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Question
In an experiment on the photoelectric effect, the slope of the cut-off voltage versus the 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 (v) of incident light is given as:
`"V"/"v"` = 4.12 × 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 × 10−19 × 4.12 × 10−15
= 6.592 × 10−34 Js
Therefore, the value of Planck’s constant is 6.592 × 10−34 Js.
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