Q 12-11-157JEE MainJEE Main 2019 (10 Jan, Shift 1)Medium
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 $7.5\times10^{-12}\ \text{m}$, the minimum electron energy required is close to:
Answer: (B) $25\ \text{keV}$
We need $\lambda = 7.5\times10^{-12}\ \text{m}$, so $p = \dfrac h\lambda = \dfrac{6.63\times10^{-34}}{7.5\times10^{-12}} = 8.84\times10^{-23}\ \text{kg m/s}$.
$$K = \frac{p^2}{2m} = \frac{(8.84\times10^{-23})^2}{2\times9.1\times10^{-31}} = 4.3\times10^{-15}\ \text{J} \approx 27\ \text{keV}$$
This is closest to $25\ \text{keV}$.
Solution by Sreeraj P, M.Sc Physics