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Ashok0210
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Asked: 2021-04-27T07:04:51+00:00 In:

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

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Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.6
NCERT Solutions for Class 12 Physics Chapter 11 Question-6

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      The slope of the cut-off voltage (V) versus frequency (v) 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 x 10-19 C and

      h = Planck’s constant Therefore, h = e x (v/v)= 1.6 x 10-19 x 4.12 x 10–15 = 6.592 x 10–34 Js

      Hence ,the value of Plank’s constant is 6.592 x 10–34 Js

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