CBSE 2025, Set 3

Question

Answer the following, giving reasons:
(a) The maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation.
(b) Photoelectric current increases with the increase in the intensity of the incident radiation.
(c) The stopping potential $V_0$ varies linearly with the frequency $
u$ of the incident radiation for a given photosensitive surface.

Edvaya · Accepted Answer

The correct answer is Subjective. **(a)** According to the photon picture of light, a change in the intensity of incident radiation (at a constant frequency) only changes the *number* of photons incident per unit area per unit time, not the *energy* of each individual photon ($E=h
u$). Since one photon interacts with one electron, the maximum kinetic energy of emitted photoelectrons remains unchanged.

**(b)** An increase in the intensity of incident radiation means more photons strike the metal surface per second. Since each photon can eject an electron, a greater number of photons leads to a greater number of photoelectrons emitted per second, thereby increasing the photoelectric current.

**(c)** According to Einstein's photoelectric equation:
$$K_{max} = h
u - h
u_0$$
Since $K_{max} = eV_0$, we have:
$$eV_0 = h
u - h
u_0$$
$$V_0 = \left(\frac{h}{e}ight)
u - \left(\frac{h
u_0}{e}ight)$$
This is the equation of a straight line ($y = mx + c$), demonstrating that the stopping potential $V_0$ varies linearly with the frequency $
u$ of the incident radiation.

![Stopping Potential Graph](https://www.edvaya.com/images/cbse/physics/2025/set3/q24c-stopping-potential-frequency-graph-answer.webp)

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