0242 Å
If θ = 180º, then Δλ = 2λ_c = 0.0484 Å
It is the maximum value of Compton shift.

Experimental setup

Experimental setup of Compton effect

Relative intensities versus wavelength curve of Compton effect at different angles

The excellent agreement between the Compton shift obtained from Compton theory and the experimental results confirms that the elastic collision between a photon and an electron is just like an elastic collision between two particles.
The Compton effect proves the quantum theory (particle nature) of radiation.

To know more about Compton effect, please click here for English and click here for bilingual lecture (Hindi/English)

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