The graphs of stopping potential versus frequency of incident radiati…

The correct answer is Option (4) → (B) and (D) only

From Einstein's photoelectric equation, \( eV_0 = h\nu - \phi_0 \), so

    \( \displaystyle V_0 = \frac{h}{e}\,\nu - \frac{\phi_0}{e} \)

This is a straight line of slope \( h/e \), which cuts the \( \nu \)-axis at the threshold frequency \( \nu_0 = \phi_0/h \) and, when extended, cuts the \( V_0 \)-axis at \( -\phi_0/e \).

(A) The slope of the graphs is equal to the Planck's constant. — Incorrect. The slope is \( h/e \), not \( h \).

(B) Intercept on the (-y) axis is equal to (work function/e). — Correct. Putting \( \nu = 0 \) gives \( V_0 = -\phi_0/e \), so the intercept below the origin has magnitude \( \phi_0/e \).

(C) Threshold frequency for metal A is higher than metal B — Incorrect. Line A cuts the \( \nu \)-axis at \( \nu_0 \), to the left of B's \( \nu_0' \), so A has the lower threshold frequency.

(D) Work function for metal B is greater than metal A — Correct. \( \phi_0 = h\nu_0 \) and \( \nu_0' \gt \nu_0 \); the extended line B also cuts the negative \( V_0 \)-axis further down.

NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Equation (11.3) shows that the greater the work function \( \phi_0 \), the higher the minimum or threshold frequency \( \nu_0 \) needed to emit photoelectrons.”

Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Using Eq. (11.1), the photoelectric equation, Eq. (11.2), can be written as

    \( \displaystyle eV_0 = h\nu - \phi_0; \quad \text{for } \nu \ge \nu_0 \)

or

    \( \displaystyle V_0 = \frac{h}{e}\,\nu - \frac{\phi_0}{e} \qquad (11.4) \)

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.”