# The Compton Effect: Find the angles of scattered photon and electron

1. Feb 27, 2013

### amr55533

1. The problem statement, all variables and given/known data

In the Compton scattering event seen in the figure, the scattered photon has an energy of 120 keV and the recoiling electron has an energy of 40 keV.

Find:

a) The wavelength of the incident photon
b) The angle θ at which the photon is scattered
c) The recoil angle ϕ of the electron

http://img11.imageshack.us/img11/9245/328q.jpg [Broken]

2. Relevant equations

For photons:

P=E/c=hf/c=h/λ

h=6.626E-34 Js

For electron:

P=mv/√(1-v^2/c^2)

E=mc^2/√(1-v^2/c^2)

E=√(P^2c^2+m^2c^4)

Conservation of momentum and energy:

Pbefore=Pafter

Ebefore=Eafter

3. The attempt at a solution

I solved part (a) as follows:

Ebefore=Eafter

--> Ebefore=40keV+120keV=160keV=2.563E-14J

P=E/c=(2.563E-14J)/(2.998E8m/s)=8.55E-23 kgm/s

P=h/λ ---> λ=h/P=(6.626E-34Js)/(8.55E-23 kgm/s)=7.75E-12m

I have an idea on how to solve parts (b) and (c). I would simply calculate the momentum of the scattered photon and the electron, and then use conservation of momentum in the x and y directions, and conservation of energy.

Scattered photon:

P'=E'/c=(1.922E-14J)/(2.998E8m/s)=6.412E-23kgm/s

Recoiling electron:

E=√(P^2c^2+m^2c^4)

6.408E-15J=√(P^2(2.998E8m/s)^2+(9.109E-31kg)^2(2.998E8m/s)^4)

This is where I get stuck. I cannot solve for "P" because it would create a negative under the square root.

After observing the givens, I found that this is because the energy of the electron is given as 40keV which is less than the rest mass energy of an electron of 511keV. Is it possible for an electron to have a total energy less than its rest mass energy? How would you solve for this? Thanks!

Last edited by a moderator: May 6, 2017
2. Mar 1, 2013

### clamtrox

Re: The Compton Effect: Find the angles of scattered photon and electr

The energy you are given is the kinetic energy (call it K). So you have $E = K + mc^2$.

Oh, and stick to electron volts in a calculation like this. There is no reason to fool around with kilograms.

3. Mar 2, 2013

### vela

Staff Emeritus
It's useful to know that $hc = 1240\text{ nm eV}$. Don't plug in a value for $c$. Pull the factors of $c$ into the units. For example, the 120-keV photon has a momentum of 120 keV/c, and the mass of the electron is 511 keV/c2.

Sometimes, you'll want to introduce factors of $c$ to make formulas work out more easily. For instance, you found $E_\text{before} = 160\text{ keV}$. The momentum of the photon is then $p = 160\text{ keV/}c$. To find its wavelength, you used $\lambda = h/p$. If you throw in some factors of $c$, everything works out nicely:
$$\lambda = \frac{h}{p} = \frac{hc}{pc} = \frac{1240\text{ nm eV}}{160\text{ keV}} = 7.75\times 10^{-3}\text{ nm}.$$ It's worth taking a little time to learn how to work in these units. It'll save you from a lot of tedious unit conversions.