In an experiment on photoelectric effect, the slope of the cut off voltage versus frequency of incident light is found to be `4.12xx10^(-15)Vs`. Given `e=1.6xx10^(-19)C`, estimate the value of Planck's constant.

Explanation

The slope of the cut-off voltage (V) versus frequency (v) of an incident light is given as : `V/v=4.12xx10^(-15) Vs`
V is related to frequency by the equation : hv=eV
Where, e=Charge on an electron =`1.6xx10^(-19) C`
h=Planck's contant `therefore h=exxV/v`
`=1.6xx10^(-19)xx4.12xx10^(-15)=6.592xx10^(-34) Js`
Therefore , the value of Planck's constant is `6.592xx10^(-34) Js`