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.
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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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 6: In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incid....
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