Ultraviolet radiations of 6.2 eV are incident on an aluminum surface of work function 4.2 eV. The maximum kinetic energy of the emitted electrons is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2.0 eV
By Einstein's photoelectric equation, the photon energy is used first to overcome the work function, and the rest appears as the maximum kinetic energy of the electron.
\( \displaystyle K_{\max} = h\nu - \phi_0 \)
\( \displaystyle K_{\max} = 6.2\,\mathrm{eV} - 4.2\,\mathrm{eV} = 2.0\,\mathrm{eV} \)