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  • A photon of energy E ejects a photoelectron from a metal surface whose work function is . If this electron enters

A photon of energy E ejects a photoelectron from a metal surface whose work function is \phi_0. If this electron enters into a uniform magnetic field 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)
 

Option: 1

\sqrt{\frac{2 m\left(E-\phi_0\right)}{e B}}


Option: 2

\sqrt{2 \mathrm{~m}\left(\mathrm{E}-\phi_0\right) \mathrm{eB}}


Option: 3

\sqrt{\frac{2 \mathrm{e}\left(\mathrm{E}-\phi_0\right)}{\mathrm{mB}}}


Option: 4

\sqrt{\frac{2 \mathrm{~m}\left(\mathrm{E}-\phi_0\right)}{\mathrm{eB}}}


Answers (1)

best_answer

As the electron describes a circular path of radius r in the magnetic field, therefore,
\frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{evB} ; \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{eB}}=\frac{\mathrm{p}}{\mathrm{eB}}=\frac{\sqrt{2 \mathrm{mK}}}{\mathrm{eB}} \quad \quad \quad \quad (As \mathrm{~K}=\frac{\mathrm{p}^2}{2 \mathrm{~m}} )

From Einstein's photoelectric equation

\begin{aligned} & K=E-\phi_0 \\ & \therefore r=\frac{\sqrt{2 m\left(E-\phi_0\right)}}{e B} \end{aligned}

Posted by

Rishabh

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