When a light beam of frequency $v$ is incident on a metal, photo electrons are emitted. If these electrons describe a circle of radius $r$ in a magnetic field of strength $B$ , then the work function of the metal is:

h v + \frac{B e r}{2 m_{e}}

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h v - \frac{{(B e r)}^{2}}{2 m_{e}}

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h v + \frac{{(B e r)}^{2}}{2 m_{e}}

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h v - \frac{B e r}{2 m_{e}}

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Solution

The correct option is D $h v - \frac{{(B e r)}^{2}}{2 m_{e}}$
For electrons describing a circle of radius $r$ in a magnetic field of strength $B$ ,
$e v B = \frac{m_{e} v^{2}}{r}$
$⟹ v = \frac{e B r}{m_{e}}$
Thus kinetic energy of the electrons= $\frac{1}{2} m_{e} v^{2} = \frac{(B e r)^{2}}{2 m_{e}}$
We know that in photoelectric effect,
$h ν = W_{0} + K E$
$⟹ W_{0} = h ν - \frac{(B e r)^{2}}{2 m_{e}}$

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When a light beam of frequency v is incident on a metal, photo electrons are emitted. If these electrons describe a circle of radius r in a magnetic field of strength B, then the work function of the metal is:

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When a light beam of frequency $v$ is incident on a metal, photo electrons are emitted. If these electrons describe a circle of radius $r$ in a magnetic field of strength $B$ , then the work function of the metal is: