Question

A photoelectron ejected from a metal surface of work function $\text{W}_{0}$ by a photon of energy $E$ enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$ which is given by (Symbols have the usual meaning)

• A. $\sqrt{\frac{2 m \left( E + W _{0}\right)}{ eB }}$
• B. $\sqrt{2 m \left( E - W _{0}\right) eB }$
• C. $\sqrt{\frac{2 m \left( E - W _{0}\right)}{ mB }}$
• D. $\frac{\sqrt{2 m \left( E - W _{0}\right)}}{ eB }$

Answer 1

Answer: (D)

From Einstein's equation
$\text{E} = \text{W}_{0} + \frac{1}{2} \text{m} v ^{2}$
$\sqrt{\frac{2\left( E - W _{0}\right)}{ m }}= v$
and a charged particle placed in uniform magnetic field will experience a force,
$F = \frac{m v^{2}}{r}$
or $e v B = \frac{m v^{2}}{r}$
or $r = \frac{m v}{e B}$
or $r =\frac{\sqrt{\frac{2\left( E - W _{0}\right)}{ m }} \times m }{ eB }$
$ r=\frac{\sqrt{2 m \left( E - W _{0}\right)}}{ eB } $