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 × 10–15 V s. Calculate the value of Planck’s constant.

Answer 1

Given: The slope of the cut off voltage versus frequency of incident light is 4.12× 10 −15  Vs.

The Planck’s constant is given as,

h= eV ν

Where, the frequency of the light is ν, Planck’s constant is h, the cut off voltage is V and the charge on an electron is e.

By substituting the given values in the above equation, we get

h=1.6× 10 −19 ( 4.12× 10 −15 ) =6.592× 10 −34  Js

Thus, the value of Plank’s constant is 6.592× 10 −34  Js.