Light of wavelength 3500 Å is incident on two metals A and B. The work function of A is 4.2 eV and that of B is 1.19 eV. Photoelectrons will be emitted by
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → metal B
By Einstein’s photoelectric equation, \( K_{\mathrm{max}}=h\nu-\phi_0 \). So a metal emits photoelectrons only if the photon energy is greater than its work function.
Take \( h=6.63\times10^{-34}\,\mathrm{J\,s} \), \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \). Since 1 Å = \( 10^{-10}\,\mathrm{m} \), the wavelength is \( \lambda=3.5\times10^{-7}\,\mathrm{m} \). The photon energy is:
\( \begin{aligned} E &= \frac{hc}{\lambda} \\ &= \frac{6.63\times10^{-34}\,\mathrm{J\,s}\times3\times10^{8}\,\mathrm{m\,s^{-1}}}{3.5\times10^{-7}\,\mathrm{m}} \\ &\approx 5.68\times10^{-19}\,\mathrm{J} \\ &\approx 3.55\,\mathrm{eV} \end{aligned} \)
Option (1) metal A — Incorrect. The work function of A, 4.2 eV, is more than the photon energy of about 3.55 eV, so no electron can escape from A.
Option (2) metal B — Correct. The work function of B, 1.19 eV, is less than the photon energy. So each photon can free an electron from B, and photoelectrons are emitted.
Option (3) both A and B — Incorrect. Metal A needs at least 4.2 eV per photon, which this light does not supply.
Option (4) neither metal A nor metal B — Incorrect. Each photon of about 3.55 eV can free an electron from metal B, whose work function is only 1.19 eV.
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”
\( \displaystyle \nu_0=\frac{\phi_0}{h} \qquad (11.3) \)