A beam of gamma-ray having a photon of energy 510 keV is incident on a foil of aluminum. Calculate the wavelength of the radiation scattered at 90 degrees

Bruno Thompson
2022-07-22
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eishale2n

Answered 2022-07-23
Author has **15** answers

Given the beam of gamma-ray of energy 510 keV is incident on the aluminum foil. The wavelength will be given by,

${\lambda}_{0}=\frac{hc}{E}$

${\lambda}_{0}=\frac{(6.63\times {10}^{-34}\text{}J\cdot s)(3\times {10}^{8}\text{}m\cdot {s}^{-1})}{(510\times {10}^{3}\times 1.6\times {10}^{-19}\text{}J)}\phantom{\rule{0ex}{0ex}}=2.426\times {10}^{-12}\text{}m$

Now wavelength of the scattered radiation is given by,

$\lambda ={\lambda}_{0}+\frac{h}{{m}_{0}c}(1-\mathrm{cos}\theta )$

$\lambda =4.85\times {10}^{-12}\text{}m$

${\lambda}_{0}=\frac{hc}{E}$

${\lambda}_{0}=\frac{(6.63\times {10}^{-34}\text{}J\cdot s)(3\times {10}^{8}\text{}m\cdot {s}^{-1})}{(510\times {10}^{3}\times 1.6\times {10}^{-19}\text{}J)}\phantom{\rule{0ex}{0ex}}=2.426\times {10}^{-12}\text{}m$

Now wavelength of the scattered radiation is given by,

$\lambda ={\lambda}_{0}+\frac{h}{{m}_{0}c}(1-\mathrm{cos}\theta )$

$\lambda =4.85\times {10}^{-12}\text{}m$

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Is there a mathematical way to prove this?

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Wikipedia states that the radiation of quasars is partially nonthermal (i.e. not blackbody).

Well, what percentage of the total radiation isn't black body?

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I was wondering if using the De Broglie equation

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for object traveling at really high speeds would result in a significant error. For example if an object travelled at $0.02c$ would the error be negligible? How can I calculate the uncertainty in the result?

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${u}_{\lambda}(\lambda ,T)=\frac{2h{c}^{2}}{{\lambda}^{5}}\frac{1}{{e}^{\frac{hc}{\lambda kT}}-1}$

${u}_{\nu}(\nu ,T)=\frac{2h{\nu}^{3}}{{c}^{2}}\frac{1}{{e}^{\frac{h\nu}{kT}}-1}$

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Planck's Law is commonly stated in two different ways:

${u}_{\lambda}(\lambda ,T)=\frac{2h{c}^{2}}{{\lambda}^{5}}\frac{1}{{e}^{\frac{hc}{\lambda kT}}-1}$

${u}_{\nu}(\nu ,T)=\frac{2h{\nu}^{3}}{{c}^{2}}\frac{1}{{e}^{\frac{h\nu}{kT}}-1}$

We can find the maximum of those functions by differentiating those equations with respect to $\lambda $ and to $\nu $, respectively. We get two ways to write Wien's Displacement Law:

${\lambda}_{\text{peak}}T=2.898\cdot {10}^{-3}m\cdot K$

$\frac{{\nu}_{\text{peak}}}{T}=5.879\cdot {10}^{10}Hz\cdot {K}^{-1}$

We see that ${\lambda}_{\text{peak}}\ne \frac{c}{{\nu}_{\text{peak}}}$. So what frequency or wavelength is actually detected by an optical instrument most intensely when analyzing a black body? If they are ${\lambda}_{\text{peak}}$ and ${\nu}_{\text{peak}}$, how is ${\lambda}_{\text{peak}}\ne \frac{c}{{\nu}_{\text{peak}}}$?

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But I am stuck here because if I integrate the L.H.S.

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I know this might be a silly question, but is it necessary to know Planck's Law in order to show that ${\lambda}_{max}\propto \frac{1}{T}$? If you set

$u(\lambda ,T)=\frac{f(\lambda T)}{{\lambda}^{5}}$

then

$\frac{\mathrm{\partial}u}{\mathrm{\partial}\lambda}=\frac{1}{{\lambda}^{5}}\frac{\mathrm{\partial}f}{\mathrm{\partial}\lambda}-\frac{5}{{\lambda}^{4}}f=0$

$\frac{\frac{\mathrm{\partial}f}{\mathrm{\partial}\lambda}}{f}=5\lambda $

But I am stuck here because if I integrate the L.H.S.

$\int \frac{\frac{\mathrm{\partial}f}{\mathrm{\partial}\lambda}}{f}d\lambda =\mathrm{log}(f(\lambda T))=\frac{5}{2}{\lambda}^{2}$

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