Answer
$9768$ $V$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$, ......................$(1)$
where $h$ is the planck's constant and $p$ is the momentum of the moving particle.
For gamma rays, the relation between momentum $p$ and energy $E$ is given by
$E=pc$
Thus, $p=\frac{E}{c}$ ......................$(2)$
Substituting eq. $2$ in eq. $1$, we get
$λ=\frac{hc}{E}$ ......................$(3)$
For electron, the relation between momentum $p$ and energy $E$ is given by
$E=\frac{p^2}{2m}$
or, $p=\sqrt {2mE}$ ......................$(4)$
where $m$ is the mass of the electron.
Substituting eq. $4$ in eq. $1$, we get
$λ=\frac{h}{\sqrt {2mE}}$ ......................$(5)$
If the electron is accelerated with voltage $V$, then $E=eV$, where $e$ is the charge of an electron.
Thus eq. $5$ becomes
$λ=\frac{h}{\sqrt {2meV}}$ ......................$(6)$
As the resolving power of two microscopes are same,
$\frac{h}{\sqrt {2meV}}=\frac{hc}{E}$
or, $V=\frac{E^2}{2emc^2}$
or, $V=\frac{(100\times 10^{3}\times 1.6\times 10^{-19})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (3\times 10^{8})^2}$ $V$
or, $V=9768$ $V$
$\therefore$ The required accelerating voltage is $9768$ $V$.
