The work function of a metal is 4.2 eV. Will this metal emit photoelectrons when light of wavelength 330 nm falls on it? Choose the answer with the correct reason.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → No, because \( \nu\lt\nu_0 \)
Photoelectrons are emitted only if the frequency of the light is above the threshold frequency \( \nu_0=\phi_0/h \) of the metal, whatever the intensity of the light.
Taking \( h=6.63\times10^{-34}\,\mathrm{J\,s} \), \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \), the frequency of the light is:
\( \begin{aligned} \nu &= \frac{c}{\lambda} \\ &= \frac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{330\times10^{-9}\,\mathrm{m}} \\ &\approx 9.1\times10^{14}\,\mathrm{Hz} \end{aligned} \)
The threshold frequency of the metal is:
\( \begin{aligned} \nu_0 &= \frac{\phi_0}{h} \\ &= \frac{4.2\times1.6\times10^{-19}\,\mathrm{J}}{6.63\times10^{-34}\,\mathrm{J\,s}} \\ &\approx 1.01\times10^{15}\,\mathrm{Hz} \end{aligned} \)
From the above, \( \nu\lt\nu_0 \).
Option (1) Yes, because \( \nu\gt\nu_0 \) — Incorrect. The frequency of the 330 nm light is below the threshold frequency. So no photoelectrons are emitted.
Option (2) No, because \( \nu\lt\nu_0 \) — Correct. Since \( \nu\lt\nu_0 \), each photon carries only about 3.77 eV. This is less than the work function 4.2 eV, so no electron can escape from the metal.
Option (3) Yes, because \( \nu\lt\nu_0 \) — Incorrect. Below the threshold frequency there is no photoelectric emission at all, however intense the light.
Option (4) No, because \( \nu\gt\nu_0 \) — Incorrect. The frequency is below \( \nu_0 \). Light above the threshold frequency would eject photoelectrons, so it cannot be a reason for No.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Since Kmax must be non-negative, … photoelectric emission is possible only if hν > φ0 or ν > ν0, where”
\( \displaystyle \nu_0=\frac{\phi_0}{h} \qquad (11.3) \)
“… 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.”