Q. 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) Ultra-violet region
B) Infra-red region
C) Visible region
D) X-ray region
Explanation
(a) $hv_{0} = 6.2\ eV,\ \ eV_{0} = 5\ eV$
from Einstein’s photoelectric equation
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ hv = hv_{0} + eV_{0} = \ 6.2 + 5\ = \ 11.2\ eV$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{hc}{\lambda} = 11.2$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{hc}{11.2} = 1108.9\mathring{\mathrm{A}}$
Which belongs to ultra-violet region.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.