In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of the incident light gives

$h/e$

The correct answer is Option (2) → $h/e$

Concept:

The photoelectric equation is given by:

$eV_0 = hν - φ$

where $V_0$ is the cut-off voltage, $ν$ is the frequency, $h$ is Planck’s constant, and $φ$ is the work function.

Comparing with the linear form $y = mx + c$:

$V_0 = \frac{h}{e}ν - \frac{φ}{e}$

Hence, the slope of the graph of $V_0$ versus $ν$ is:

$\text{slope} = \frac{h}{e}$

Final Answer:

Slope = $\frac{h}{e}$ (ratio of Planck’s constant to charge of electron)