The work functions of silver and sodium are 4.6 eV and 2.3 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that for sodium is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 : 1
Einstein’s photoelectric equation is \( eV_0=h\nu-\phi_0 \). So the stopping potential is
\( \displaystyle V_0=\frac{h}{e}\,\nu-\frac{\phi_0}{e} \)
The \( V_0 \) versus \( \nu \) plot is a straight line of slope \( h/e \). This slope has only universal constants, so it is the same for every metal. The work function decides only where the line meets the frequency axis. That point is the threshold frequency \( \nu_0=\phi_0/h \).
\( \begin{aligned} \frac{\text{slope for silver}}{\text{slope for sodium}} &= \frac{h/e}{h/e} \\ &= 1 \end{aligned} \)
Therefore, the ratio of the slopes is 1 : 1. The different work functions, 4.6 eV and 2.3 eV, only shift the two lines along the frequency axis.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… the photoelectric equation … can be written as
\( e\,V_0=h\,\nu-\phi_0;\ \text{for}\ \nu\ge\nu_0 \)
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.”