Exercises — Quantum numbers n, l, mₗ, mₛ

WHAT: we list $l$ from the rule $l=0,1,\dots ,(n-1)$. WHY: $l$ stops one short of $n$ because the radial wave needs at least one unit of "node budget" to be a bound state. With $n=4$: $l=0,1,2,3$ — that is the s, p, d, f sub-shells. Four values.

$m_l$ runs $-l$ to $+l$ in integer steps: $m_l=-3,-2,-1,0,+1,+2,+3$. Count $=2l+1=2(3)+1=\boxed{7}$ values. WHY negatives appear: $m_l$ is the projection $L_z/\hslash$ of a vector on the $z$-axis — a projection can point down (negative) just as easily as up.

$m_s=+\frac{1}{2}$ and $m_s=-\frac{1}{2}$. These do not come from the Schrödinger equation. They were forced on us by the Stern-Gerlach Experiment (a beam of silver atoms split into exactly two spots) and later derived by Dirac's relativistic equation. See Angular Momentum in Quantum Mechanics.

WHAT: sum $2(2l+1)$ over $l=0,1,2$. $l=0$: $2(1)=2$. $l=1$: $2(3)=6$. $l=2$: $2(5)=10$. Total $=2+6+10=18$. Check: $2n^2=2(3{)}^2=18$. ✓ WHY the factor 2: each $(n,l,m_l)$ orbital holds two electrons, one $m_s=+\frac{1}{2}$ and one $m_s=-\frac{1}{2}$ (Pauli Exclusion Principle).

$4f\Rightarrow n=4,l=3$. $|\vec{L}|=\sqrt{l(l+1)}\hslash =\sqrt{3\cdot 4}\hslash =\sqrt{12}\hslash =2\sqrt{3}\hslash \approx 3.464\hslash .$ WHY $\sqrt{l(l+1)}$ and not $l$: Bohr's old $L=l\hslash$ is wrong (that is Mistake B in the parent). Solving the eigenvalue equation gives $|\vec{L}{|}^2=l(l+1){\hslash }^2$.

$3d\Rightarrow l=2$, so $m_l=-2,-1,0,+1,+2$. $L_z=m_l\hslash$, so $L_z\in \{-2\hslash ,-\hslash ,0,+\hslash ,+2\hslash \}$. Note the largest $L_z=2\hslash$ is smaller than $|\vec{L}|=\sqrt{6}\hslash \approx 2.449\hslash$ — the vector never points fully along $z$ (space quantization). See the figure below.

(a) Invalid. $l=2$ but max is $n-1=1$. Rule broken: $l\le n-1$. (b) Valid. $l=2\le 2$ ✓, $m_l=-2$ lies in $[-2,2]$ ✓, $m_s=-\frac{1}{2}$ ✓. (c) Invalid. $l=0\Rightarrow m_l$ can only be $0$, but here $m_l=-1$. Rule broken: $|m_l|\le l$. (d) Invalid. $m_s$ must be $\pm \frac{1}{2}$, never $0$. WHY these caps exist: $l\le n-1$ and $|m_l|\le l$ both come from demanding the wavefunction not blow up (at $r\to \infty$ and at the poles $\theta =0,\pi$ respectively) — the boundary conditions from solving Schrödinger's equation.

WHAT: solve $\sqrt{l(l+1)}=\sqrt{6}$, so $l(l+1)=6\Rightarrow l=2$. WHY smallest $n$: we need $l\le n-1$, i.e. $n\ge l+1=3$. Smallest is $n=3$. $l=2$ is the d sub-shell, so the answer is $3d$.

WHY this formula: $L_z$ is the adjacent side (projection onto $z$) and $|\vec{L}|$ is the hypotenuse of the right triangle formed by $\vec{L}$ and the $z$-axis — see the figure. So $\cos \theta =\text{adjacent}/\text{hypotenuse}=L_z/|\vec{L}|$. $\cos \theta =\frac{m_l\hslash }{\sqrt{l(l+1)}\hslash }=\frac{2}{\sqrt{6}}\approx \mathrm{0.8165.}$ $\theta =\arccos (0.8165)\approx 35.26^{\circ }.$ Sanity check: $\theta \ne 0$, confirming the vector cannot lie exactly on the axis even at maximum $m_l$.

