The voltage applied to an electron microscope to produce electrons of wavelength $0.50 Å$ is
The voltage applied to an electron microscope to produce electrons of wavelength $0.50 Å$ is
$602 \mathrm{~V}$
$50 \mathrm{~V}$
$138 \mathrm{~V}$
$812 \mathrm{~V}$
Solution
Given, wavelength of electron, $\lambda=0.5 Å$
By using de-Broglie wavelength,
$
\lambda=\frac{h}{\sqrt{2 m e V}}
$
where, $h$ is Planck's constant $=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$, $m$ is mass of electron,
$e$ is charge of electron
and $V$ is applied voltage.
$
\begin{aligned}
\therefore \quad V & =\frac{h^2}{2 m e \lambda^2} \\
& =\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(0.5 \times 10^{-10}\right)^2} \\
& =602 \mathrm{~V}
\end{aligned}
$