In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of the incident light gives
Answer & explanation
Correct answer: option 2
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)