Question

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:

• A. \( \sqrt{\frac{2m \left( \frac{hc}{\lambda} - \phi \right)}{eB}} \)
• B. \( \frac{m \left( \frac{hc}{\lambda} - \phi \right)}{eB} \)
• C. \( \sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)} \div eB \)
• D. \( 2 \frac{m \left( \frac{hc}{\lambda} - \phi \right)}{eB} \)

Hint:

In problems involving magnetic fields, the radius of the electron’s path is related to its momentum and the magnetic field. The formula for the distance is derived by equating the magnetic force to the centripetal force.

Answer 1

The Correct Option is C

The maximum kinetic energy \( K_E \) of the electron is given by: \[ K_E = \frac{hc}{\lambda} - \phi \] where \( p \) is the momentum of the electron, and the relation for momentum is: \[ p = \sqrt{2m K_E} = \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} \] Since the motion is in a magnetic field, the radius of the circular path is: \[ d_{A-B} = 2R = \frac{p}{qB} \] Thus, the distance between A and B becomes: \[ d_{A-B} = \frac{2}{eB} \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} = \frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB} \]