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Physics Question on Dual nature of radiation and matter

In an experiment on photoelectric effect,the slope of the cut-off voltage versus frequency of incident light is found to be 4.12×10–15Vs4.12×10^{–15} V s. Calculate the value of Planck’s constant.

Answer

The correct answer is: 6.592×10−34Js.6.592×10^{-34}Js.
The slope of the cut-off voltage(V) versus frequency(ν) of an incident light is given as:
Vv=4.12×10−15Vs\frac{V}{v}=4.12×10^{-15}Vs
V is related to frequency by the equation:
hv=eVhv=eV
Where,
e=Charge on an electron=1.6×10−19C=1.6×10^{-19}C
h=Planck's constant
∴h=e×Vv∴h=e×\frac{V}{v}
=1.6×10−19×4.12×10−15=6.592×10−34Js=1.6×10^{-19}×4.12×10^{-15}=6.592×10^{-34}Js
Therefore,the value of Planck’s constant is 6.592×10−34Js.6.592×10^{-34}Js.