A 100 MeV photon collides with a resting prion. Calculate the maximum energy loss that the photon can suffer.

Baladdaa9
2022-07-17
Answered

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autosmut6p

Answered 2022-07-18
Author has **8** answers

For the photon:

${E}_{1}=\frac{hc}{{\lambda}_{1}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{1}=\frac{hc}{{E}_{1}}$

Substitute $100\times {10}^{6}\text{eV}$ for ${E}_{1},\text{}3\times {10}^{8}\text{m/s}$ for c, and $6.63\times {10}^{-34}\text{}J\cdot s$ for h

${\lambda}_{1}=\frac{(6.63\times {10}^{-34}\text{}J\cdot s)(3\times {10}^{8}\text{m/s})}{(100\times {10}^{6}\text{eV})}\phantom{\rule{0ex}{0ex}}=\frac{(6.63\times {10}^{-34}\text{}J\cdot s)(3\times {10}^{8}\text{m/s})}{(100\times {10}^{6}\text{eV})\times (\frac{1.6\times {10}^{-19}\text{}J}{1\text{eV}})}\phantom{\rule{0ex}{0ex}}=\frac{(6.63\times {10}^{-34}\text{}J\cdot s)(3\times {10}^{8}\text{m/s})}{(100\times {10}^{6}\times 1.6\times {10}^{-19})J}\phantom{\rule{0ex}{0ex}}{\lambda}_{1}=12.43125\times {10}^{-15}\text{}m$

By using the Compton's effect:

${\lambda}_{2}-{\lambda}_{1}=\frac{h}{{m}_{p}c}(1-\mathrm{cos}\theta )$

Here, the maximum possible change of energy takes place when the photon recoils in the opposite direction to that of the incident direction.

When $\mathrm{cos}\theta =-1$ therefore,

$\theta ={\mathrm{cos}}^{-1}(-1)\phantom{\rule{0ex}{0ex}}\theta ={180}^{\circ}$

Substitute $12.43125\times {10}^{-15}\text{}m$ for ${\lambda}_{1},\text{}{180}^{\circ}$ for $\theta ,1.6\times {10}^{-27}\text{}kg$ for ${m}_{p},3\times {10}^{8}\text{m/s}$ for c, and $6.63\times {10}^{-34}\text{}J\cdot s$ for h into equation.

${\lambda}_{2}-12.4\times {10}^{-15}\text{}m=\frac{6.63\times {10}^{-34}\text{}J\cdot s}{(1.6\times {10}^{-27}\text{kg})(3\times {10}^{8}\text{m/s})}(1-\mathrm{cos}{180}^{\circ})\phantom{\rule{0ex}{0ex}}{\lambda}_{2}-12.4\times {10}^{-15}\text{}m=2.76\times {10}^{-15}\phantom{\rule{0ex}{0ex}}{\lambda}_{2}=15.16\times {10}^{-15}\text{}m$

The lose of energy is,

$\mathrm{\u25b3}E={E}_{2}-{E}_{1}\phantom{\rule{0ex}{0ex}}=\frac{hc}{{\lambda}_{2}}-\frac{hc}{{\lambda}_{1}}$

So:

$\mathrm{\u25b3}E=\frac{6.63\times {10}^{-34}J\cdot s\times 3\times {10}^{8}\text{m/s}}{15.16\times {10}^{-15}\text{}m}-\frac{6.63\times {10}^{-34}J\cdot s\times 3\times {10}^{8}\text{m/s}}{12.4\times {10}^{-15}\text{}m}\phantom{\rule{0ex}{0ex}}=1.312\times {10}^{-11}\text{}J-1.604\times {10}^{-11}\text{}J\phantom{\rule{0ex}{0ex}}\mathrm{\u25b3}E=2.916\times {10}^{-11}\text{}J$

Hence, the maximum energy loss that the photon can suffer is $2.916\times {10}^{-11}\text{}J$

${E}_{1}=\frac{hc}{{\lambda}_{1}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{1}=\frac{hc}{{E}_{1}}$

Substitute $100\times {10}^{6}\text{eV}$ for ${E}_{1},\text{}3\times {10}^{8}\text{m/s}$ for c, and $6.63\times {10}^{-34}\text{}J\cdot s$ for h

${\lambda}_{1}=\frac{(6.63\times {10}^{-34}\text{}J\cdot s)(3\times {10}^{8}\text{m/s})}{(100\times {10}^{6}\text{eV})}\phantom{\rule{0ex}{0ex}}=\frac{(6.63\times {10}^{-34}\text{}J\cdot s)(3\times {10}^{8}\text{m/s})}{(100\times {10}^{6}\text{eV})\times (\frac{1.6\times {10}^{-19}\text{}J}{1\text{eV}})}\phantom{\rule{0ex}{0ex}}=\frac{(6.63\times {10}^{-34}\text{}J\cdot s)(3\times {10}^{8}\text{m/s})}{(100\times {10}^{6}\times 1.6\times {10}^{-19})J}\phantom{\rule{0ex}{0ex}}{\lambda}_{1}=12.43125\times {10}^{-15}\text{}m$

By using the Compton's effect:

${\lambda}_{2}-{\lambda}_{1}=\frac{h}{{m}_{p}c}(1-\mathrm{cos}\theta )$

Here, the maximum possible change of energy takes place when the photon recoils in the opposite direction to that of the incident direction.

