Question

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5\times {10}^{-12}\text{m}$, the minimum electron energy required is close to

• A. $1\text{keV}$
• B. $25\text{keV}$
• C. $500\text{keV}$
• D. $100\text{keV}$

Answer 1

The correct option is B $25\text{keV}$

Given:
$\lambda =7.5\times {10}^{-12}$
We know that,
$\lambda =\frac{h}{P}\Rightarrow P=\frac{h}{\lambda }$

Minimum energy required,
$KE=\frac{P^2}{2m}=\frac{(h/\lambda {)}^2}{2m}=\frac{{\left\{\frac{6.6\times {10}^{-34}}{7.5\times {10}^{-12}}\right\}}^2}{2\times 9.1\times {10}^{-31}}$

$KE\approx 25\text{keV}$

Hence, option $\left(D\right)$ is correct.