When photon of energy \( h\nu \) falls on a photosensitive metallic surface (work function \( h\nu_0 \)) electrons are emitted from the metallic surface. It is possible to say that :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → The ejected electrons have a distribution of KE, the most energetic ones having KE equal to \( (h\nu - h\nu_0) \)
Einstein's photoelectric equation gives the largest kinetic energy an emitted electron can have.
\( \displaystyle K_{\max} = h\nu - h\nu_0 \)
The work function \( h\nu_0 \) is the minimum energy needed to free an electron from the surface. Electrons from deeper inside the metal lose extra energy on the way out, so the emitted electrons have a range of kinetic energies from zero up to \( K_{\max} \).
Option (1) All ejected electrons have the same KE equal to \( (h\nu - h\nu_0) \) — Incorrect. Only the least tightly bound electrons get this value; the others come out with less.
Option (2) The ejected electrons have a distribution of KE, the most energetic ones having KE equal to \( (h\nu - h\nu_0) \) — Correct. The energies are spread out, and the maximum is \( h\nu - h\nu_0 \).
Option (3) The most energetic electrons having KE equal to \( h\nu \) — Incorrect. At least \( h\nu_0 \) is used up in escaping.
Option (4) The KE of ejected electrons is \( h\nu_0 \) — Incorrect. \( h\nu_0 \) is the work function, not the kinetic energy.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… In photoelectric effect, an electron absorbs a quantum of energy (hν) of radiation. If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function φ0), the electron is emitted with maximum kinetic energy
\( \displaystyle K_{max} = h\nu - \phi_0 \qquad (11.2) \)
More tightly bound electrons will emerge with kinetic energies less than the maximum value.”