In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of the incident light gives
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( h/e \)
Einstein's photoelectric equation written for the cut-off (stopping) voltage \( V_0 \) is \( eV_0 = h\nu - \phi_0 \). Dividing by \( e \):
\( \displaystyle V_0 = \frac{h}{e}\,\nu - \frac{\phi_0}{e} \)
This is a straight line in \( \nu \), so the graph of \( V_0 \) against \( \nu \) has slope \( h/e \), the same for every metal.
Why the other options are incorrect
Option (1) \( h \) — Incorrect. \( h \) is the slope of the graph of \( eV_0 \) (the maximum kinetic energy) against \( \nu \), not of \( V_0 \) against \( \nu \).
Option (3) \( e/h \) — Incorrect. This is the reciprocal of the slope.
Option (4) \( \nu_0 \) — Incorrect. The threshold frequency \( \nu_0 \) is the intercept on the frequency axis, not the slope.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Using Eq. (11.1), the photoelectric equation, Eq. (11.2), can be written as
\( \displaystyle e V_0 = h\nu - \phi_0;\ \text{for}\ \nu \geq \nu_0 \)
\( \displaystyle \text{or}\quad V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \qquad (11.4) \)
This is an important result. It predicts that the V0 versus ν curve is a straight line with slope = (h/e), independent of the nature of the material.”