The slope of the stopping potential $V_0$ versus frequency $v$ of the incident radiation plot is

(Where symbols have their usual meaning)

$h/e$

The correct answer is Option (3) → $h/e$

Photoelectric effect equation:

$e V_0 = h \nu - \phi$

Rewriting:

$V_0 = \frac{h}{e} \nu - \frac{\phi}{e}$

This is in the form of a straight line: $V_0 = \text{slope} \cdot \nu + \text{intercept}$

Therefore, slope of $V_0$ versus $\nu$ plot:

Slope $= \frac{h}{e}$