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

2.0 eV

The correct answer is Option (3) → 2.0 eV

Given:

Energy of incident photons $E = 6.2\ \text{eV}$

Work function of aluminum $\phi = 4.2\ \text{eV}$

Maximum kinetic energy of emitted electrons:

$K_{\text{max}} = E - \phi$

$K_{\text{max}} = 6.2 - 4.2 = 2.0\ \text{eV}$

Answer: $2.0\ \text{eV}$