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Cut-off Wavelength

A beam of rad...

Question

A beam of radiation of wavelength 0.0242 $˚ A$ is scattered from a target at an angle of $90^{o}$ . Find the energy and direction of recoil electron.

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Solution

$\begin{matrix} Δ P = \frac{2 E}{C} E = \frac{Δ P C}{2} E = \frac{h c}{λ 2} = \frac{h c}{0.0242 \times 2} H e n c e, E = 1.642 \times 10^{- 23} J \end{matrix}$

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