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:
Answer & explanation
Correct answer: option 2
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)
$m=\frac{h}{e}$
$∴h=slope×e$