WHAT: fix $n=3,l=1$ (the $3p$ sub-shell). Free labels left: $m_l$ and $m_s$. $m_l\in \{-1,0,+1\}$ → 3 orbitals. Each holds 2 spins. $3\times 2=\boxed{6}$ electrons. WHY: by Pauli, each of the 6 distinct $(m_l,m_s)$ combinations is one unique seat.

$1s\Rightarrow n=1,l=0,m_l=0$ for both electrons. Electron A: $(1,0,0,+\frac{1}{2})$. Electron B: $(1,0,0,-\frac{1}{2})$. Only $m_s$ differs. If it did not, all four numbers would match — forbidden by Pauli. This is exactly why an orbital caps at 2 electrons.

WHAT: solve $2n^2=32\Rightarrow n^2=16\Rightarrow n=4$. Sub-shells: $l=0,1,2,3$ → $4s,4p,4d,4f$. Check the count: $2(1)+2(3)+2(5)+2(7)=2+6+10+14=32$ ✓ (see Electron Configuration & Periodic Table for why filling order differs from this raw capacity).

WHAT: impose the loop-matching condition. $\Phi (\varphi +2\pi )=e^{i{m}_l(\varphi +2\pi )}=e^{i{m}_l\varphi }{e}^{i{m}_{l}2\pi }=\Phi (\varphi )\cdot e^{i{m}_{l}2\pi }.$ For this to equal $\Phi (\varphi )$ we need $e^{i{m}_{l}2\pi }=1$. WHY: by Euler's relation $e^{ix}=\cos x+i\sin x$, we have $e^{i{m}_{l}2\pi }=1$ only when $\cos (2\pi m_l)=1$ and $\sin (2\pi m_l)=0$, i.e. when $2\pi m_l$ is a whole number of full turns: $m_l\in \{\dots ,-1,0,1,\dots \}$. Physical nonsense of non-integer $m_l$: the electron wave would come back to the same point in space with a different value, so the cloud's density would be two-valued at one location — impossible.

WHAT: compute $\sum _{l=0}^{n-1}2(2l+1)=2\sum _{l=0}^{n-1}(2l+1)$. The inner sum is the sum of the first $n$ odd numbers $1+3+5+\cdots +(2n-1)$, which is a classic identity equal to $n^2$. WHY $n^2$: stacking odd numbers builds a perfect square, layer by layer (an $L$-shaped shell of $2k+1$ dots added to a $k\times k$ square gives $(k+1)\times (k+1)$). See the figure. Therefore total $=2\cdot n^2=2n^2$. $■$

Bohr: $L=n\hslash =1\cdot \hslash =\hslash$ (non-zero). Schrödinger: $|\vec{L}|=\sqrt{l(l+1)}\hslash =\sqrt{0\cdot 1}\hslash =0$. Which is right: Schrödinger. An $s$-electron has a perfectly spherical probability cloud with no preferred rotation axis, so it cannot carry orbital angular momentum. Spectroscopic fine-structure and the hydrogen spectrum confirm $l=0$ states with zero orbital angular momentum. Bohr's $n\hslash$ rule was a lucky approximation for energies but wrong for angular momentum.

• Max electrons in $n=4$? ::: $2n^2=32$.
• $|\vec{L}|$ for a $3d$ electron? ::: $\sqrt{6}\hslash$.
• Angle of the $m_l=+2$ state ($l=2$) from $z$? ::: $\arccos (2/\sqrt{6})\approx 35.26^{\circ }$.
• Why is $(4,0,-1,+\frac{1}{2})$ invalid? ::: $l=0$ forces $m_l=0$, so $m_l=-1$ is forbidden.
• What boundary condition quantises $m_l$? ::: $\Phi (\varphi +2\pi )=\Phi (\varphi )$ ⇒ $e^{i{m}_{l}2\pi }=1$ ⇒ integer $m_l$.