Question

If $I_1,I_2\text{and}{I}_3$ are the intensities of three radiations incident on the same metal surface with the same frequency such that $(I_1>I_3>I_2)$ and following graph of variation of photo current with respect to anode potential is observed. Find which curve represents the three intensities.

• A. $A-I_1,B-I_2,C-I_3$
• B. $A-I_1,B-I_3,C-I_2$
• C. $A-I_3,B-I_2,C-I_1$
• D. $A-I_2,B-I_3,C-I_1$

Answer 1

The correct option is B $A-I_1,B-I_3,C-I_2$

Since the frequency of all three incident radiation is same, stopping potential (that is the $-ve$ anode potential for zero photo current) is also same.
As intensity increases, photo current increases

$\therefore (A-I_1)>(B\to I_3)>(C\to I_2)$
$\therefore$ Option $\left(B\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question?}\\ \text{This is very typical question testing the concept of more the intensity more is the photo current.}\end{array}$