Question

In compton effect, when scatterring angle of x-rays is zero,hence there is no change in frequency and energy.
but in this case elecron recoils at 90 degrees why?
From where does it get the energy to recoil?

Answer 1

Answer on Question #46943, Physics, Quantum Mechanics

So we have three equations:

$$\begin{array}{l} E _ {g} = \frac {\sqrt {E _ {e} ^ {2} - m ^ {2} c ^ {4}}}{c} \cdot \cos \varphi + \frac {E _ {g} - E _ {e}}{c} \cos \theta \\ \frac {\sqrt {E _ {e} ^ {2} - m ^ {2} c ^ {4}}}{c} \cdot \sin \varphi = \frac {E _ {g} - E _ {e}}{c} \sin \theta \\ E _ {g} - E _ {g} ^ {\prime} = E _ {e} \\ \end{array}$$

We will introduce everything as functions of $\varphi$

$E_{e} = \frac{1}{4 E_{g}} \cdot \left(m^{2} c^{4} + 4 E_{g}^{2} \cdot \cos^{2}(\varphi)\right)$ - final energy of the electron

if $\varphi = 0 \Rightarrow E_{e} = \frac{m^{2}c^{4}}{4E_{g}}$ - this is minimal possible energy, that electron can obtain via Compton scattering.

$$E _ {g} ^ {\prime} = E _ {g} - \frac {m ^ {2} c ^ {4} + 4 E _ {g} ^ {2} \cdot \cos^ {2} (\varphi)}{4 E _ {g}}$$

if $\varphi = 0 \Rightarrow E_g' = E_g - \frac{m^2c^4}{4E_g}$ - so we will have non zero energy, so non zero momentum along Y axis.

We can also calculate angle $\theta$

$\sin \theta = \frac{\sqrt{E_e^2 - m^2c^4}}{E_g'}$ - also non zero angle.

Yes, $\theta$ won't be equal to 90 degrees.

ANSWER:

Electron will get it's energy from the photon.

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