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)

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

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

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\frac{\sqrt{2 m (E - W_{0})}}{m B}

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

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Solution

The correct option is C $\frac{\sqrt{2 m (E - W_{0})}}{e B}$
From Einstein's equation,
energy of incident photon
$E = W_{0} + \frac{1}{2} m v^{2}$
We get, velocity of photoelectron
$v = \sqrt{\frac{2 (E - W_{0})}{m}}$
A charged particle placed in uniform magnetic field experience a magnetic force which makes the charge to trace a circular path.
Centripetal force $F = \frac{m v^{2}}{r}$

Magnetic force $F_{m} = e v B$

or $e v B = \frac{m v^{2}}{r}$

or $r = \frac{m v}{e B}$

or $r = \frac{m \sqrt{\frac{2 (E - W_{0})}{m}}}{e B}$

$⟹ r = \frac{\sqrt{2 m (E - W_{0})}}{e B}$

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