Question

Figure shows the graph of photocurrent versus potential of the anode with respect to the cathode. The intensity of the incident radiation is halved and frequency doubled. The stopping potential and the saturation current will now be respectively.

• A. $-6.0\text{V}$ and $2.0\mu \text{A}$
• B. $-6.0\text{V}$ and $4.0\mu \text{A}$
• C. $-6.0\text{V}$ and $1.0\mu \text{A}$
• D. None of these

Answer 1

The correct option is D None of these

Since intensity is increased the saturation current will also increase , so the new saturation current will be more than $4\mu A$
$v_s=\frac{hf}{e}-\frac{\varphi }{e}$
$3=\frac{hf}{e}-\frac{\varphi }{e}$
$v^{\prime }=\frac{2hf}{e}-\frac{\varphi }{e}=2\times 3+\frac{\varphi }{e}=6+\frac{\varphi }{e}>6$
The magnitude of stopping potential will be more than $6\text{V}$