When a radiation is incident on a metal surface of work function ($\phi_0$), photo electrons are emitted with some maximum kinetic energy ($k$). If the frequency of the incident radiation is doubled, then the maximum kinetic energy of the emitted photo electrons will become

$2k + \phi_0$

The correct answer is Option (3) → $2k + \phi_0$

From Einstein’s photoelectric equation:

$k = h\nu - \phi_0$

If the frequency is doubled, new kinetic energy becomes:

$k' = h(2\nu) - \phi_0 = 2h\nu - \phi_0$

Subtracting the first equation from the second:

$k' - k = (2h\nu - \phi_0) - (h\nu - \phi_0) = h\nu$

∴ $k' = k + h\nu$

$= 2k + \phi_0$

Hence, the maximum kinetic energy becomes $k + h\nu$ when the frequency of radiation is doubled.