The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in
Answer & explanation
Correct answer: option 3
$\text{Work function of the material is }\phi = 6.2eV$
$ \text{ Stopping Potential } V_s = 5V$
$ \text{ Energy of the incident radiation is } = \phi + eV_s = 11.2eV$
$ \lambda = \frac{12400}{E} = \frac{12400A^o}{11.2} = 1107A^o$
It lies in the ultra-violet region.