The slope of the cut off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} ~Vs$. The value of Planck's constant is:

e times the slope of cut off voltage versus frequency

The correct answer is Option (2) → e times the slope of cut off voltage versus frequency

According to photoelectric equation -

$eV_0=hv+\phi_{0}$

$V_0=\left(\frac{h}{e}\right)v+\phi_{0}$  ....(1)

and,

$y=mx+c$   ....(2)

On comparing (1) and (2)

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

$∴h=slope×e$