Question

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 5V. The incident radiation lies in :

• A. X-ray region
• B. Ultra-violet region
• C. Infra-red region
• D. Visible region

Answer 1

The correct option is B Ultra-violet region

Given,

Work function, $\varphi =6.2eV$

Stopping potential = S

i.e. $5eV=h\nu -6.2$

$h\nu =11.2eV$

$\therefore \frac{hc}{\lambda }=11.2eV$

$\frac{1240}{11.2}nm=\lambda$

$\lambda =110.7nm$

i.e. this lies in UV light region.

So, the answer is option (B).