Aur beam having photon of energy 480 ev is incident on a foil of aluminium . If the photon is scattered at ${45}^{\circ}$ then calculate the energy of recall electron

kituoti126
2022-11-18
Answered

Aur beam having photon of energy 480 ev is incident on a foil of aluminium . If the photon is scattered at ${45}^{\circ}$ then calculate the energy of recall electron

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metodikkf6z

Answered 2022-11-19
Author has **14** answers

The problem is based on the concept of the Compton effect.

When the photon of energy incident on rest electron, they collide elastically. So their energy before scattering is the same as the energy after scattering. After collision incident photon gets scattered by an angle $\theta $ whereas rest electron gets recoiled.

Due to elastic collision; the energy of the incident photon $(E=hv)$ is equaled to the energy of scattered electron $({E}^{\prime}=h{v}^{\prime})$ and energy of the recoiled electron $({E}_{e})$

$E={E}^{\prime}+{E}_{e}$

According to the Compton effect, the shift in the wavelength is ;

$({\lambda}^{\prime}-\lambda )=2{\lambda}_{c}{\mathrm{sin}}^{2}(\frac{\theta}{2})$

As given, energy of incident photon is,

$E=hv=\frac{hc}{\lambda}=480\text{keV}=480\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}\lambda =\frac{hc}{480}\phantom{\rule{0ex}{0ex}}\lambda =\frac{(4.1356\times {10}^{-15}\text{eV}\times 3\times {10}^{8}\text{m/s})}{480\times {10}^{3}}\phantom{\rule{0ex}{0ex}}\lambda =0.0258475\times {10}^{-10}m\phantom{\rule{0ex}{0ex}}\lambda =2.58475\times {10}^{-12}m\phantom{\rule{0ex}{0ex}}({\lambda}^{\prime}-\lambda )={\lambda}_{c}(1-\mathrm{cos}\theta ))\phantom{\rule{0ex}{0ex}}=2.426\times {10}^{-12}\text{m}\times (0.2929)\phantom{\rule{0ex}{0ex}}=0.7106\times {10}^{-12}\phantom{\rule{0ex}{0ex}}{\lambda}^{\prime}=\lambda +0.7106\times {10}^{-12}\phantom{\rule{0ex}{0ex}}{\lambda}^{\prime}=3.29535\times {10}^{-12}\text{}m$

Now we can calculate the energy of scattered photon;

${E}^{\prime}=\frac{hc}{{\lambda}^{\prime}}$

We have

${\lambda}^{\prime}=3.29535\times {10}^{-12}\text{}m\phantom{\rule{0ex}{0ex}}h=4.1356\times {10}^{-15}\text{eV}\phantom{\rule{0ex}{0ex}}c=3\times {10}^{8}\text{m/s}\phantom{\rule{0ex}{0ex}}{E}^{\prime}=\frac{(4\xb71356\times {10}^{-15}\text{eV}\times 3\times {10}^{8}\text{m/s})}{3.29535\times {10}^{-12}m}\phantom{\rule{0ex}{0ex}}{E}^{\prime}=3.765\times {10}^{5}\text{eV}\phantom{\rule{0ex}{0ex}}{E}^{\prime}=376.5\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}{E}_{e}=\text{energy of recoil electron}\phantom{\rule{0ex}{0ex}}{E}_{e}=E-{E}^{\prime}$

we have,

$E=480\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}{E}^{\prime}=376.5\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}{E}_{e}=(480-376.5)\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}=103.5\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}{E}_{e}=103.5\text{keV}\phantom{\rule{0ex}{0ex}}=165.6\times {10}^{-16}\text{J}\phantom{\rule{0ex}{0ex}}{E}_{e}=1.656\times {10}^{-14}\text{J}$

Answer: The energy of recoil electron is $1.656\times {10}^{-14}\text{J}$ or $103.5\text{keV}$

When the photon of energy incident on rest electron, they collide elastically. So their energy before scattering is the same as the energy after scattering. After collision incident photon gets scattered by an angle $\theta $ whereas rest electron gets recoiled.

Due to elastic collision; the energy of the incident photon $(E=hv)$ is equaled to the energy of scattered electron $({E}^{\prime}=h{v}^{\prime})$ and energy of the recoiled electron $({E}_{e})$

$E={E}^{\prime}+{E}_{e}$

According to the Compton effect, the shift in the wavelength is ;

$({\lambda}^{\prime}-\lambda )=2{\lambda}_{c}{\mathrm{sin}}^{2}(\frac{\theta}{2})$

As given, energy of incident photon is,

$E=hv=\frac{hc}{\lambda}=480\text{keV}=480\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}\lambda =\frac{hc}{480}\phantom{\rule{0ex}{0ex}}\lambda =\frac{(4.1356\times {10}^{-15}\text{eV}\times 3\times {10}^{8}\text{m/s})}{480\times {10}^{3}}\phantom{\rule{0ex}{0ex}}\lambda =0.0258475\times {10}^{-10}m\phantom{\rule{0ex}{0ex}}\lambda =2.58475\times {10}^{-12}m\phantom{\rule{0ex}{0ex}}({\lambda}^{\prime}-\lambda )={\lambda}_{c}(1-\mathrm{cos}\theta ))\phantom{\rule{0ex}{0ex}}=2.426\times {10}^{-12}\text{m}\times (0.2929)\phantom{\rule{0ex}{0ex}}=0.7106\times {10}^{-12}\phantom{\rule{0ex}{0ex}}{\lambda}^{\prime}=\lambda +0.7106\times {10}^{-12}\phantom{\rule{0ex}{0ex}}{\lambda}^{\prime}=3.29535\times {10}^{-12}\text{}m$

Now we can calculate the energy of scattered photon;

${E}^{\prime}=\frac{hc}{{\lambda}^{\prime}}$

We have

${\lambda}^{\prime}=3.29535\times {10}^{-12}\text{}m\phantom{\rule{0ex}{0ex}}h=4.1356\times {10}^{-15}\text{eV}\phantom{\rule{0ex}{0ex}}c=3\times {10}^{8}\text{m/s}\phantom{\rule{0ex}{0ex}}{E}^{\prime}=\frac{(4\xb71356\times {10}^{-15}\text{eV}\times 3\times {10}^{8}\text{m/s})}{3.29535\times {10}^{-12}m}\phantom{\rule{0ex}{0ex}}{E}^{\prime}=3.765\times {10}^{5}\text{eV}\phantom{\rule{0ex}{0ex}}{E}^{\prime}=376.5\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}{E}_{e}=\text{energy of recoil electron}\phantom{\rule{0ex}{0ex}}{E}_{e}=E-{E}^{\prime}$

we have,

$E=480\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}{E}^{\prime}=376.5\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}{E}_{e}=(480-376.5)\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}=103.5\times {10}^{3}\text{eV}\phantom{\rule{0ex}{0ex}}{E}_{e}=103.5\text{keV}\phantom{\rule{0ex}{0ex}}=165.6\times {10}^{-16}\text{J}\phantom{\rule{0ex}{0ex}}{E}_{e}=1.656\times {10}^{-14}\text{J}$

Answer: The energy of recoil electron is $1.656\times {10}^{-14}\text{J}$ or $103.5\text{keV}$

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