A monochromatic light is incident on a metallic plate having work function $\phi$. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field perpendicular to the initial velocity of the electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is:
(Given: the magnitude of charge of an electron is $e$ and mass is $m$, $h$ is Planck's constant and $c$ is velocity of light. Take the magnetic field to exist throughout the path of the electron.)
Answer: (C) $\dfrac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$
Maximum kinetic energy: $K = \dfrac{hc}{\lambda} - \phi$, so the momentum is $p = \sqrt{2mK}$.
The field is perpendicular to the velocity, so the electron moves on a circle of radius $r = \dfrac{p}{eB}$. The velocity at A is along the normal to the plate, so the centre of the circle lies on the plate, and the electron returns to the plate after a semicircle. Hence
$$AB = 2r = \frac{2\sqrt{2mK}}{eB} = \frac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$$
Solution by Sreeraj P, M.Sc Physics