After a 0.800-nm x-ray photon scatters from a free electron, the electron recoils at $1.40\times {10}^{6}\text{}m/s$. What is the scattering angle?

Liberty Page
2022-09-25
Answered

After a 0.800-nm x-ray photon scatters from a free electron, the electron recoils at $1.40\times {10}^{6}\text{}m/s$. What is the scattering angle?

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Davian Nguyen

Answered 2022-09-26
Author has **9** answers

We know Compton shift is given by,

$\mathrm{\u25b3}\lambda =\frac{h}{{m}_{e}c}(1-\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}$

Solving for $\theta $ we get,

$(1-\mathrm{cos}\theta )=\frac{{m}_{e}c\mathrm{\u25b3}\lambda}{h}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{cos}}^{-1}(1-\frac{{m}_{e}c\mathrm{\u25b3}\lambda}{h})$

Substitute $\mathrm{\u25b3}\lambda $ as 0.003 nm, h as $6.626\times {10}^{-34}\text{}J\cdot s$ as $9.1\times {10}^{-31}\text{}kg$, and c as $3\times {10}^{8}\text{}m\cdot {s}^{-1}$

$\theta ={\mathrm{cos}}^{-1}(1-\frac{(9.1\times {10}^{-31}\text{}kg)(3\times {10}^{8}m\cdot {s}^{-1})(0.003nm)}{(6.626\times {10}^{-34}\text{}J\cdot s)})\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}(-0.236)\phantom{\rule{0ex}{0ex}}={103.3}^{\circ}$

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

$\mathrm{\u25b3}\lambda =\frac{h}{{m}_{e}c}(1-\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}$

Solving for $\theta $ we get,

$(1-\mathrm{cos}\theta )=\frac{{m}_{e}c\mathrm{\u25b3}\lambda}{h}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{cos}}^{-1}(1-\frac{{m}_{e}c\mathrm{\u25b3}\lambda}{h})$

Substitute $\mathrm{\u25b3}\lambda $ as 0.003 nm, h as $6.626\times {10}^{-34}\text{}J\cdot s$ as $9.1\times {10}^{-31}\text{}kg$, and c as $3\times {10}^{8}\text{}m\cdot {s}^{-1}$

$\theta ={\mathrm{cos}}^{-1}(1-\frac{(9.1\times {10}^{-31}\text{}kg)(3\times {10}^{8}m\cdot {s}^{-1})(0.003nm)}{(6.626\times {10}^{-34}\text{}J\cdot s)})\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}(-0.236)\phantom{\rule{0ex}{0ex}}={103.3}^{\circ}$

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

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