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## Homework Statement

A photon of wavelength ##\lambda_i = 200## pm hits an electron at rest, and is scattered exactly backwards. Find the approximate recoil velocity v of the electron using momentum conservation.

## Homework Equations

Comptons Scattering:

$$\lambda_f = \lambda_i +\frac{h}{mc}(1-cos(\theta)$$

Momentum of photon:

$$p_\gamma = \frac{h}{\lambda}$$

Momentum conservation:

$$p_1+p_2 = p_1'+p_2'$$

## The Attempt at a Solution

Since it recoils exactly backwards, ##\theta = \pi##, so ##(1-cos(\theta) = 1-cos(\pi) = 2##.

So,

$$\lambda_f = \lambda_i +\frac{2\pi \hbar c}{mc^2}$$

Using ##\hbar c = 197.33## MeV fm, and ##mc^2 = E_e = 937## MeV,

$$\lambda_f = 200 pm + \frac{2*2\pi (197.33 MeV fm \frac{10^{-3} pm}{fm})}{937 MeV} = 200.001$$

So for the momentum,

$$p_\gamma = \frac{h}{\lambda_i} = \frac{2\pi\hbar}{\lambda_i}$$

and

$$p_\gamma' = -\frac{h}{\lambda_f} = -\frac{2\pi\hbar}{\lambda_f}$$

and for the electron, ##p_e = 0## and ##p_e' = mv = \frac{mc^2}{c^2}v = \frac{E_e}{c^2}v##

Using conservation of momentum,

$$p_\gamma + p_e = p_\gamma' + p_e' \Rightarrow p_e' = p_\gamma - p_\gamma' \Rightarrow \frac{E_e}{c^2}v = \frac{2\pi\hbar}{\lambda_i} - (-\frac{2\pi\hbar}{\lambda_f}) = 2\pi\hbar(\frac{1}{\lambda_i}+\frac{1}{\lambda_f})$$

$$\Rightarrow v = \frac{2\pi\hbar c*c}{E_e}(\frac{1}{\lambda_i}+\frac{1}{\lambda_f})$$

Converting ##\hbar c## from MeV fm to MeV pm and plugging in all variables gave me ##v = 3969.65 \frac{m}{s}##.

The multiple choice answers were in km/s, and all had over 1000 km/s. My answer would give me about 3.97 km/s.

I have gone through trying to figure out what I did wrong. I though that I might need to use relativistic velocity; however, I found it negligible even with the largest answer choice.