Compton effect, experimental setup, working, observations, explanation, practice problems, FAQs

Many scientists who studied the electromagnetic waves proposed that they are composed of particles known as photons. It was also proposed that these photons carry energy and momentum although they have no mass. One of the early direct experimental evidence was given by the phenomenon called Compton effect.

Table of contents

Compton effect
Experimental apparatus
Working
Observations
Explanation for the shift
Mathematical formulation for Compton shift

Compton effect

Arthur H. Compton carried out an experiment in which he observed that when the X-rays are incident on a target having loosely bound electrons, the energy of the scattered rays decreases and there occurs a shift in the wavelength of the scattered rays as compared to the incident rays. This scattering phenomena was studied by the Arthur H. Compton which was later called Compton effect.

Experimental apparatus

The arrangement used by Arthur H. Compton to prove the photon concept proposed by Einstein is shown in the image below. The apparatus consists of an X-ray source which is placed in front of the collimating slits as shown in the image. The purpose of collimating slits is to block the unwanted protons. The X-rays source incident monochromatic rays on a graphite target. A movable detector is placed in the path of the scattered rays. This detector is used to measure the intensity of radiation in whatever direction scattering is taking place.

(Recreate the image replace by 1 and ' by 2 )
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Working

Monochromatic X-rays from the X-rays source are made to incident on the graphite target via collimating slits. The incident rays get scattered in different directions by the atoms of the graphite target. The intensity of the scattered rays in any given direction with respect to the incident rays direction, is found by rotating the movable collector in a circular arc. The intensity and wavelength of the radiation incident is known whereas that for the scattered radiations are measured and plotted against the parameters of incident radiations at different scattering angles.

Observations

The variation of intensity versus wavelength of the scattered rays is plotted at different scattering angles as shown in the image.

It can be seen that at different scattering angles (except at 0o), two peaks are observed in the intensity. One peak appears at the wavelength of the incident ray whereas the other peak was observed at some other wavelength 2. The difference between these wavelengths represents the Compton shift. The shift depends upon the scattering angle which is just an angular position of the detector.

Explanation for the shift

The deviation in the wavelength of the scattered rays as compared to the incident rays is in disent with the wave character of the X-rays. As per Compton, the incident X-rays consists of a stream of photons which when incident upon the graphite target inelastically collides with the valence electrons that are loosely bound in the graphite atoms. After the collision with the valence electrons, the scattered photons have less energy and momentum as compared to the incident one because it loses some energy and momentum during the collision with the electrons. Due to loss in energy, the photons emerge at high frequency or larger wavelengths. The Compton scattering experiment gave a convincing argument that the electromagnetic waves do behave like streams of photons.

Mathematical formulation for Compton shift

Compton used his model to derive the expression for Compton shift. In this model he assumed the electrons and photons as relativistic particles which follow two common principles:

Principle of conservation of linear momentum
Principle of conservation of total relativistic energy

Initial momentum of the photons,

$p_{1} = \frac{E_{1}}{c} = \frac{h f_{1}}{c} = \frac{h}{λ_{1}} [S i n c e c = f λ]$

Final momentum of the photons,

$p_{2} = \frac{E_{2}}{c} = \frac{h f_{2}}{c} = \frac{h}{λ_{2}} [S i n c e c = f λ]$

Let pe be the momentum of the electron after the inelastic collision with the incident photons.

From the principle of conservation of linear momentum,

${\vec{p}}_{1} = {\vec{p}}_{2} + {\vec{p}}_{e}$

$p_{e}^{2} = p_{1}^{2} + p_{2}^{2} - 2 p_{1} . p_{2}$

$p_{e}^{2} = p_{1}^{2} + p_{2}^{2} - 2 p_{1} p_{2} c o s θ . . . . . . . . . . . . . . . . (i)$

The electron was having the rest mass energy before the collision with the photon i.e. $E_{o} = m c^{2}$

After the collision, energy $E^{2} = E_{o}^{2} + p_{e}^{2} c^{2}$

Using the conservation of energy,

Initial energy of incident photon + electrons rest mass energy = energy of scattered photon + energy of electron after collision

$E_{1} + E_{o} = E_{2} + E$

$p_{1} c + E_{o} = p_{2} c + \sqrt{E_{o}^{2} + p_{e}^{2} c^{2}}$

$\Rightarrow (p_{1} - p_{2}) c + E_{o} = \sqrt{E_{o}^{2} + p_{e}^{2} c^{2}}$

Squaring both sides,

$(p_{1} - p_{2})^{2} c^{2} + E_{o}^{2} + 2 (p_{1} - p_{2}) c E_{o} = E_{o}^{2} + p_{e}^{2} c^{2}$

