Light of a certain frequency falls on a metal of work function 2.3 eV, and the stopping potential is 1.6 V. The same light then falls on another metal of work function 2.9 eV. The stopping potential now is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.0 V
The light is the same, so the energy \( h\nu \) of each photon does not change. Einstein’s equation \( eV_0=h\nu-\phi_0 \) gives this photon energy from the first metal. The same equation then gives the new stopping potential.
\( \begin{aligned} h\nu &= eV_{01}+\phi_{01} \\ &= 1.6+2.3 \\ &= 3.9\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} eV_{02} &= h\nu-\phi_{02} \\ &= 3.9-2.9 \\ &= 1.0\ \mathrm{eV} \end{aligned} \)
Therefore, \( V_{02}=1.0\ \mathrm{V} \). For the same light, a metal with a larger work function gives a smaller stopping potential.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… the photoelectric equation … can be written as”
\( eV_0=h\nu-\phi_0;\ \text{for } \nu\ge\nu_0 \)\( \qquad (11.4) \)