Question

A photon of energy \(E\) ejects a photoelectron from a metal surface whose work function is \(W_0\). If this electron enters into a uniform magnetic field of induction \(B\) in a direction perpendicular to the field and describes a circular path of radius \(r\), then the radius \(r\) is given by (in the usual notation)

• A. \(\dfrac{\sqrt{2m(E-W_0)}}{eB}\)
• B. \(\dfrac{\sqrt{2m(E-W_0)eB}}{mB}\)
• C. \(\dfrac{\sqrt{2e(E-W_0)}}{mB}\)
• D. \(\dfrac{\sqrt{2m(E-W_0)}}{eB}\)

Hint:

Use \(r=\frac{mv}{eB}\) and \(v=\sqrt{\frac{2K}{m}}\). Here \(K=E-W_0\).

Answer 1

The Correct Option is D

Step 1: Photoelectron kinetic energy.
\[ K = E - W_0 \]
Step 2: Relate kinetic energy with velocity.
\[ K = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2(E-W_0)}{m}} \]
Step 3: Radius of circular motion in magnetic field.
For particle moving perpendicular to magnetic field:
\[ r = \frac{mv}{eB} \]
Step 4: Substitute \(v\).
\[ r = \frac{m}{eB}\sqrt{\frac{2(E-W_0)}{m}} = \frac{\sqrt{2m(E-W_0)}}{eB} \]
Final Answer:
\[ \boxed{\dfrac{\sqrt{2m(E-W_0)}}{eB}} \]