An electron and a photon each have a wavelength of $\mathbf{0}.\mathbf{20}\mathbf{nm}$. What is the momentum (in $\raisebox{1ex}{${\displaystyle \text{kg}\cdot \text{m}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$) of the (a) electron and (b) photon? What is the energy (in $\mathbf{\text{eV}}$) of the (c) electron and (d) photon?

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

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

$p=\frac{h}{\lambda }$

Here, $\lambda$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 $9.11\times {10}^{-31}\text{kg}$.

Kinetic energy of photon is,

$K=pc$

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

The momentum of electron is,

$p=\frac{h}{\lambda }$

The wavelength is given as,

$\lambda =0.20\mathrm{nm}=0.2\times {10}^{-9}\text{m}$

So momentum is,

$\begin{array}{rcl}p& =& \frac{6.63\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{0.2\times {10}^{-9}\text{\hspace{0.17em}m}}\\ & =& 3.3\times {10}^{-24}\raisebox{1ex}{$\text{kg}\cdot \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 $3.3\times {10}^{-24}\raisebox{1ex}{$\text{kg}\cdot \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" $\begin{array}{rcl}{K}_e& =& \frac{{{\displaystyle \left(3.3\times {10}^{-24}\raisebox{1ex}{$\text{kg}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)}}^2}{2(9.11\times {10}^{-31}\text{\hspace{0.17em}kg})}\\ & =& 6\times {10}^{-18}\text{J}\\ & =& 38\text{eV}\end{array}$

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

The kinetic energy of photon is

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

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