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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. - CBSE (Science) Class 12 - Physics

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Question

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

Solution

The slope of the cut-off voltage (V) versus frequency (ν) of an incident light is given as:

`V/v = 4.12 xx 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 xx V/v`

`= 1.6 xx 10^(-19) xx 4.12 xx 10^(-15) = 6.592 xx 10^(-34) Js`

Therefore, the value of Planck’s constant is `6.592 xx 10^(-34) Js`

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APPEARS IN

 NCERT Solution for Physics Textbook for Class 12 (2018 to Current)
Chapter 11: Dual Nature of Radiation and Matter
Q: 6 | Page no. 407
Solution 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. Concept: Einstein’S Photoelectric Equation: Energy Quantum of Radiation.
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