Q. 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 $7.5 \times 10^{-12}m$, the minimum electron energy required is close to :


$\lambda = \frac{h}{p} \left\{\lambda = 7.5 \times10^{-12}\right\}$
$ P = \frac{h}{\lambda} $
$ KE = \frac{P^{2}}{2m} = \frac{\left(h/\lambda\right)^{2}}{2m} = \frac{\left\{\frac{6.6 \times10^{-34}}{7.5 \times10^{-12}}\right\}}{2 \times9.1 \times10^{-31}} $
KE = 25 Kev

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions