According to Bohr's model, the orbital angular momentum of electrons in third excited state is [h = 6.63 × 10⁻³⁴ Js]

Question

Accepted Answer

(1) 4.2 × 10⁻³⁴ kg m²s⁻¹. Given: h=6.63×10⁻³⁴ J s. Counting the ground state as n=1, the first excited state is n=2, so the third excited state is n=4. Idea: Bohr's second postulate lets the electron carry only whole multiples of h/(2π) of orbital angular momentum, L=(nh)/(2π). L=(4×6.63×10⁻³⁴)/(2π)=(2×6.63×10⁻³⁴)/(3.14). L=4.22×10⁻³⁴ J s. A joule second is a kg m² s⁻¹, so this is the 4.2×10⁻³⁴ kg m² s⁻¹ of the first option. Beware the labelling: third excited state means n=4, not n=3.

According to Bohr's model, the orbital angular momentum of electrons in third excited state is ______ [h = 6.63 × 10⁻³⁴ Js]

Step-by-step solution

Why the other options are wrong