The cathode of a photoelectric cell is changed so that the work function changes from \( W_1 \) to \( W_2 \) \( (W_1\gt W_2) \). Light of the same frequency (above both threshold frequencies) is used. If the stopping potentials before and after are \( V_1 \) and \( V_2 \) respectively, then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( V_1\lt V_2 \)
By Einstein’s photoelectric equation, \( eV_0=h\nu-\phi_0 \). For the same frequency, a smaller work function leaves more energy for the electron and so needs a larger stopping potential.
For the two cathodes:
\( eV_1=h\nu-W_1 \)
\( eV_2=h\nu-W_2 \)
Subtracting the first from the second:
\( e\,(V_2-V_1)=W_1-W_2 \)
Since \( W_1\gt W_2 \), the right side is positive, so \( V_2\gt V_1 \).
The new cathode holds its electrons less tightly, so they leave with a larger maximum kinetic energy. Hence, \( V_1\lt V_2 \).
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “Using … the photoelectric equation … can be written as”
\( eV_0=h\nu-\phi_0 \)\( ;\ \text{for } \nu\ge\nu_0 \)
\( \displaystyle \text{or}\quad V_0=\left(\frac{h}{e}\right)\nu-\frac{\phi_0}{e} \qquad (11.4) \)