Question

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

Answer 1

Slope of the cut off voltage versus frequency of incident light graph

= 4.12 x 10^-15 Vs

but the slope of the graph = \(\frac{h}{e}\)

h = e x slope of the graph

= 1.6 x 10¹⁹ x 4.12 x 10^-15 

= 6.592 x 10^-34Js