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Q : 11.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of  incident light is found to be  4.12\times10^-^1^5Vs. Calculate the value of Planck’s constant. 

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The slope of the cut-off voltage versus frequency of incident light is given by h/e where h is Plank's constant and e is an electronic charge.

h=slope\times e

h=4.12\times10^{-15}\times1.6\times10^{-19}

h=6.59210^{-34} Js

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