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 \( K_E = \frac{hc}{\lambda} - \phi \). The momentum \( p \) of the electron is related by \( p = \sqrt{2m K_E} = \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} \). In a magnetic field, the electron follows a circular path with radius \( R \). The distance between A and B, \( d_{A-B} \), is twice the radius, \( d_{A-B} = 2R \), and is given by \( d_{A-B} = \frac{p}{qB} \). Substituting the expression for momentum, we get \( d_{A-B} = \frac{2}{eB} \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} = \frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB} \).