For a photosensitive metal, the graph of stopping potential \(V_0\) against frequency \(\nu\) is a straight line. The point where this line meets the frequency axis gives
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → the threshold frequency of the metal
Einstein’s equation for the stopping potential is \( V_0=(h/e)\nu-\phi_0/e \). This is a straight line of slope \( h/e \). On the frequency axis \( V_0=0 \), which gives \( h\nu=\phi_0 \):
\( \begin{aligned} \nu &= \frac{\phi_0}{h} \\ &= \nu_0 \end{aligned} \)
Option (1) the maximum kinetic energy of the photoelectrons — Incorrect. \( K_{\max}=eV_0 \) is read on the \( V_0 \) axis for a chosen frequency. At the point on the frequency axis it is zero.
Option (2) the work function of the metal, in volts — Incorrect. \( \phi_0/e \) is read where the extended line cuts the \( V_0 \) axis, below the origin.
Option (3) the value of Planck’s constant — Incorrect. Planck’s constant comes from the slope of the line, \( h/e \), multiplied by \( e \).
Option (4) the threshold frequency of the metal — Correct. The stopping potential falls to zero at \( \nu_0=\phi_0/h \). This is the threshold frequency. Below it, no photoelectrons are emitted.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… the photoelectric equation … can be written as”
\( \displaystyle eV_0=h\nu-\phi_0;\ \text{for } \nu\ge\nu_0 \quad \text{or} \quad V_0=\frac{h}{e}\nu-\frac{\phi_0}{e} \qquad (11.4) \)
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Thus, there exists a threshold frequency ν0 (= φ0/h) for the metal surface, below which no photoelectric emission is possible …”