91Ó°ÊÓ

The photons scatters at the angle,$\mathrm{θ}=90°$$\mathrm{θ}=90°$

Initial wavelength,$\mathrm{λ}=3×{10}^{−12}\text{″¾}$

The expression to calculate the initial energy is given as follows.

$E=\frac{hc}{\mathrm{λ}}$ …… (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.

$\mathit{Δ}\mathit{λ}\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{c}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{θ}\mathbf{)}$…… (ii)

Here, m is the mass of the electron.

The expression to calculate the new wavelength is given as follows.

${\mathit{λ}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{Δ}\mathit{λ}\mathbf{+}\mathit{λ}$…… (iii)

The expression to calculate the kinetic energy of the electron is given as follows.

$\mathit{K}\mathbf{=}\mathit{E}\mathbf{-}{\mathit{E}}^{\mathbf{\text{'}}}$ …… (iv)

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

The value of plank’s constant is $6.62×{10}^{-34}\text{ }m\text{ }kg/s$.

The value of mass of electron is$9.11×{10}^{−31}\text{ k²µ}$.

Calculate the initial energy.

Substitute $6.62×{10}^{−34}{\text{″¾}}^2\text{kg}/\text{s}$for h, $3×{10}^8\text{″¾}/\text{s}$for c and$3×{10}^{−12}$for$\mathrm{λ}$into equation (i).

$\begin{array}{l}E=\frac{6.62×{10}^{−34}×3×{10}^8}{3×{10}^{−12}}\\ E=\frac{6.62×{10}^{−26}}{{10}^{−12}}\\ E=6.62×{10}^{−14}\text{ J}\end{array}$

Calculate the change in the wavelength,

Substitute$6.62×{10}^{−34}{\text{″¾}}^2\text{kg}/\text{s}$for h , $3×{10}^8\text{″¾}/\text{s}$for C, $9.11×{10}^{−31}\text{ k²µ}$for m and $90°$for $\mathrm{θ}$into equation (ii).

$\begin{array}{c}\mathrm{Δ}\mathrm{λ}=\frac{6.62×{10}^{−34}}{9.11×{10}^{−31}×3×{10}^8}(1−\cos 90°)\\ \mathrm{Δ}\mathrm{λ}=\frac{6.62×{10}^{−34}}{27.33×{10}^{−23}}(1−0)\\ \mathrm{Δ}\mathrm{λ}=0.243×{10}^{−11}\\ \mathrm{Δ}\mathrm{λ}=2.43×{10}^{−12}\text{″¾}\end{array}$

Calculate the value of the new wavelength,

Substitute $2.43×{10}^{−12}\text{″¾}$for$\mathrm{Δ}\mathrm{λ}$and $3×{10}^{−12}\text{″¾}$for$\mathrm{λ}$into equation (iii).

$\begin{array}{l}{\mathrm{λ}}^{\text{'}}=2.43×{10}^{−12}+3×{10}^{−12}\\ {\mathrm{λ}}^{\text{'}}=(2.43+3){10}^{−12}\\ {\mathrm{λ}}^{\text{'}}=5.43×{10}^{−12}\text{″¾}\end{array}$

Calculate the energy after the scattering,

Substitute $6.62×{10}^{−34}{\text{″¾}}^2\text{kg}/\text{s}$for h , $3×{10}^8\text{″¾}/\text{s}$for c and $5.43×{10}^{−12}$for$\mathrm{λ}$ into equation (i).

$\begin{array}{l}{E}^{\text{'}}=\frac{6.62×{10}^{−34}×3×{10}^8}{5.43×{10}^{−12}}\\ E^{\text{'}}=\frac{19.86×{10}^{−26}}{5.43×{10}^{−12}}\\ E^{\text{'}}=3.65×{10}^{−14}\text{ J}\end{array}$

The kinetic energy of the electron would be the difference of the energy of before and after the scattering.

Calculate the kinetic energy of electron.

Substitute$3.65×{10}^{-14}$ for $E^{\text{'}}$and$6.62×{10}^{−14}\text{ J}$ for E intoequation (iv).

$\begin{array}{l}K=6.62×{10}^{−14}−3.65×{10}^{−14}\\ K=(6.62−3.65)×{10}^{−14}\\ K=2.97×{10}^{−14}\text{ J}\end{array}$

Hence the kinetic energy of the electron is $2.97×{10}^{−14}\text{ J}$.