The smallest dimensions (resolving power) that can be resolved by an electron microscope is equal to the de Broglie wavelength of its electrons. What accelerating voltage would be required for the electrons to have the same resolving power as could be obtained using $\mathbf{\text{100keV}}$ gamma rays?

Momentum, the product of a particle's mass and its velocity. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle. See Newton's laws of motion.

Consider the given data as below.

The smallest dimensions (resolving power) that can be resolved by an electron microscope is equal to the de Broglie wavelength of its electrons.

The energy, is

The momentum is defined by,

$p=\frac{E}{c}$

Here, $E$ is energy and $c$is speed of light having a value $3\times {10}^8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

Write the equation for kinetic energy of electron as below.

$K=\frac{p^2}{2m}$

Here, $m$is mass of electron having a value$\begin{array}{rcl}& & 9.11\times {10}^{-31}\text{\hspace{0.17em}kg}\end{array}$.

The accelerating potential is,

$V=\frac{K}{e}$

Here, $K$ is kinetic energy and $e$ is charge of electron $1.6\times {10}^{-19}\text{C}$.

The momentum of $100\mathrm{keV}$photon is,

$\begin{array}{rcl}p& =& \frac{E}{c}\\ & =& \frac{(100\times {10}^3{\displaystyle \text{eV}})(1.6\times {10}^{-19}{\displaystyle \raisebox{1ex}{${\displaystyle \text{J}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{eV}}$}\right.})}{(3\times {10}^8\text{\hspace{0.17em}m}/\text{s})}\\ & =& 5.33\times {10}^{-23}\raisebox{1ex}{${\displaystyle \text{kg}\cdot \text{m}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

This is momentum of electron.

The kinetic energy of electron is,

$\begin{array}{rcl}K& =& \frac{p^2}{2m}\\ & =& \frac{{(5.33\times {10}^{-23}{\displaystyle \raisebox{1ex}{$\text{kg}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.})}^2}{2(9.11\times {10}^{-31}\text{\hspace{0.17em}kg})}\\ & =& 1.56\times {10}^{-15}\text{\hspace{0.17em}J}\end{array}$

The accelerating potential is,

Hence, the accelerating potential is $\begin{array}{rcl}& & 9.76\times {10}^3\text{\hspace{0.17em}V}\end{array}$.