Read the following passage and answer the questions.
The figure shows the set-up a student uses to study the photoelectric effect. An evacuated glass tube holds a photosensitive plate C (the emitter) and a metal plate A (the collector). Light from a source S enters through a quartz window and falls on plate C. A battery keeps a potential difference between C and A, and she can change its value. A commutator reverses the polarity, so plate A can be made positive or negative with respect to C. A voltmeter reads this potential difference, and a microammeter reads the photocurrent.
She chooses the frequency of the light by placing coloured filters in its path. She changes the intensity by moving the source nearer to or farther from plate C. For each setting, she records the microammeter reading as the potential of A is changed from a large positive value to negative values.
Light of the same frequency but of intensities \( I_1 \), \( I_2 \) and \( I_3 \) \( (I_1>I_2>I_3) \) falls in turn on the plate C. What can we say about the stopping potentials \( V_1 \), \( V_2 \) and \( V_3 \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( V_1=V_2=V_3 \)
The stopping potential measures the maximum kinetic energy of the photoelectrons, \( eV_0=K_{\mathrm{max}} \), and by Einstein’s photoelectric equation \( K_{\mathrm{max}}=h\nu-\phi_0 \). The intensity does not appear in either relation.
Option (1) \( V_1\lt V_2\lt V_3 \) — Incorrect. This would make the stopping potential fall as the light gets brighter. But the energy of each photon, and so \( K_{\mathrm{max}} \), does not change with intensity.
Option (2) \( V_1\gt V_2\gt V_3 \) — Incorrect. Brighter light ejects more electrons per second. This raises the saturation current. The maximum kinetic energy of each electron stays the same.
Option (3) \( V_1=V_2=V_3 \) — Correct. The frequency is the same, so every photon has the same energy \( h\nu \). The fastest photoelectrons from plate C have the same \( K_{\mathrm{max}} \) in all three cases. Hence the stopping potential does not depend on intensity.
Option (4) cannot say — Incorrect. The frequency and the plate material are the same in all three cases, and these two fix \( K_{\mathrm{max}} \) and hence \( V_0 \).
NCERT (2026-27):
Effect of Potential on Photoelectric Current – “… We can now repeat this experiment with incident radiation of the same frequency but of higher intensity I2 and I3 (I3 > I2 > I1). We note that the saturation currents are now found to be at higher values. This shows that more electrons are being emitted per second, proportional to the intensity of incident radiation. … Thus, for a given frequency of the incident radiation, the stopping potential is independent of its intensity.”