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.

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#### 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`

Concept: Einstein’S Photoelectric Equation: Energy Quantum of Radiation

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