Physical observables determined by quantum numbers: n, l, ml, ms

[indie452]

exam practise question:

what are the physical observables determined by the quantum numbers n, l, m_l and m_s of the electron in a hydrogen atom.

most places just give the equation or the name not the physical observable determined.
so is this right?:-

n = principal number determines the energy state that th atom is in
l = azimuthal number i know is to do with the orbital angular momentum, does this determine the angular momentum state?
m_l = magnetic quantum number associated with angular momentum
m_s = magnetic quantum number associated with spin and if we have spin +1/2 then the electron will have a higher energy then if it was -1/2

and that's all i know.
any help would be great.

 

[DrClaude]

This considers that the hydrogen atom is in a simultaneous eigenstate of operators $\hat{H}$, $\hat{L}^2$, $\hat{L}_z$ and $\hat{S}_z$.

• $n$ is the principle quantum number and is related to the eigenvalue of $\hat{H}$, as $E_n = - \frac{ m_e e^4}{2 ( 4 \pi \epsilon_0)^2 \hbar^2 } \frac{1}{n^2}$.
• $l$ is the (orbital) angular momentum quantum number, as the eigenvalue of $\hat{L}^2$ is $\sqrt{l (l+1)} \hbar$. For a given $n$, the possible (integer) values are $0 \leq l < n$.
• $m_l$ is the (orbital) angular momentum magnetic quantum number and comes from the eigenvalue of $\hat{L}_z$, which is $m_l \hbar$. It is an integer $-l \leq m_l \leq l$.
• $m_s$ is the spin magnetic quantum number and comes from the eigenvalue of $\hat{S}_z$, which is $m_s \hbar$, with values $m_s = \pm 1/2$.

if we have spin +1/2 then the electron will have a higher energy then if it was -1/2

No. In the absence of an external (magnetic) field, there is no preferred direction for spin and the eigenstates are degenerate.