What is the slope of the graph drawn between frequency of incident light and stopping potential for a given surface?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( he^{-1} \)
By Einstein's photoelectric equation, the maximum kinetic energy of the photoelectrons, \( eV_0 \), equals the photon energy minus the work function \( \phi_0 \):
\( \displaystyle eV_0 = h\nu - \phi_0 \)
\( \displaystyle V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \)
With the stopping potential on the y-axis and the frequency on the x-axis, this is a straight line \( y = mx + c \) with slope
\( \displaystyle m = \frac{h}{e} = he^{-1} \)
The slope is the same for every metal; only the intercept (the threshold frequency) depends on the surface.
Why the other options are incorrect
Option (1) \( h \) — Incorrect. \( h \) is the slope of the graph of maximum kinetic energy \( eV_0 \) against frequency, not of the stopping potential.
Option (3) \( eh \) — Incorrect. Here \( e \) multiplies instead of dividing; the potential is the energy divided by \( e \).
Option (4) \( e \) — Incorrect. The slope must have the unit of volt second, which \( e \) (in coulomb) does not have.
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
\( \displaystyle eV_0 = h\nu - \phi_0; \quad \text{for } \nu \geq \nu_0 \quad \text{or} \quad V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \qquad (11.4) \)
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.”