A given photosensitive metal has a work function equal to 0.75 J. When a radiation of frequency \( \nu \) is incident on it, the maximum kinetic energy of the emitted photoelectrons is 30 J. If the frequency of incident radiation is doubled, then the maximum kinetic energy of the emitted photoelectron would be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 60.75 J
By Einstein's photoelectric equation, the photon energy goes partly into freeing the electron (the work function \( \phi_0 \)) and the rest appears as the maximum kinetic energy.
\( \displaystyle K_{\max} = h\nu - \phi_0 \)
For the first radiation:
\( \displaystyle h\nu = K_{\max} + \phi_0 = 30 + 0.75 = 30.75\,\mathrm{J} \)
Doubling the frequency doubles the photon energy, but the work function stays the same:
\( \displaystyle K'_{\max} = 2h\nu - \phi_0 = 2 \times 30.75 - 0.75 = 61.5 - 0.75 = 60.75\,\mathrm{J} \)