Question

The angular momentum of an electron in the 3^rd Bohr orbit of a Hydrogen atom is 3.165 × 10^-34 kg m²/s. Calculate Plank’s constant h.    

Answer 1

Given: L₃ = 3.165 × 10^-34 kg m²/s, n = 3

To find: Planck’s constant (h) 

Formula: L_n = `"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.