When $\mathrm{cos}\theta =-1$ therefore,

$\theta ={\mathrm{cos}}^{-1}(-1)\phantom{\rule{0ex}{0ex}}\theta ={180}^{\circ}$

Substitute $12.43125\times {10}^{-15}\text{}m$ for ${\lambda}_{1},\text{}{180}^{\circ}$ for $\theta ,1.6\times {10}^{-27}\text{}kg$ for ${m}_{p},3\times {10}^{8}\text{m/s}$ for c, and $6.63\times {10}^{-34}\text{}J\cdot s$ for h into equation.

${\lambda}_{2}-12.4\times {10}^{-15}\text{}m=\frac{6.63\times {10}^{-34}\text{}J\cdot s}{(1.6\times {10}^{-27}\text{kg})(3\times {10}^{8}\text{m/s})}(1-\mathrm{cos}{180}^{\circ})\phantom{\rule{0ex}{0ex}}{\lambda}_{2}-12.4\times {10}^{-15}\text{}m=2.76\times {10}^{-15}\phantom{\rule{0ex}{0ex}}{\lambda}_{2}=15.16\times {10}^{-15}\text{}m$

The lose of energy is,

$\mathrm{\u25b3}E={E}_{2}-{E}_{1}\phantom{\rule{0ex}{0ex}}=\frac{hc}{{\lambda}_{2}}-\frac{hc}{{\lambda}_{1}}$

So:

$\mathrm{\u25b3}E=\frac{6.63\times {10}^{-34}J\cdot s\times 3\times {10}^{8}\text{m/s}}{15.16\times {10}^{-15}\text{}m}-\frac{6.63\times {10}^{-34}J\cdot s\times 3\times {10}^{8}\text{m/s}}{12.4\times {10}^{-15}\text{}m}\phantom{\rule{0ex}{0ex}}=1.312\times {10}^{-11}\text{}J-1.604\times {10}^{-11}\text{}J\phantom{\rule{0ex}{0ex}}\mathrm{\u25b3}E=2.916\times {10}^{-11}\text{}J$

Hence, the maximum energy loss that the photon can suffer is $2.916\times {10}^{-11}\text{}J$

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I am a layman in Physics and I stumbled on this question while understanding electron microscopy

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$m$ would increase with $v$. Thus the $\lambda $ value should change doubly...first for the rise of $v$ and second for the rise of $m$

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So I was learning de Broglie wavelength today in my physics class, and I started playing around with it. I wondered if it was possible to calculate the energy of a light wave given its wavelength and speed. After rearranging a bit, I plugged it into $E=m{c}^{2}$, and realized I had found the equation for the energy of a photon that I learned at the beginning of my quantum mechanics unit, $E=hf$

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Secondly, as I was doing those steps, I thought I would be calculating the energy of the entire light ray. I now realize I was finding the energy of a single photon. In hindsight, this makes sense, because the energy of the entire light ray must depend on some length value, correct?

This led me to two other questions: do light rays have some finite length? how do I calculate the energy of a light ray, not just a single photon?

So I was learning de Broglie wavelength today in my physics class, and I started playing around with it. I wondered if it was possible to calculate the energy of a light wave given its wavelength and speed. After rearranging a bit, I plugged it into $E=m{c}^{2}$, and realized I had found the equation for the energy of a photon that I learned at the beginning of my quantum mechanics unit, $E=hf$

$\lambda =\frac{h}{p}$

$p=\frac{h}{\lambda}$

$E=m{c}^{2}$

$E=\frac{p}{c}\cdot {c}^{2}$

$E=\frac{hc}{\lambda}=hf$

I have a few questions about this. Firstly, I do not understand how it can make sense to do $\frac{p}{c}$ in this context, because, as I understand it, light has no mass. How can I come to $E=hf$ using the mass of a massless object?

Secondly, as I was doing those steps, I thought I would be calculating the energy of the entire light ray. I now realize I was finding the energy of a single photon. In hindsight, this makes sense, because the energy of the entire light ray must depend on some length value, correct?

This led me to two other questions: do light rays have some finite length? how do I calculate the energy of a light ray, not just a single photon?