The work function of a metal is 2.3 eV. When light falls on it, the photoelectric current becomes zero at a stopping potential of 0.8 V. The wavelength of the light is (Take \(hc=1240\ \mathrm{eV\,nm}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 400 nm
The stopping potential just stops the fastest photoelectrons, so \( eV_0 \) is their maximum kinetic energy. By Einstein’s equation, the photon energy is the work function plus this kinetic energy. The wavelength then follows from \( E=hc/\lambda \).
Given:
\( \phi_0=2.3\ \mathrm{eV} \)
\( eV_0=0.8\ \mathrm{eV} \)
\( \begin{aligned} E &= \phi_0+eV_0 \\ &= 2.3+0.8 \\ &= 3.1\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} \lambda &= \frac{hc}{E} \\ &= \frac{1240}{3.1} \\ &= 400\ \mathrm{nm} \end{aligned} \)
Hence, the wavelength of the incident light is 400 nm.
NCERT (2026-27):
Example 11.2 – “(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy e V0 by the retarding potential V0. Einstein’s Photoelectric equation is”
\( \displaystyle eV_0=h\nu-\phi_0=\frac{hc}{\lambda}-\phi_0 \quad \text{or,} \quad \)\( \lambda=hc/(eV_0+\phi_0) \)