Baladdaa9

2022-07-17

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

autosmut6p

Beginner2022-07-18Added 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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