91Ó°ÊÓ

An electron and a photon each have a wavelength of$\mathrm{λ}=0.20\mathrm{nm}$.

The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength.

The momentum of electron is given by

$p=\frac{h}{\mathrm{λ}}$

Here, $\mathrm{λ}$is wavelength and$h$is Plank’s constant having a value .

The kinetic energy of electron is,

$K_e=\frac{p^2}{2m_e}$

Here, $p$is momentum and $m_e$is mass of electron having a value $9.11×{10}^{−31}\text{kg}$.

Kinetic energy of photon is,

$K=pc$

Here, $p$is momentum and $c$is speed of light having a value role="math" localid="1663143010431" $\begin{array}{rcl}& & 2.998\end{array}\begin{array}{rcl}& & ×\end{array}\begin{array}{rcl}& & 10\end{array}\begin{array}{rcl}& & 8\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$.

The momentum of electron is,

$p=\frac{h}{\mathrm{λ}}$

The wavelength is given as,

$\mathrm{λ}=0.20\mathrm{nm}=0.2×{10}^{−9}\text{m}$

So momentum is,

$\begin{array}{rcl}p& =& \frac{6.63×{10}^{−34}\text{ J}â‹\dots \text{s}}{0.2×{10}^{−9}\text{″¾}}\\ & =& 3.3×{10}^{−24}\raisebox{1ex}{$\text{kg}â‹\dots \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

The momentum of photon is same as the momentum of electron.

Hence, the momentum of photon is $3.3×{10}^{−24}\raisebox{1ex}{$\text{kg}â‹\dots \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

The energy of electron is.

$K_e=\frac{p^2}{2m_e}$

Substitute the known values in the above equation.

role="math" localid="1663142990892" $\begin{array}{rcl}{K}_e& =& \frac{{{\displaystyle \left(3.3×{10}^{−24}\raisebox{1ex}{$\text{kg}â‹\dots \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)}}^2}{2(9.11×{10}^{−31}\text{ k²µ})}\\ & =& 6×{10}^{−18}\text{J}\\ & =& 38\text{eV}\end{array}$

Therefore, the energy of electron is $\begin{array}{rcl}& & 38\text{eV}\end{array}$.

The kinetic energy of photon is

$\begin{array}{rcl}K& =& pc\\ & =& \left(3.3×{10}^{−24}\raisebox{1ex}{$\text{kg}â‹\dots \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)\left(2.998×{10}^8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)\\ & =& 9.9×{10}^{−16}\text{J}\\ & =& 6.2\text{keV}\end{array}$

Hence, the energy of photon is $\begin{array}{rcl}& & 6.2\text{keV}\end{array}$.