Q.
In a photoelectric experiment the relation between applied potential difference between cathode and anode V and the photoelectric current I was found to be shown in graph below. If Planck’s constant $h = 6.6 \times 10^{- 34}Js$, the frequency of incident radiation would be nearly $(in\ s^{- 1})$
A) $0.436 \times 10^{18}$
B) $0.436 \times 10^{17}$
C) $0.775 \times 10^{15}$
D) $0.775 \times 10^{16}$
Explanation
(c) For photoelectric effect,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ eV_{0} = hv \Rightarrow v = \frac{eV_{0}}{h}$
$v = \frac{1.6 \times 10^{- 19} \times 3.2}{6.6 \times 10^{- 34}}$
$= 0.775 \times 10^{15}Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.