Modern Physics: Mastering Physics: Compton Scattering

Learning Goal: To understand the derivation of the Compton scattering formula.

In 1923, Arthur H. Compton conducted a series of experiments that demonstrated the particle-like nature of electromagnetic radiation. Compton studied the collision of photons from a uniform wavelength beam with a stationary target. He observed that the radiation that exited from the target showed two characteristic wavelengths, one at the original wavelength lambda

and another at a new wavelength lambda'

shifted by an amount Delta lambda

from the original. In this problem we will use the principles of conservation of energy and conservation of momentum, coupled with some basic ideas in quantum physics and relativity, to derive a statement for the shifted wavelength lambda'

Part A

Consider a photon with initial momentum p_i

on a collision course with a stationary electron of mass

. What is the total energy E_i

of these two particles before the collision? Don't forget to include the rest mass energy of the electron.

Express your answer in terms of the p_i

, the speed of light

, and

ANSWER:

$p_{i}c+mc^{2}$

Part B

After the collision, the photon exits with a momentum p_f

at an angle phi

from its initial momentum vector. The electron scatters off with a momentum p_e

. What is the total energy after the collision? In this case, do not forget to include the relativistic energy of a particle.

Express your answer in terms of p_f

, and

ANSWER:

$p_{f}c+\sqrt{\left(mc^{2}\right)^{2}+\left(p_{e}c\right)^{2}}$

Part C

The conservation of energy requires that the energy of the system before the collision E_i

and the energy after the collision E_f

be equal to one another. We want to manipulate this conservation equation to arrive at the result for Compton scattering. The first step is to take E_i

as one side of the equation and E_f

as the other side of the equation. Divide both sides by

and isolate the square root on one side by subtracting the term p_f

from both sides of the equation. You should now be left with a trinomial on one side. Complete the right-hand side of the equation with these three terms.

Express your answer in terms of p_i

, and

ANSWER:

$\sqrt{(mc)^2 + p_e^2}$ =

$p_{i}-p_{f}+mc$

Part D

Square both sides of the equation obtained in Part C and choose the result of squaring the right-hand side (the one you entered in the last part) as your answer. Your unreduced answer should have nine terms in it. If you further reduce your answer you should have at least six terms.

Part E

You can use conservation of momentum to eliminate the term p_e

from the equation. If you recall that p_e_vec

, and

are all vectors you can use vector addition to state one of the vectors in terms of the other two. Give the value for p_e_vec

in terms of p_i_vec

and

Express your answer in terms of p_i_vec

and

ANSWER:

$\vec{p_{i}}-\vec{p_{f}}$

Part F
You now wish to get a value for in terms of the scalar quantities , , and the angle between the two vectors. Recall that when squaring a vector it is necessary to use the dot product: $p_e^2 = (\vec{p}_e)^2= \vec{p}_e \cdot \vec{p}_e$ .

Express your answer in terms of p_i

, and

ANSWER:

$p_{i}{^{2}}+p_{f}{^{2}}-2p_{i}p_{f}{\cos}{\phi}$

Part G

Substitute the value you obtained for p_e^2

in Part F into the equation obtained in Part D. Eliminate terms, and divide by 2 so that one side of the equation reads $p_ip_f - p_ip_f \cos(\phi)$ . Supply the other side of the equation in the answer box.

Express your answer in terms of p_i

, and

ANSWER:

$p_ip_f - p_ip_f \cos(\phi)$ =

$p_{i}mc-p_{f}mc$

Part H

In this final step, multiply both parts of the equation in Part E by h/(mp_ip_fc)

. By recalling the relationship between photon wavelength and momentum, $p = h/\lambda$ , you can now arrive at a statement for the shifted wavelength $\lambda^{\prime} - \lambda$ . Give the remaining part of the equation that summarizes the Compton scattering relationship.