You use x-ray photons of wavelength 0.251 nm in a Compton-scattering experiment. At a scattering angle of ${14.5}^{\circ}$, what is the increase in the wavelength of the scattered x rays compared to the incident x rays? Express your results in nm.

Arendrogfkl
2022-11-05
Answered

You use x-ray photons of wavelength 0.251 nm in a Compton-scattering experiment. At a scattering angle of ${14.5}^{\circ}$, what is the increase in the wavelength of the scattered x rays compared to the incident x rays? Express your results in nm.

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Frances Dodson

Answered 2022-11-06
Author has **17** answers

We have the expression of the Compton effect,

${\lambda}_{f}-{\lambda}_{i}=\frac{h}{{m}_{e}c}(1-\mathrm{cos}\theta )$

So, we easily calculate the final wavelength,

${\lambda}_{f}-0.251\times {10}^{-9}=\frac{6.626\times {10}^{-34}}{9.1\times {10}^{-31}\times 3\times {10}^{8}}(1-\mathrm{cos}({14.5}^{\circ}))\phantom{\rule{0ex}{0ex}}{\lambda}_{f}-0.251\times {10}^{-9}=\frac{6.626\times {10}^{-34}}{27.3\times {10}^{-23}}(1-0.9681)\phantom{\rule{0ex}{0ex}}{\lambda}_{f}-0.251\times {10}^{-9}=0.2427\times {10}^{-11}\times 0.0319\phantom{\rule{0ex}{0ex}}0.251\times {10}^{-9}=0.0077\times {10}^{-11}\phantom{\rule{0ex}{0ex}}{\lambda}_{f}=0.251077\times {10}^{-9}\text{}m$

So, the increasing wavelength,

$\mathrm{\u25b3}\lambda ={\lambda}_{f}-{\lambda}_{i}\phantom{\rule{0ex}{0ex}}\mathrm{\u25b3}\lambda =0.000077\times {10}^{-9}\text{}m$

But we have to calculate in nm

so, we change m into nm

$\mathrm{\u25b3}\lambda =0.000077\text{}nm$

${\lambda}_{f}-{\lambda}_{i}=\frac{h}{{m}_{e}c}(1-\mathrm{cos}\theta )$

So, we easily calculate the final wavelength,

${\lambda}_{f}-0.251\times {10}^{-9}=\frac{6.626\times {10}^{-34}}{9.1\times {10}^{-31}\times 3\times {10}^{8}}(1-\mathrm{cos}({14.5}^{\circ}))\phantom{\rule{0ex}{0ex}}{\lambda}_{f}-0.251\times {10}^{-9}=\frac{6.626\times {10}^{-34}}{27.3\times {10}^{-23}}(1-0.9681)\phantom{\rule{0ex}{0ex}}{\lambda}_{f}-0.251\times {10}^{-9}=0.2427\times {10}^{-11}\times 0.0319\phantom{\rule{0ex}{0ex}}0.251\times {10}^{-9}=0.0077\times {10}^{-11}\phantom{\rule{0ex}{0ex}}{\lambda}_{f}=0.251077\times {10}^{-9}\text{}m$

So, the increasing wavelength,

$\mathrm{\u25b3}\lambda ={\lambda}_{f}-{\lambda}_{i}\phantom{\rule{0ex}{0ex}}\mathrm{\u25b3}\lambda =0.000077\times {10}^{-9}\text{}m$

But we have to calculate in nm

so, we change m into nm

$\mathrm{\u25b3}\lambda =0.000077\text{}nm$

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