A photon of energy E ejects a photoelectron from a metal surface … - Vidyadip

MCQ

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\,\left( {E - {W_0}} \right)}}{{eB}}} $
B
$\sqrt {2m\,\left( {E - {W_0}} \right)eB} $
C
$\frac{{\sqrt {2e\,\left( {E - {W_0}} \right)} }}{{mB}}$
✓
$\frac{{\sqrt {2m\,\left( {E - {W_0}} \right)} }}{{eB}}$

✓

Answer

Correct option: D.

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

d
From Einstein equation $\mathrm{E}=\mathrm{W}_{0}+\frac{1}{2} \mathrm{mv}^{2}$

$\sqrt{\frac{2\left(\mathrm{E}-\mathrm{W}_{0}\right)}{\mathrm{m}}}=\mathrm{v}$

and a charged particle placed in uniform magnetic field experience a force

$\mathrm{F}=\frac{\mathrm{mv}^{2}}{\mathrm{r}} \Rightarrow \mathrm{evB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}} \Rightarrow \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{eB}}$

$\Rightarrow \mathrm{r}=\frac{\sqrt{2 \mathrm{m}\left(\mathrm{E}-\mathrm{W}_{0}\right)}}{\mathrm{eB}}$

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