Light of wavelength 496 nm ejects photoelectrons from a metal surface, but light of wavelength 620 nm does not. The work function of the metal lies between (Take \(hc=1240\ \mathrm{eV\,nm}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.0 eV and 2.5 eV
A photon can eject an electron only if its energy is more than the work function. So the 496 nm photon has enough energy, and the 620 nm photon does not. The photon energies of the two lights set the limits.
\( \begin{aligned} E_1 &= \frac{hc}{\lambda_1} \\ &= \frac{1240}{496} \\ &= 2.5\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} E_2 &= \frac{hc}{\lambda_2} \\ &= \frac{1240}{620} \\ &= 2.0\ \mathrm{eV} \end{aligned} \)
The 620 nm light fails, so \( \phi_0 \) is at least 2.0 eV. The 496 nm light works, so \( \phi_0 \) is less than 2.5 eV. Hence, the work function lies between 2.0 eV and 2.5 eV.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Since Kmax must be non-negative, … photoelectric emission is possible only if h ν > φ0 …”