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. $\frac{\sqrt{2m\left(E-W_{0}\right)}}{mB}$
• B. $\sqrt{2m\left(E-W_{0}\right)eB}$
• C. $\frac{\sqrt{2e\left(E-W_{0}\right)}}{mB}$
• D. $\frac{\sqrt{2m\left(E-W_{0}\right)}}{eB}$

Answer 1

Answer: (D)

From Einstein's equation
$E=W_{0}+\frac{1}{2} m v^{2} $
$\sqrt{\frac{2\left(E-W_{0}\right)}{m}}=v$
or A charged particle placed in uniform magnetic field 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 $r =\frac{m \sqrt{\frac{2\left(E-W_{0}\right)}{m}}}{e B}$
$\Rightarrow r=\frac{\sqrt{2 m\left(E-W_{0}\right)}}{e B}$