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A photon of energy 4 eV is incident on a metal surface whose work function is 2 eV. The minimum reverse potential to be applied for stopping the emission of electrons is

$\begin {array} {1 1} (a)\;2\: V & \quad (b)\;4\: V \\ (c)\;6\: V & \quad (d)\;8\: V \end {array}$

 

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A)
Ans : (a)
$ \large\frac{1}{2} mv^2_{max} = hv - \phi_{\circ} = 4\: eV - 2 eV = 2 eV$
Now, $eV_{\circ} = 2 eV$
$ \Rightarrow V_{\circ} = 2V$

 

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