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
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: $2.0\ \text{eV}$