91Ó°ÊÓ

The energy of the photon is,$E=200\text{ k±ð³Õ}$

The photon losses is$10\%$,.

The expression to calculate the energy is given as follows.

$E=\frac{hc}{\mathrm{λ}}$ …(¾±)

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.

$\mathrm{Δ}\mathrm{λ}=\frac{h}{mc}(1-cos\mathrm{θ})$ …(¾±¾±)

Here,m is the mass of the electron.

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

$\mathrm{Δ}E\%=\frac{E-E^{\text{'}}}{E}$ …(¾±¾±¾±)

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,

$E^{\text{'}}=\frac{hc}{{\mathrm{λ}}^{\text{'}}}$

The energy of the incident photon is given by,

$E=\frac{hc}{\mathrm{λ}}$

Calculate the change in the photon energy.

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

$\begin{array}{l}0.10=\frac{\frac{hc}{\mathrm{λ}}−\frac{hc}{{\mathrm{λ}}^{\text{'}}}}{\frac{hc}{\mathrm{λ}}}\\ 0.10=\frac{\frac{1}{\mathrm{λ}}−\frac{1}{{\mathrm{λ}}^{\text{'}}}}{\frac{1}{\mathrm{λ}}}\\ 0.10=\frac{{\mathrm{λ}}^{\text{'}}−\mathrm{λ}}{{\mathrm{λ}}^{\text{'}}}\end{array}$

Substitute $\mathrm{Δ}\mathrm{λ}$for ${\mathrm{λ}}^{\text{'}}−\mathrm{λ}$into above equation.

$\begin{array}{c}0.10=\frac{\mathrm{Δ}\mathrm{λ}}{{\mathrm{λ}}^{\text{'}}}\\ \mathrm{Δ}\mathrm{λ}=0.10{\mathrm{λ}}^{\text{'}}\\ \mathrm{Δ}\mathrm{λ}=0.10(\mathrm{λ}+\mathrm{Δ}\mathrm{λ})\\ \mathrm{Δ}\mathrm{λ}=\frac{\mathrm{λ}}{9}\end{array}$

Substitute $\mathrm{λ}/9$for $\mathrm{Δ}\mathrm{λ}$into equation (ii).

$\begin{array}{l}\frac{\mathrm{λ}}{9}=\frac{h}{mc}(1−\cos \mathrm{θ})\\ \mathrm{λ}=\frac{9h}{mc}(1−\cos \mathrm{θ})\end{array}$

Substitute$hc/E$for $\mathrm{λ}$into above equation.

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

Solve further as,

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

Substitute $511\text{ k±ð³Õ}$for $m{c}^2$and $200\text{ k±ð³Õ}$for E into above equation.

$\begin{array}{l}\mathrm{θ}={\cos }^{−1}\left(1−\frac{511\text{‿é±ð±¹}}{9×200\text{‿é±ð±¹}}\right)\\ \mathrm{θ}={\cos }^{−1}\left(0.7161\right)\\ \mathrm{θ}=44.2°\end{array}$

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