Question

Calculate the magnitude of the maximum orbital angular momentum L for an electron in a hydrogen atom for states with a principal quantum number of 96. Answer in terms of ℏ

Answer 1

Answer:

The magnitude of the maximum orbital angular momentum L for an electron in a hydrogen atom for states with a principal quantum number of 96 is 9120ℏ

Explanation:

• The angular momentum $(L)$ is the quantity of rotation possessed by a rotating body

• orbital angular momentum quantum number $(l)$ quantizes the magnitude of the angular momentum as  

        $/L/=\sqrt{l(l+1)} h/2\pi$

• The principal quantum number $(n)$, that quantizes energy and orbital angular momentum quantum number related as

       $l=0,1,2,(n-1)$

• use this conversion of Planck's constant $\hbar =h/2\pi$

from the question, we have

the principal quantum number $n=96$

⇒

orbital angular momentum quantum number can have values,

$l=0,1 upto 95$

the maximum value is $l=95$

so maximum angular momentum $/L/=\sqrt{95(95+1)} h/2\pi =9120\hbar$