Consider the given graph between the stopping potential and frequency of incident light in an experiment of photoelectric effect. The slope of the graph gives:
(where symbols have their used meanings)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{h}{e} \)
Einstein's photoelectric equation gives the maximum kinetic energy of the photoelectrons, which is measured by the stopping potential \( V_0 \):
\( \displaystyle eV_0 = h\nu - \phi_0 \)
Dividing by \( e \):
\( \displaystyle V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \)
This is a straight line of \( V_0 \) against \( \nu \) with slope \( h/e \) and intercept \( -\phi_0/e \) on the \( V_0 \) axis.
Check with the value in the graph:
\( \displaystyle \frac{h}{e} = \frac{6.63\times10^{-34}\,\mathrm{J\,s}}{1.6\times10^{-19}\,\mathrm{C}} = 4.14\times10^{-15}\,\mathrm{V\,s} \)
which agrees with the slope \( 4.15\times10^{-15}\,\mathrm{V\,s} \) marked on the line.
Shortcut: The slope of a \( V \)-against-\( \nu \) graph has the unit \( \mathrm{V\,s} \), and \( h/e \) is the only option with this unit: \( h \) is in \( \mathrm{J\,s} \), \( \Phi/e \) is in volt and \( eh \) is in \( \mathrm{C\,J\,s} \).
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.”