JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Modern Physics

The resolving power of the microscope is set by the de Broglie wavelength of the electrons, so I want the accelerating voltage that squeezes the electron wavelength down to $7.5\times 10^{-12}$ m. The handy non-relativistic shortcut is $\lambda =\sqrt{150/V}$ angstrom with $V$ in volts, and since $7.5\times 10^{-12}$ m is $7.5\times 10^{-2}$ angstrom I square to get $V=150/(7.5\times 10^{-2}{)}^2=150/(5.625\times 10^{-3})\approx 2.7\times 10^4$ V. The kinetic energy an electron picks up across that potential is therefore about $27$ keV, and at these speeds the electron is already mildly relativistic, which trims the honest value a little so that the nearest listed energy is $25$ keV.