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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 ‘r’ then the radius ‘r’ is given by ( in the usual notion)

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$\sqrt{\frac{2 m (E - w_{0})}{e B}}$

$\frac{\sqrt{2 m (E - w_{o})}}{e B}$

$\sqrt{2 m (E - w_{0})}$

$\sqrt{2 e (E - w_{0}) m B}$

answer is A.

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Detailed Solution

$\begin{array}{l} E - w_{0} = \frac{1}{2} m v^{2}; v = \sqrt{\frac{2 (E - w_{0})}{m}}; r = \frac{m v}{q B} \\ \Rightarrow r = \frac{m \sqrt{\frac{2 (E - w_{0})}{m}}}{e B} = \frac{\sqrt{2 m (E - w_{0})}}{e B} \end{array}$

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