Question

A metal surface having work function \( w_0 \) emits photoelectrons when photons of energy \( E \) are incident on it. The electron enters a uniform magnetic field \( B \) in perpendicular direction and moves in a circular path of radius \( r \). Then \( r \) is equal to ( \( m \) and \( e \) are the mass and charge of electron respectively)

• A. \( \dfrac{\sqrt{m(E-w_0)}}{eB} \)
• B. \( \dfrac{m(E-w_0)}{eB} \)
• C. \( \dfrac{\sqrt{2m(E-w_0)}}{eB} \)
• D. \( \dfrac{2m(E-w_0)}{eB} \)

Hint:

Always combine photoelectric equation with magnetic force relation for such problems.

Answer 1

The Correct Option is C

Step 1: Write Einstein’s photoelectric equation.
The maximum kinetic energy of emitted photoelectrons is \[ K_{\max} = E - w_0 \]
Step 2: Express kinetic energy in terms of velocity.
\[ \frac{1}{2}mv^2 = E - w_0 \Rightarrow v = \sqrt{\frac{2(E-w_0)}{m}} \]
Step 3: Use magnetic force as centripetal force.
\[ e v B = \frac{mv^2}{r} \]
Step 4: Solve for radius \( r \).
\[ r = \frac{mv}{eB} = \frac{m}{eB}\sqrt{\frac{2(E-w_0)}{m}} = \frac{\sqrt{2m(E-w_0)}}{eB} \]
Step 5: Conclusion.
The radius of circular path is \( \dfrac{\sqrt{2m(E-w_0)}}{eB} \).