In an experiment of photoelectric effect the slope of cutoff voltage v/s frequency of incident light is found to be \(4.12 \times 10^{-15} Vs. Calculate the value of planks constant.

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Solution

The slope of the cut - off voltage(V) versus frequency(v) of an incident light is given as:
$\frac{V}{v} = 4.12 \times 10^{- 15} V s$
V is related to frequency by the equation: hv = eV
So, $h = e \times \frac{V}{v} = 1.6 \times 10^{- 19} \times 4.12 \times 10^{- 15} = 6.592 \times 10^{- 34} J s$

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In an experiment of photoelectric effect the slope of cutoff voltage v/s frequency of incident light is found to be \(4.12 \times 10^{-15} Vs. Calculate the value of planks constant.

The slope of the cut - off voltage(V) versus frequency(v) of an incident light is given as: Vv=4.12×10−15Vs V is related to frequency by the equation: hv = eV So, h=e×Vv=1.6×10−19×4.12×10−15=6.592×10−34Js

The slope of the cut - off voltage(V) versus frequency(v) of an incident light is given as:
$\frac{V}{v} = 4.12 \times 10^{- 15} V s$
V is related to frequency by the equation: hv = eV
So, $h = e \times \frac{V}{v} = 1.6 \times 10^{- 19} \times 4.12 \times 10^{- 15} = 6.592 \times 10^{- 34} J s$