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

Answer

The slope of the cut-off voltage (V) versus frequency (ν) of an incident light is given as: V/v = 4.12 x 10-15 Vs

V is related to frequency by the equation: hv = eV

Where,

e = Charge on an electron = 1.6 × 10−19 C

h = Planck’s constant

h = e x V/v = 1.6 x 10-19 x 4.12 x 10-15 = 6.592 x 10-34 Js

Therefore, the value of Planck’s constant is 6.592 x 10-34 Js

 

 

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