The threshold wavelength of a metal is \(\lambda_0\), and its work function is 1.8 eV. Light of wavelength \(\lambda_0/4\) falls on the metal. The stopping potential is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5.4 V
At the threshold wavelength, the photon energy just equals the work function, \( hc/\lambda_0=\phi_0 \). Photon energy varies as \( 1/\lambda \). So light of one-fourth the wavelength has photons of four times this energy.
\( \begin{aligned} h\nu &= \frac{hc}{\lambda_0/4} \\ &= 4\phi_0 \\ &= 7.2\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} eV_0 &= h\nu-\phi_0 \\ &= 7.2-1.8 \\ &= 5.4\ \mathrm{eV} \end{aligned} \)
Therefore, \( V_0=5.4\ \mathrm{V} \).
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 …”
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… the photoelectric equation … can be written as”
\( eV_0=h\nu-\phi_0;\ \text{for } \nu\ge\nu_0 \)\( \qquad (11.4) \)