# Quantum Mechanics Question - Photon-Proton Collision

1. Sep 30, 2004

### AKG

Quantum Mechanics Question -- Photon-Proton Collision

A 100-MeV photon collides with a proton that is at rest. What is the maximum possible energy loss for the photon?

The subscript "p" refers to proton, and the subscript "$\gamma$" refers to photon.

The maximum energy loss occurs in a head-on collision, so we can treat the problem as a 1-d one.

By conservation of momentum:

$$p_{\gamma} + p_{p} = p'_{\gamma} + p'_{p}$$

$$p_{\gamma} = p'_{\gamma} + p'_{p}$$

$$p'_{p} = p_{\gamma} - p'_{\gamma}$$

By conservation of energy:

$$p_{\gamma}c + m_{p}c^2 = |p'_{\gamma}|c + \sqrt{(m_{p}c^2)^2 + (p'_{p})^2c^2}$$

$$p_{\gamma}c + m_{p}c^2 = |p'_{\gamma}|c + \sqrt{(m_{p}c^2)^2 + (p_{\gamma} - p'_{\gamma})^2c^2}$$

Now, if $p'_{\gamma} > 0$, then:

$$p_{\gamma}c + m_{p}c^2 = p'_{\gamma}c + \sqrt{(m_{p}c^2)^2 + (p_{\gamma} - p'_{\gamma})^2c^2}$$

$$p_{\gamma}c - p'_{\gamma}c + m_{p}c^2 = \sqrt{(m_{p}c^2)^2 + (p_{\gamma} - p'_{\gamma})^2c^2}$$

$$(p_{\gamma}c - p'_{\gamma}c + m_{p}c^2)^2 = (m_{p}c^2)^2 + (p_{\gamma} - p'_{\gamma})^2c^2$$

$$[(p_{\gamma}c - p'_{\gamma}c) + m_{p}c^2]^2 = (m_{p}c^2)^2 + (p_{\gamma} - p'_{\gamma})^2c^2$$

$$(p_{\gamma}c - p'_{\gamma}c)^2 + (m_{p}c^2)^2 + 2(m_{p}c^2)(p_{\gamma}c - p'_{\gamma}c) = (m_{p}c^2)^2 + (p_{\gamma} - p'_{\gamma})^2c^2$$

$$2(m_{p}c^2)(p_{\gamma}c - p'_{\gamma}c) = 0$$

$$2m_{p}c^3p'_{p} = 0[/itex] Which implies the final momentum of the proton is zero, so it stays at rest, and the electron loses no energy. This case clearly doesn't lead to the greatest possible energy loss, so we should consider the other case, where $p'_{\gamma} < 0$. Going back to the equation we had and adding in that $p'_{\gamma} < 0$: [tex]p_{\gamma}c + m_{p}c^2 = |p'_{\gamma}|c + \sqrt{(m_{p}c^2)^2 + (p_{\gamma} + |p'_{\gamma}|)^2c^2}$$

$$p_{\gamma}c - |p'_{\gamma}|c + m_{p}c^2 = \sqrt{(m_{p}c^2)^2 + (p_{\gamma} + |p'_{\gamma}|)^2c^2}$$

$$c(p_{\gamma} - |p'_{\gamma}|) + m_{p}c^2 = \sqrt{(m_{p}c^2)^2 + (p_{\gamma} + |p'_{\gamma}|)^2c^2}$$

$$[c(p_{\gamma} - |p'_{\gamma}|) + m_{p}c^2]^2 = (m_{p}c^2)^2 + (p_{\gamma} + |p'_{\gamma}|)^2c^2$$

$$c^2(p_{\gamma} - |p'_{\gamma}|)^2 + 2c^3m_{p}(p_{\gamma} - |p'_{\gamma}|) + (m_{p}c^2)^2 = (m_{p}c^2)^2 + (p_{\gamma} + |p'_{\gamma}|)^2c^2$$

$$(p_{\gamma} - |p'_{\gamma}|)^2 + 2cm_{p}(p_{\gamma} - |p'_{\gamma}|) + (m_{p}c)^2 = (m_{p}c)^2 + (p_{\gamma} + |p'_{\gamma}|)^2$$

$$(p_{\gamma} - |p'_{\gamma}|)^2 + 2cm_{p}(p_{\gamma} - |p'_{\gamma}|) = (p_{\gamma} + |p'_{\gamma}|)^2$$

$$2cm_{p}(p_{\gamma} - |p'_{\gamma}|) = (p_{\gamma} + |p'_{\gamma}|)^2 - (p_{\gamma} - |p'_{\gamma}|)^2$$

$$2cm_{p}(p_{\gamma} - |p'_{\gamma}|) = (p_{\gamma} + |p'_{\gamma}| + p_{\gamma} - |p'_{\gamma}|)(p_{\gamma} + |p'_{\gamma}| - p_{\gamma} + |p'_{\gamma}|)$$

$$2cm_{p}(p_{\gamma} - |p'_{\gamma}|) = (2p_{\gamma})(2|p'_{\gamma}|)$$

$$2cm_{p}(p_{\gamma} - |p'_{\gamma}|) = 4p_{\gamma}|p'_{\gamma}|$$

$$|p'_{\gamma}| = \frac{p_{\gamma}}{1 + \frac{2p_{\gamma}}{m_{p}c}}$$

The energy loss of the photon is:

$$\Delta E = (p_{\gamma} - |p'_{\gamma}|)c = \frac{2p_{\gamma}^2}{m_{p}(1 + \frac{2p_{\gamma}}{m_{p}c})}$$

From the initial energy of the photon (given) I can find the inital momentum ($p_{gamma}$) and I can look up values for $m_{p}$ and $c$, so I'm done, right?

2. Oct 2, 2014

### carlesyoo

Hi AKG,
I don't understand that step after saying proton final momentum is 0, when you replace proton final momentum with the sum of photon momentum.
$$\sqrt{(mc^2)^2 + (p_p' c)^2}$$
and you replace $$p_p'$$ with $$p_\gamma +|p_\gamma'|$$
Wouldn't the final momentum be just $p_\gamma$?

Carles,
Thanks