$\Rightarrow \frac{2 (p_{1} - p_{2}) E_{o}}{c} + (p_{1} - p_{2})^{2} = p_{e}^{2}$

$\frac{2 (p_{1} - p_{2}) E_{o}}{c} + p_{1}^{2} + p_{2}^{2} - 2 p_{1} p_{2} = p_{e}^{2}$

Using equation (i),

$\frac{2 (p_{1} - p_{2}) E_{o}}{c} + p_{1}^{2} + p_{2}^{2} - 2 p_{1} p_{2} = p_{1}^{2} + p_{2}^{2} - 2 p_{1} p_{2} c o s θ$

$\Rightarrow \frac{2 (p_{1} - p_{2}) E_{o}}{c} = 2 p_{1} p_{2} (1 - c o s θ)$

$\Rightarrow \frac{(p_{1} - p_{2})}{p_{1} p_{2}} = \frac{c (1 - c o s θ)}{E_{o}}$

$\Rightarrow \frac{1}{p_{2}} - \frac{1}{p_{1}} = \frac{c (1 - c o s θ)}{E_{o}}$

$\Rightarrow \frac{λ_{2}}{h} - \frac{λ_{1}}{h} = \frac{c (1 - c o s θ)}{E_{o}}$

$\Rightarrow λ_{2} - λ_{1} = \frac{h c (1 - c o s θ)}{m c^{2}} = \frac{h (1 - c o s θ)}{m c}$

$\Rightarrow λ_{2} - λ_{1} = \frac{h}{m c} (1 - c o s θ)$

$\frac{h}{m c} = λ_{c} = 0.00243 n m \to C o m p t o n w a v e l e n g t h$

$\Rightarrow λ_{2} - λ_{1} = λ_{c} (1 - c o s θ)$

Practice problem

Q. An X-ray having wavelength 71 pm is incident on a calcite target. What will be the wavelength of the ray scattered at 30o?

A. Wavelength of the incident ray, 1=71 pm

Wavelength of the scattered ray =2

Scattering angle, =30o

Shift in wavelength according to Compton scattering model,

$λ_{2} - λ_{1} = λ_{c} (1 - c o s θ)$

$λ_{2} - λ_{1} = 2.43 (1 - c o s 30^{o}) = 2.43 \times 0.134 = 0.325$

$\Rightarrow λ_{2} = 0.325 + λ_{1} = 0.325 + 71 = 71.325 p m$

Q. What is the smallest shift possible in the wavelength of the scattered rays when an incident ray having wavelength 60 pm falls on a graphite target?

A. Shift in wavelength according to Compton scattering model,

$λ_{2} - λ_{1} = λ_{c} (1 - c o s θ)$

$θ = 0^{o}, λ_{2} - λ_{1} = 0$

Therefore, the smallest shift possible is 0 pm.

Q. A visible photon has a wavelength of 500 nm. What will be the velocity of an electron having momentum equal to that of the photon?

A. Wavelength of photon, $λ = 500 n m = 5 \times 10^{- 7} m$

Momentum of the photon, $p = \frac{h}{λ} = \frac{6.63 \times 10^{- 34}}{5 \times 10^{- 7}} = 1.33 \times 10^{- 27} k g m / s$

Momentum of an electron, =mv

$m v = 1.33 \times 10^{- 27}$

$\Rightarrow v = \frac{1.33 \times 10^{- 27}}{9.1 \times 10^{- 31}} \approx 1460 m / s$

Q. Calculate the scattering angle at which the shift in wavelength will be equal to the Compton wavelength.

A. Shift in wavelength according to Compton scattering model,

$λ_{2} - λ_{1} = λ_{c} (1 - c o s θ)$

Where = Scattering angle

c= Compton wavelength

According to the question, $λ_{2} - λ_{1} = λ_{c}$

$\Rightarrow 1 - c o s θ = 1$

$\Rightarrow θ = 90^{o}$

FAQs

Q. Can we observe the Compton effect using visible light?
A. No, we can’t observe the Compton effect using visible light since the energy of the visible light is too small to knock electrons from the atoms. Also the wavelength of the visible light is large enough that the shift in its wavelength won’t be easily detectable.

Q. What kind of collision takes place between the photon and loosely bound electrons of the target in the Compton scattering experiment?
A. Inelastic collision

Q. How does a photon having zero mass still have momentum?
A. The momentum of a photon comes from its frequency and energy as shown by the relation E=hf (Plank Einstein relation). The momentum equation p=mv is not applicable to the particles moving at speed of light.

Q. What are the dimensions of Compton wavelength?
A. The dimensions of compton wavelength is same as that of normal wavelength i.e. [L].