Assertion (A): When ultraviolet light is incident on a photocell, its…

The correct answer is Option (4) → (A) is false but (R) is true.

(A) When ultraviolet light is incident on a photocell, its stopping potential is \( V_0 \) and the maximum kinetic energy of the photoelectrons is \( K_{max} \). When the ultraviolet light is replaced by infrared light, both \( V_0 \) and \( K_{max} \) increase. — False. By Einstein’s photoelectric equation, \( K_{max}=h\nu-\phi_0 \), and \( eV_0=K_{max} \). Infrared light has a lower frequency than ultraviolet light, so both \( K_{max} \) and \( V_0 \) decrease. If the infrared frequency is below the threshold frequency, no photoelectrons are emitted at all.

(R) Photoelectrons are emitted with speeds ranging from zero to a maximum value. — True. The photon energy \( h\nu \) is the same for every electron, but electrons that are more tightly bound use up more of it in escaping. So the emitted electrons have kinetic energies, and speeds, spread from nearly zero up to the maximum value fixed by \( K_{max} \).

NCERT (2026-27):
Effect of Frequency of Incident Radiation on Stopping Potential – “(iii) For a given photosensitive material, there exists a certain minimum cut-off frequency of the incident radiation, called the threshold frequency, below which no emission of photoelectrons takes place, no matter how intense the incident light is. Above the threshold frequency, the stopping potential or equivalently the maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of the incident radiation…”

Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… the electron is emitted with maximum kinetic energy”

    \( K_{\mathrm{max}}=h\nu-\phi_0 \qquad (11.2) \)

“More tightly bound electrons will emerge with kinetic energies less than the maximum value.”