In an experiment on photoelectric effect,
Question:

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} \mathrm{Vs}$. 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:

$\frac{V}{V}=4.12 \times 10^{-15} \mathrm{Vs}$

$V$ is related to frequency by the equation:

$h v=e V$

Where,

e = Charge on an electron = 1.6 × 10−19 C

 

h = Planck’s constant

$\therefore h=e \times \frac{V}{v}$

$=1.6 \times 10^{-19} \times 4.12 \times 10^{-15}=6.592 \times 10^{-34} \mathrm{JS}$

Therefore, the value of Planck’s constant is $6.592 \times 10^{-34} \mathrm{Js}$.

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