When a radiation is incident on a metal surface of work function \( (\phi_0) \), photoelectrons 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 photoelectrons will become
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2k + \phi_0 \)
By Einstein's photoelectric equation, the photon energy is shared between the work function and the maximum kinetic energy.
\( \displaystyle h\nu = \phi_0 + k \)
When the frequency is doubled, the photon energy doubles but the work function stays the same.
\( \displaystyle k' = 2h\nu - \phi_0 = 2(\phi_0 + k) - \phi_0 = 2k + \phi_0 \)