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. $\sqrt{\frac{2m(E-W_0)}{eB}}$
• B. $\sqrt{2m(E-W_0)eB}$
• C. $\frac{\sqrt{2m(E-W_0)}}{mB}$
• D. $\frac{\sqrt{2m(E-W_0)}}{eB}$

Answer 1

Answer: (D)

$\frac{\sqrt{2m(E-W_0)}}{eB}$

From Einstein's equation,
energy of incident photon
$E=W_0+\frac{1}{2}m{v}^2$

We get, velocity of photoelectron

$v=\sqrt{\frac{2(E-W_0)}{m}}$
A charged particle placed in uniform magnetic field experience a magnetic force which makes the charge to trace a circular path.

Centripetal force $F=\frac{m{v}^2}{r}$

Magnetic force $F_m=evB$

or $evB=\frac{m{v}^2}{r}$

or $r=\frac{mv}{eB}$

or $r=\frac{m\sqrt{\frac{2(E-W_0)}{m}}}{eB}$

$⟹r=\frac{\sqrt{2m(E-W_0)}}{eB}$