- #1

- 476

- 30

- Homework Statement
- Prove that the scattered photon is fixed at a constant value with only the condition that both angles from the photon and electron sum up to 90 degrees after the collision.

- Relevant Equations
- -

Hi all,

Currently given a problem to prove that the scattered photon is fixed at a constant value with only the condition that both angles from the photon and electron sum up to 90 degrees after the collision. I can't seem to prove it and listed my steps below, was wondering if anyone can guide/correct my mistakes. Cheers.

For x-direction:

$$

P_{photon}=P'_{photon}cos\theta +P_{e}cos\phi

$$

$$

\frac{h}{\lambda_{0}}= \dfrac {h}{\lambda ^{'}}\cos \theta +P_{e}\cos \left( 90-\theta \right)

$$

$$\frac{h}{\lambda_{0}}= \dfrac {h}{\lambda ^{'}}\cos \theta +P_{e}\sin \left( \theta \right)$$

$$\begin{aligned}\dfrac {\lambda ^{0}}{h}=\dfrac {\lambda ^{'}}{h\cos \theta }+\dfrac {1}{P_{e}\sin \theta }\\ \lambda ^{0}=\dfrac {\lambda ^{'}}{\cos \theta }+\dfrac {h}{P_{e}\sin \theta }\end{aligned}$$

For y-direction:

$$

p_{e}\sin \phi =P'_{photon}\sin \theta

$$

$$

p_{e}\sin \left( 90-\theta \right) =P'_{photon}\sin \theta

$$

$$

p_{e}\cos \left(\theta \right) = \dfrac {h}{\lambda '}\sin \theta

$$

$$p_{e}\cot \left(\theta \right) = \dfrac {h}{\lambda '}$$

$$\lambda '=\dfrac {h}{P_{e}\cot \theta } $$

From the Compton equation,

$$\lambda '-\lambda _{0}=\dfrac {h}{m_{e}c}\left( 1-\cos \theta \right)$$

I was wondering if this is correct because I can't seem to solve it moving on from these steps. Any advice help will be appreciated. Cheers

Currently given a problem to prove that the scattered photon is fixed at a constant value with only the condition that both angles from the photon and electron sum up to 90 degrees after the collision. I can't seem to prove it and listed my steps below, was wondering if anyone can guide/correct my mistakes. Cheers.

For x-direction:

$$

P_{photon}=P'_{photon}cos\theta +P_{e}cos\phi

$$

$$

\frac{h}{\lambda_{0}}= \dfrac {h}{\lambda ^{'}}\cos \theta +P_{e}\cos \left( 90-\theta \right)

$$

$$\frac{h}{\lambda_{0}}= \dfrac {h}{\lambda ^{'}}\cos \theta +P_{e}\sin \left( \theta \right)$$

$$\begin{aligned}\dfrac {\lambda ^{0}}{h}=\dfrac {\lambda ^{'}}{h\cos \theta }+\dfrac {1}{P_{e}\sin \theta }\\ \lambda ^{0}=\dfrac {\lambda ^{'}}{\cos \theta }+\dfrac {h}{P_{e}\sin \theta }\end{aligned}$$

For y-direction:

$$

p_{e}\sin \phi =P'_{photon}\sin \theta

$$

$$

p_{e}\sin \left( 90-\theta \right) =P'_{photon}\sin \theta

$$

$$

p_{e}\cos \left(\theta \right) = \dfrac {h}{\lambda '}\sin \theta

$$

$$p_{e}\cot \left(\theta \right) = \dfrac {h}{\lambda '}$$

$$\lambda '=\dfrac {h}{P_{e}\cot \theta } $$

From the Compton equation,

$$\lambda '-\lambda _{0}=\dfrac {h}{m_{e}c}\left( 1-\cos \theta \right)$$

I was wondering if this is correct because I can't seem to solve it moving on from these steps. Any advice help will be appreciated. Cheers