# Compton Scattering and Recoil Velocity

## 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 $v = 3969.65 \frac{m}{s}$.

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.

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nrqed
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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.
The mistake is right there.

The mistake is right there.
Are you referring to the missing factor of 2 in the equation for $\lambda_f$? If so, when. I plugged in the values in the next equation, I added the factor. The first equation missed the factor because of a typo.

nrqed
Are you referring to the missing factor of 2 in the equation for $\lambda_f$? If so, when. I plugged in the values in the next equation, I added the factor. The first equation missed the factor because of a typo.