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A metal surface having work function ' $\mathrm{w}_{0}{ }^{\prime}$ emits photoelectrons when photons of energy 'E' are incident on it. The electron enters the uniform magnetic field (B) in perpendicular direction and moves in circular path of radius 'r'. Then 'r' is equal to (m and e be the mass and charge of electron respectively).
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The correct answer is:
$\frac{\sqrt{2 m\left(E-W_{0}\right)}}{\mathbf{e B}}$
Kinetic energy of the electron $\mathrm{K}=\mathrm{E}-\omega_{0}$
Momentum $P=\sqrt{2 m k}=\sqrt{2 m\left(E-\omega_{\mathrm{o}}\right)}$
Radius of the circular path $r=\frac{P}{e B}=\frac{\sqrt{2 m\left(E-\omega_{0}\right)}}{e B}$
Momentum $P=\sqrt{2 m k}=\sqrt{2 m\left(E-\omega_{\mathrm{o}}\right)}$
Radius of the circular path $r=\frac{P}{e B}=\frac{\sqrt{2 m\left(E-\omega_{0}\right)}}{e B}$
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