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Question
The angular momentum of an electron in the 3rd Bohr orbit of a Hydrogen atom is 3.165 × 10-34 kg m2/s. Calculate Plank’s constant h.
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Solution
Given: L3 = 3.165 × 10-34 kg m2/s, n = 3
To find: Planck’s constant (h)
Formula: Ln = `"n""h"/(2pi)`
Calculation:
From formula,
h = `(2pi"L"_"n")/"n"`
= `(2 xx 3.142 xx 3.165 xx 10^-34)/3`
= 6.284 × 1.055 × 10−34
= antilog {log(6.284) + log(1.055)} × 10−34
= antilog {0.7982 + 0.0232} × 10−34
= antilog{0.8214} × 10−34
= 6.628 × 10−34 Js
The value of Planck’s constant (h) is 6.628 × 10–34 Js.
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