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

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 $evB=\frac{m{v}^2}{r}$
or $r=\frac{mv}{eB}$
or $r=\frac{\sqrt{\frac{2\left(E-W_0\right)}{m}}\times m}{eB}$
$r=\frac{\sqrt{2m\left(E-W_0\right)}}{eB}$