A monochromatic light is incident on a metallic plate having work function φ. 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 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 exists throughout the path of electron)

Solution

Physics sir · 4 steps

The electron traces half a circle before it lands again, so AB is the diameter, not the radius.

1. Photoelectric energy. Einstein's equation gives the maximum kinetic energy of the emitted electron: K = hc/λ - φ. Its momentum is then p = √2mK.
2. Circular path. The field is perpendicular to the velocity, so the magnetic force evB acts as the centripetal force mv²/r. This gives the radius of the circle.

   r = mv/eB = √2m(hc/λ-φ)/eB

3. Half a circle. The electron leaves the plate normally, so the plate is tangent to its circle at A. It curves round and meets the plate again at B, the point diametrically opposite A, after a semicircle. So AB = 2r.
4. So. AB = 2√2m(hc/λ-φ)/eB = √8m(hc/λ-φ)/eB, which is option (b).

The idea to keep. For a charge in a magnetic field, write r = √2mK/qB. If it starts perpendicular to a surface, it lands back one diameter, 2r, away.

Answer from the NTA final answer key. Solution by PrepWiser.