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A photon of energy E ejects a photoelectron from a metal surface whose work function is \mathrm{W_0} If an electron having maximum kinetic energy enters into a uniform magnetic field of induction \mathrm{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):

Option: 1

\frac{\sqrt{2 m\left(E-W_0\right)}}{e B}


Option: 2

\sqrt{2 \mathrm{~m}\left(\mathrm{E}-\mathrm{W}_0\right) \mathrm{eB}}


Option: 3

\frac{\sqrt{2 \mathrm{~m}\left(\mathrm{E}-\mathrm{W}_0\right)}}{\mathrm{mB}}


Option: 4

 None of these


Answers (1)

\begin{aligned} &\text{As }\frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{qvB} \Rightarrow \frac{\mathrm{mv}}{\mathrm{qB}}=\mathrm{r}\\ &\text{Also, }\mathrm{mv}=\mathrm{P}=\mathrm{P}=\sqrt{2 \mathrm{mk}}\\ &=\mathrm{r}=\frac{\sqrt{2 \mathrm{Km}}}{\mathrm{Bq}}=\frac{\sqrt{2\left(\mathrm{E}-\mathrm{W}_0\right) \mathrm{m}}}{\mathrm{eB}} \end{aligned}

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Sumit Saini

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