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

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Solution

The slope of the cut-off voltage versus frequency of incident light is given as,

\frac{∆ V}{∆ ν} = 4.12 \times 10^{- 15} Vs = 4.12 \times 10^{- 15} \frac{J . s .}{C}

When we multiply this result with the charge of an electron, which is the fundamental charge (e = 1.6 x 10^–19 C) we get,

E = hv

\Rightarrow

h = \frac{E}{v} = J . s .

\Rightarrow

h = 4.12 \times 10^{- 15} \times 1.6 \times 10^{- 19}

i.e.,

h = 6.592 \times 10^{- 34} Js .

is the value of the planck's constant.

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