In an experiment on photo electric effect, the slope of cut - off voltage versus frequency, incident light is found to be `4.12 xx 10^(-15)` Vs. Calculate the of Planck's constant.

Explanation

Given, slope `= 4. 12 xx 10^(-15)`Vs
From PEE, `(hv)/e= (hv)_(0)/e + V_(s)`
or `V_(s) = (hv)/ e - (hv)_(0)/e`
Here h/e represents the slope
`h//e = 4.12 xx 10^(-15)"Js" C^(-1)`
`therefore` Planck.s constant ` h = 4.12 xx 10^(-15)xx1.6 xx 10^(-19) Js = 6.592 xx 10^(-34)`