The slope of the stopping potential \( V_0 \) versus frequency \( \nu \) of the incident radiation plot is
(Where symbols have their usual meaning)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( h/e \)
Einstein's photoelectric equation gives the maximum kinetic energy of the emitted electrons, and the stopping potential \( V_0 \) is the potential that just stops them: \( eV_0 = K_{\max} \).
\( \displaystyle h\nu = \phi_0 + eV_0 \;\Rightarrow\; V_0 = \frac{h}{e}\,\nu - \frac{\phi_0}{e} \)
This is a straight line \( y = mx + c \) in \( V_0 \) against \( \nu \), so the slope is \( h/e \).
Shortcut: The slope has the unit of volt per hertz, that is \( \mathrm{V\,s} \). Only \( h/e \) has this unit: \( \mathrm{J\,s/C} = \mathrm{V\,s} \).
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
\( \displaystyle eV_0 = h\nu - \phi_0; \quad \text{for } \nu \ge \nu_0 \)
\( \displaystyle \text{or}\ \ V_0 = \frac{h}{e}\,\nu - \frac{\phi_0}{e} \qquad (11.4) \)
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.”