When light of different frequencies is incident on a metal surface, the graph between stopping potential and frequency of incident radiation is a straight line. The slope of the graph represents
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Planck’s constant divided by electronic charge
Einstein’s photoelectric equation gives the stopping potential \( V_{0} \) as a straight-line function of frequency:
\( eV_{0} = h\nu - \phi_{0} \)
\( \displaystyle V_{0} = \left(\frac{h}{e}\right)\nu - \frac{\phi_{0}}{e} \)
Comparing with \( y = mx + c \), the slope is \( h/e \). It is the same for every metal, which is how Millikan measured \( h \).
Why the other options are incorrect
Option (1) Work function of the metal — Incorrect. The work function gives the intercept \( -\phi_{0}/e \) on the \( V_{0} \)-axis, not the slope.
Option (2) Threshold frequency of the metal — Incorrect. The threshold frequency is the intercept on the frequency axis, where \( V_{0} = 0 \).
Option (3) Planck’s constant — Incorrect. \( h \) is the slope of the graph of maximum kinetic energy against frequency. Stopping potential is that energy divided by \( e \), so its slope is \( h/e \).
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
\( \displaystyle e V_0 = h\nu - \phi_0; \quad \text{for } \nu \ge \nu_0 \)
or
\( \displaystyle V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \qquad (11.4) \)
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.”