Light of wavelength \( \lambda \) falls on a metal having work function \( hc/\lambda_0 \). Photoelectric emission will take place only if
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \lambda\le\lambda_0 \)
By Einstein’s photoelectric equation, \( K_{\mathrm{max}}=h\nu-\phi_0 \), and \( K_{\mathrm{max}} \) cannot be negative. A photon can free an electron only if its energy \( hc/\lambda \) is at least the work function \( hc/\lambda_0 \).
\( \displaystyle \frac{hc}{\lambda}\ge\frac{hc}{\lambda_0} \)
\( \Rightarrow\; \lambda\le\lambda_0 \)
Option (1) \( \lambda\ge\lambda_0 \) — Incorrect. A wavelength longer than \( \lambda_0 \) means a frequency below the threshold frequency. The photon energy is then less than the work function, so no electron comes out.
Option (2) \( \lambda\ge2\lambda_0 \) — Incorrect. Each photon then carries at most \( hc/2\lambda_0 \). This is only half the work function, so there is no emission.
Option (3) \( \lambda\le\lambda_0 \) — Correct. A shorter wavelength means a higher photon energy. For \( \lambda\le\lambda_0 \) the photon energy is at least the work function, and electrons are emitted.
Option (4) \( \lambda=4\lambda_0 \) — Incorrect. The photon energy is then \( hc/4\lambda_0 \), a quarter of the work function. It cannot free an electron, however intense the light is.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Thus, there exists a threshold frequency ν0 (= φ0/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.”