# After a 0.800-nm x-ray photon scatters from a free electron, the electron recoils at 1.40 xx 10^6 m/s. (b) What is the scattering angle?

After a 0.800-nm x-ray photon scatters from a free electron, the electron recoils at . What is the scattering angle?
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Davian Nguyen
We know Compton shift is given by,
$\mathrm{△}\lambda =\frac{h}{{m}_{e}c}\left(1-\mathrm{cos}\theta \right)\phantom{\rule{0ex}{0ex}}$
Solving for $\theta$ we get,
$\left(1-\mathrm{cos}\theta \right)=\frac{{m}_{e}c\mathrm{△}\lambda }{h}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{cos}}^{-1}\left(1-\frac{{m}_{e}c\mathrm{△}\lambda }{h}\right)$
Substitute $\mathrm{△}\lambda$ as 0.003 nm, h as as , and c as

Thus the angle of scattering will be ${103.3}^{\circ }$.