Through what angle must a$\mathbf{200}\mathbf{\text{\hspace{0.17em}keV}}$photon be scattered by a free electron so that the photon loses $\mathbf{10}\mathit{\%}$of its energy?

The energy of the photon is,$E=200\text{\hspace{0.17em}keV}$

The photon losses is$10\%$,.

The expression to calculate the energy is given as follows.

$E=\frac{hc}{\lambda }$ …(i)

Here,h is the plank’s constant and c is the speed of light.

The expression to calculate the change in the wavelength is given as follows.

$\Delta \lambda =\frac{h}{mc}(1-cos\theta )$ …(ii)

Here,m is the mass of the electron.

The expression to calculate the changes in the photon’s energy is given as follows.

$\Delta E\%=\frac{E-E^{\text{'}}}{E}$ …(iii)

Here,$E^{\text{'}}$ is the energy after the scattering and is the energy before the scattering.

The energy of the scattered photon is given by,

$E^{\text{'}}=\frac{hc}{{\lambda }^{\text{'}}}$

The energy of the incident photon is given by,

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

Calculate the change in the photon energy.

Substitute$0.10$for$\Delta E\%$, $\frac{hc}{{\lambda }^{\text{'}}}$for $E^{\text{'}}$and$\frac{hc}{\lambda }$for E into equation (iii).

$\begin{array}{l}0.10=\frac{\frac{hc}{\lambda }-\frac{hc}{{\lambda }^{\text{'}}}}{\frac{hc}{\lambda }}\\ 0.10=\frac{\frac{1}{\lambda }-\frac{1}{{\lambda }^{\text{'}}}}{\frac{1}{\lambda }}\\ 0.10=\frac{{\lambda }^{\text{'}}-\lambda }{{\lambda }^{\text{'}}}\end{array}$

Substitute $\Delta \lambda$for ${\lambda }^{\text{'}}-\lambda$into above equation.

$\begin{array}{c}0.10=\frac{\Delta \lambda }{{\lambda }^{\text{'}}}\\ \Delta \lambda =0.10{\lambda }^{\text{'}}\\ \Delta \lambda =0.10(\lambda +\Delta \lambda )\\ \Delta \lambda =\frac{\lambda }{9}\end{array}$

Substitute $\lambda /9$for $\Delta \lambda$into equation (ii).

$\begin{array}{l}\frac{\lambda }{9}=\frac{h}{mc}(1-\cos \theta )\\ \lambda =\frac{9h}{mc}(1-\cos \theta )\end{array}$

Substitute$hc/E$for $\lambda$into above equation.

$\begin{array}{c}\frac{hc}{E}=\frac{9h}{mc}(1-\cos \theta )\\ \frac{c}{E}=\frac{9}{mc}(1-\cos \theta )\\ 1=\frac{9E}{m{c}^2}(1-\cos \theta )\\ 1-\cos \theta =\frac{m{c}^2}{9E}\end{array}$

Solve further as,

role="math" localid="1663146814795" $\begin{array}{c}\cos \theta =1-\frac{m{c}^2}{9E}\\ \theta ={\cos }^{-1}\left(1-\frac{m{c}^2}{9E}\right)\end{array}$

Substitute $511\text{\hspace{0.17em}keV}$for $m{c}^2$and $200\text{\hspace{0.17em}keV}$for E into above equation.

$\begin{array}{l}\theta ={\cos }^{-1}\left(1-\frac{511\text{\hspace{0.17em}Kev}}{9\times 200\text{\hspace{0.17em}Kev}}\right)\\ \theta ={\cos }^{-1}\left(0.7161\right)\\ \theta =44.2°\end{array}$

Hence the required value of the angle is.$44.2°$