Key idea: dependence of photocurrent on wavelength

For a photocell, when the anode voltage is kept fixed, the plate current depends on how many photoelectrons are emitted per second and collected by the anode.

The photoelectric current is mainly proportional to the number of incident photons per second (provided the frequency is above threshold and the applied anode voltage is sufficient to collect emitted electrons).

What happens when wavelength changes?

The energy of one photon is $E = h\nu = \frac{hc}{\lambda}$

So as $\lambda$ increases, photon energy decreases.

• If $\lambda < \lambda_0$ (where $\lambda_0$ is the threshold wavelength), photoemission occurs.
• If $\lambda > \lambda_0$, photon energy is less than work function, so no photoelectrons are emitted and current becomes zero.

Behavior of photocurrent before threshold

In such standard photocell questions, intensity is taken fixed. Then the number of photons incident per second is $N \propto \frac{\text{intensity}}{E} \propto \frac{1}{hc/\lambda} \propto \lambda$

Hence, for fixed intensity and fixed anode voltage, the photocurrent $I \propto N \propto \lambda$ as long as photoemission is possible.

Therefore:

• For $\lambda < \lambda_0$, current increases linearly with $\lambda$.
• At $\lambda = \lambda_0$, emission just stops.
• For $\lambda > \lambda_0$, current is zero.

Required graph

So the graph of $I$ versus $\lambda$ is:

• a rising straight line up to threshold wavelength,
• then suddenly drops to zero and remains zero beyond threshold.

Matching with options

This corresponds to Option B.

Comparison with stored answer

Derived answer: B
Stored correct answer: B

Hence, they agree.