Question

For given graph (showing variation of photo current with respect to anode potential)


If $f_1,f_2\text{and}{f}_3$ are the frequencies of incident radiations. Then, which of the following is true.

• A. $f_1=f_2=f_3$
• B. $f_1>f_2>f_3$
• C. $f_1<f_2<f_3$
• D. $f_1=f_2<f_3$

Answer 1

The correct option is B $f_1>f_2>f_3$

Here, the saturation current is same, thus, indicating same intensity of all three radiations.
But, $-ve$ anode potential (stopping potential is different).
As we know,
$1.$ More the frequency of incident radiation, more is the energy $\because E=hf$
$2.$ More the energy, more is the maximum possible kinetic energy, also, more the $-ve$ anode potential i.e. stopping potential.
So, here $-V_1$ is the largest $-ve$ potential, followed by $-V_2$ and $-V_3.$
$\therefore f_1>f_2>f_3$
Hence, option $\left(B\right)$ is correct.

$\begin{array}{c}\text{Why this question?}\\ \text{This question tests understanding of dependency of stopping potential on energy \& hence frequency of the incident radiation.}\end{array}$