Paloma Sanford

2022-10-23

Through what angle must a 0.2 MeV photon be scattered by a free electron in order for it to lose 10% of its energy?

na1p1a2pafr

Beginner2022-10-24Added 16 answers

The fractional change in energy of a photon in Compton effect,

Energy of the photon, $h{v}^{\prime}=0.2\text{MeV}$

Energy lost, $\mathrm{\u25b3}E=10\mathrm{\%}$ of its original energy

$\mathrm{\u25b3}E=10\mathrm{\%}E$

rest mass of the electron, ${m}_{0}=0.511{\text{MeV/c}}^{2}$

$\theta =$ angle of scattering of the electron

$\frac{\mathrm{\u25b3}E}{E}=\frac{10}{100}=0.1$

using the formula, we get

$\frac{\mathrm{\u25b3}E}{E}=\frac{h{v}^{\prime}}{{m}_{0}{c}^{2}}(1-\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}0.1=\frac{0.2}{(0.511{\text{MeV/c}}^{2}){c}^{2}}(1-\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}0.1=\frac{0.2}{0.511}(1-\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}1-0.2555=\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =0.7445\phantom{\rule{0ex}{0ex}}\theta ={41.88}^{\circ}$

therefore the angle of scattering, $\theta ={41.88}^{\circ}$

Answer: required angle of scattering, $\theta ={41.88}^{\circ}$

Energy of the photon, $h{v}^{\prime}=0.2\text{MeV}$

Energy lost, $\mathrm{\u25b3}E=10\mathrm{\%}$ of its original energy

$\mathrm{\u25b3}E=10\mathrm{\%}E$

rest mass of the electron, ${m}_{0}=0.511{\text{MeV/c}}^{2}$

$\theta =$ angle of scattering of the electron

$\frac{\mathrm{\u25b3}E}{E}=\frac{10}{100}=0.1$

using the formula, we get

$\frac{\mathrm{\u25b3}E}{E}=\frac{h{v}^{\prime}}{{m}_{0}{c}^{2}}(1-\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}0.1=\frac{0.2}{(0.511{\text{MeV/c}}^{2}){c}^{2}}(1-\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}0.1=\frac{0.2}{0.511}(1-\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}1-0.2555=\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =0.7445\phantom{\rule{0ex}{0ex}}\theta ={41.88}^{\circ}$

therefore the angle of scattering, $\theta ={41.88}^{\circ}$

Answer: required angle of scattering, $\theta ={41.88}^{\circ}$

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