In an experiment on photoelectric effect...

In an experiment on photoelectric effect.the slope of the cutoff voltage versus frequency incident light is found to be `4.12xx10^(-15)` V sec calculate the value of Planck's constant.

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Here `(DeltaV_(0))/(Deltav)=4.12xx10^(-15)Vs`
`e=1.6xx10^(-19)J`
`implieshv=eV_(0)`
h and e are constant
`therefore hDeltav=eDeltaV_(0)`
`therefore=e(DeltaV_(0))/(DeltaV)`
`=1.6xx10^(-19)xx4.12xx10^(-15)`
`therefore h=6.592xx10^(-34)Js`

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