Question

What is the angular momentum of the electron in the third Bohr orbit in the hydrogen atom ?

[ \(\cfrac{h}{2π}\) = 1.055 × 10^-34 kg∙m² /s

Answer 1

Angular momentum, L = mvr = \(\cfrac{nh}{2π}\)

For n = 3, 

L = 3 \(\left(\cfrac{h}{2π}\right)\) = 3 (1.055 × 10^-34)

= 3.165 × 10^-34 kg∙m² /s