The threshold frequency for a metallic surface corresponds to an energy of $6.2 \mathrm{eV}$, and the…

The threshold frequency for a metallic surface corresponds to an energy of $6.2 \mathrm{eV}$, and the stopping potential for a radiation incident on this surface $5 \mathrm{~V}$. The incident radiation lies in
  1. X-ray region
  2. ultra-violet region
  3. infra-red region
  4. visible region

Solution

$\lambda=\frac{1242 \mathrm{e} V \mathrm{~nm}}{11.2} \approx 1100 Å$ Ultraviolet region

Asked in: JEE Main 2006

Practice more Dual Nature of Matter and Radiation questions on Aicharya