Light of frequency \( \nu \) falls on a metal whose work function is \( h\nu_0 \). Choose the most appropriate condition from the following under which the photoelectric effect will take place:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \nu \gt \nu_0 \)
By Einstein's photoelectric equation, a photon of energy \( h\nu \) can eject an electron only if its energy exceeds the work function \( h\nu_0 \); the excess appears as the kinetic energy of the electron.
\( \displaystyle K_{\mathrm{max}} = h\nu - h\nu_0 \gt 0 \;\Rightarrow\; \nu \gt \nu_0 \)
So emission takes place for every frequency above the threshold frequency \( \nu_0 \), however low the intensity.
Option (1) \( \nu \gt \nu_0 \) — Correct. This is the general threshold condition: any frequency above \( \nu_0 \) ejects electrons.
Option (2) \( \nu \lt \nu_0 \) — Incorrect. Below the threshold the photon energy is less than the work function, so no electron is emitted, whatever the intensity.
Option (3) \( \nu \gt 200\nu_0 \) — Incorrect. Such light does eject electrons, but the condition is far too narrow: emission also takes place for all frequencies between \( \nu_0 \) and \( 200\nu_0 \), so it is not the condition under which the effect occurs.
Option (4) \( \nu \lt 0.5\nu_0 \) — Incorrect. This lies below the threshold frequency, so no emission takes place.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Since Kmax must be non-negative, Eq. (11.2) implies that photoelectric emission is possible only if \( h\nu \gt \phi_0 \) or \( \nu \gt \nu_0 \), where
\( \displaystyle \nu_0 = \frac{\phi_0}{h} \qquad (11.3) \)
Thus, there exists a threshold frequency \( \nu_0 \) (\( = \phi_0/h \)) for the metal surface, below which no photoelectric emission is possible, no matter how intense the incident radiation may be or how long it falls on the surface.”