Question

Show that when a photon of energy Eis scattered from a free electron at rest, the maximum kinetic energy of the recoiling electron is given by

$\mathit{E}\mathbf{=}\frac{{\mathbf{E}}^{\mathbf{2}}}{\mathbf{E}\mathbf{+}\mathbf{m}{\mathbf{c}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}$

Short answer

The required equation$\frac{E^2}{E+m{c}^2/2}$ for the kinetic energy of the electron is showed.

Step-by-step solution

Write the given data from the question.

The energy of the photon is $E$.

The mass of the energy is $m$.

Determine the formulas to show the maximum kinetic energy of the of recoiling electron is E2/(E+mc2/2).

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

$E=hv$ …(i)

Here,h is the plank’s constant and vis the frequency.

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 frequency is given as follows.

$v=\frac{\lambda }{c}$ …(iii)

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

$K=hv-h{v}^{\text{'}}$ …(iv)

Show the maximum kinetic energy of the of recoiling electron is E2/(E+mc2/2).

The wavelength when the photon get free from the electron is given as,

${\lambda }^{\text{'}}=\lambda +\Delta \lambda$

Substitute $c/v^{\text{'}}$for${\lambda }^{\text{'}}$, $c/v$for$\lambda$and $h/mc(1-\cos \theta )$for $\Delta \lambda$into above equation.

role="math" localid="1663144849902" $\begin{array}{l}\frac{c}{v^{\text{'}}}=\frac{c}{v}+\frac{h}{mc}\left(1-\cos \theta \right)\\ \frac{1}{v^{\text{'}}}=\frac{1}{v}+\frac{h}{m{c}^2}(1-\cos \theta )\\ \frac{1}{v^{\text{'}}}=\frac{1}{v}[1+\frac{hv}{m{c}^2}(1-\cos \theta )]\\ v^{\text{'}}=\frac{v}{1+\frac{hv}{m{c}^2}(1-\cos \theta )}\end{array}$

Calculate the maximum kinetic energy of the electron.

Substitute$\frac{v}{1+\frac{hv}{m{c}^2}(1-\cos \theta )}$for v into equation (iv).

$\begin{array}{l}K=hv-h\frac{v}{1+\frac{hv}{m{c}^2}(1-\cos \theta )}\\ K=hv\left(1-\frac{1}{1+\frac{hv}{m{c}^2}(1-\cos \theta )}\right)\\ K=hv\left(\frac{1+\frac{hv}{m{c}^2}(1-\cos \theta )-1}{1+\frac{hv}{m{c}^2}(1-\cos \theta )}\right)\\ K=hv\left(\frac{\frac{hv}{m{c}^2}(1-\cos \theta )}{1+\frac{hv}{m{c}^2}(1-\cos \theta )}\right)\end{array}$

Substitute $180°$for$\theta$ into above equation.

$\begin{array}{l}K=hv\left(\frac{\frac{hv}{m{c}^2}(1-\cos 180)}{1+\frac{hv}{m{c}^2}(1-\cos 180)}\right)\\ K=hv\left(\frac{\frac{2hv}{m{c}^2}}{1+\frac{2hv}{m{c}^2}}\right)\\ K=hv\left(\frac{2hv}{m{c}^2+2hv}\right)\\ K=\left(\frac{{\left(hv\right)}^2}{\frac{m{c}^2}{2}+hv}\right)\end{array}$

Substitute E for $hv$into above equation.

$\begin{array}{l}K=\frac{E^2}{\frac{m{c}^2}{2}+E}\\ K=\frac{E^2}{E+m{c}^2/2}\end{array}$

Hence the required equation $\frac{E^2}{E+m{c}^2/2}$for the kinetic energy of the electron is showed.