Question

A photon undergoes Compton scattering off a stationary free electron.

The photon scatters at$\mathbf{90}\mathbf{.}\mathbf{0}\mathbf{°}$ from its initial direction;

its initial wavelength is$\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathit{m}$ . What is the electron’s kinetic energy?

Short answer

The kinetic energy of the electron is $2.97\times {10}^{-14}\text{\hspace{0.17em}J}$.

Step-by-step solution

Write the given data from the question.

The photons scatters at the angle,$\theta =90°$$\theta =90°$

Initial wavelength,$\lambda =3\times {10}^{-12}\text{\hspace{0.17em}m}$

Determine the formulas to calculate the kinetic energy of the electron.

The expression to calculate the initial 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.

$\mathit{\Delta }\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{c}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{)}$…… (ii)

Here, m is the mass of the electron.

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

${\mathit{\lambda }}^{\mathbf{\text{'}}}\mathbf{=}\mathit{\Delta }\mathit{\lambda }\mathbf{+}\mathit{\lambda }$…… (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.

Calculate the electron’s kinetic energy.

The value of plank’s constant is $6.62\times {10}^{-34}mkg/s$.

The value of mass of electron is$9.11\times {10}^{-31}\text{\hspace{0.17em}kg}$.

Calculate the initial energy.

Substitute $6.62\times {10}^{-34}{\text{\hspace{0.17em}m}}^2\text{kg}/\text{s}$for h, $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$for c and$3\times {10}^{-12}$for$\lambda$into equation (i).

$\begin{array}{l}E=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3\times {10}^{-12}}\\ E=\frac{6.62\times {10}^{-26}}{{10}^{-12}}\\ E=6.62\times {10}^{-14}\text{\hspace{0.17em}J}\end{array}$

Calculate the change in the wavelength,

Substitute$6.62\times {10}^{-34}{\text{\hspace{0.17em}m}}^2\text{kg}/\text{s}$for h , $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$for C, $9.11\times {10}^{-31}\text{\hspace{0.17em}kg}$for m and $90°$for $\theta$into equation (ii).

$\begin{array}{c}\Delta \lambda =\frac{6.62\times {10}^{-34}}{9.11\times {10}^{-31}\times 3\times {10}^8}(1-\cos 90°)\\ \Delta \lambda =\frac{6.62\times {10}^{-34}}{27.33\times {10}^{-23}}(1-0)\\ \Delta \lambda =0.243\times {10}^{-11}\\ \Delta \lambda =2.43\times {10}^{-12}\text{\hspace{0.17em}m}\end{array}$

Calculate the value of the new wavelength,

Substitute $2.43\times {10}^{-12}\text{\hspace{0.17em}m}$for$\Delta \lambda$and $3\times {10}^{-12}\text{\hspace{0.17em}m}$for$\lambda$into equation (iii).

$\begin{array}{l}{\lambda }^{\text{'}}=2.43\times {10}^{-12}+3\times {10}^{-12}\\ {\lambda }^{\text{'}}=(2.43+3){10}^{-12}\\ {\lambda }^{\text{'}}=5.43\times {10}^{-12}\text{\hspace{0.17em}m}\end{array}$

Calculate the energy after the scattering,

Substitute $6.62\times {10}^{-34}{\text{\hspace{0.17em}m}}^2\text{kg}/\text{s}$for h , $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$for c and $5.43\times {10}^{-12}$for$\lambda$ into equation (i).

$\begin{array}{l}{E}^{\text{'}}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{5.43\times {10}^{-12}}\\ E^{\text{'}}=\frac{19.86\times {10}^{-26}}{5.43\times {10}^{-12}}\\ E^{\text{'}}=3.65\times {10}^{-14}\text{\hspace{0.17em}J}\end{array}$

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$3.65\times {10}^{-14}$ for $E^{\text{'}}$and$6.62\times {10}^{-14}\text{\hspace{0.17em}J}$ for E intoequation (iv).

$\begin{array}{l}K=6.62\times {10}^{-14}-3.65\times {10}^{-14}\\ K=(6.62-3.65)\times {10}^{-14}\\ K=2.97\times {10}^{-14}\text{\hspace{0.17em}J}\end{array}$

Hence the kinetic energy of the electron is $2.97\times {10}^{-14}\text{\hspace{0.17em}J}$.