Question

In a photoelectric experiment, the relation between applied potential difference between cathode and anode $\left(V\right)$ and the photoelectric current $\left(I\right)$ 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. $0.775\times {10}^{16}$

Answer 1

The correct option is D $0.775\times {10}^{16}$

The kinetic energy is given as

$K=e{V}_0$

$K=1.6\times {10}^{-19}\times \left(-3.2\right)$

$K=5.12\times {10}^{-19}$

By substituting the above value in

$5.12\times {10}^{-19}=6.6\times {10}^{34}\times f+3.2$

The frequency is $0.29\times 1^{15}$