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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 $\mathrm{r}$, then the radius $\mathrm{r}$ is given by, (in the usual notation)
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The correct answer is:
$\frac{\sqrt{2 \mathrm{~m}\left(\mathrm{E}-\mathrm{W}_{0}\right)}}{\mathrm{eB}}$
From Einstein's photoelectric equation
$E=W_{0}+\frac{1}{2} m v^{2} \quad \text { or } \quad \sqrt{\frac{2\left(E-W_{0}\right)}{m}}=v$
A charged particle placed in uniform magnetic field experience a force
$F=e v B=\frac{m v^{2}}{r}$
or $r=\frac{m v}{e B}$
or $r=\frac{m \sqrt{\frac{2\left(E-W_{0}\right)}{m}}}{e B} \Rightarrow r=\frac{\sqrt{2 m\left(E-W_{0}\right)}}{e B}$
$E=W_{0}+\frac{1}{2} m v^{2} \quad \text { or } \quad \sqrt{\frac{2\left(E-W_{0}\right)}{m}}=v$
A charged particle placed in uniform magnetic field experience a force
$F=e v B=\frac{m v^{2}}{r}$
or $r=\frac{m v}{e B}$
or $r=\frac{m \sqrt{\frac{2\left(E-W_{0}\right)}{m}}}{e B} \Rightarrow r=\frac{\sqrt{2 m\left(E-W_{0}\right)}}{e B}$
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