Exercises — Quantum numbers — n (principal), l (azimuthal), mₗ (magnetic), mₛ (spin)

WHAT the label means: the leading number is $n$, the letter is the code for $l$. The code: $s\to l=0$, $p\to l=1$, $d\to l=2$, $f\to l=3$. Read off: "4p" ⇒ $n=4$ and $p\Rightarrow l=1$. Answer: $\boxed{n=4,l=1}$.

WHY $(2l+1)$: $m_l$ runs in integer steps from $-l$ to $+l$, which is $2l+1$ values. For d: $l=2$, so count $=2(2)+1=5$. List: $m_l=-2,-1,0,+1,+2$. Answer: $\boxed{5}$ values.

WHAT $m_s$ is: the intrinsic spin projection, fixed to only two settings. Answer: $\boxed{2}$ choices, $m_s=+\frac{1}{2}$ or $-\frac{1}{2}$.

Check $l\le n-1$: $l=2$, $n-1=2$. $2\le 2$ ✓ (this is a 3d electron). Check $|m_l|\le l$: $|-2|=2\le 2$ ✓. Check $m_s$: $+\frac{1}{2}$ is allowed ✓. Answer: $\boxed{\text{Valid}}$ — a 3d electron with $m_l=-2$, spin up.

WHY sum over $l$: each subshell $l$ contributes $2l+1$ orbitals, and $l$ runs $0\to n-1$. For $n=4$: $l=0,1,2,3$ give $1+3+5+7$. $1+3+5+7=16=n^2=4^2.$ Answer: $\boxed{16}$ orbitals (a check on the "sum of first $n$ odd numbers $=n^2$" fact).

WHY factor of 2: every orbital holds two electrons (opposite spins) by Pauli exclusion principle. $\text{electrons}=2n^2=2(3{)}^2=18.$ Answer: $\boxed{18}$ electrons.

WHY this formula: the magnitude of orbital angular momentum is $|\vec{L}|=\sqrt{l(l+1)}\hslash$, not $l\hslash$ — quantum mechanics fixes the length this way. For p: $l=1$, so $\sqrt{1(1+1)}=\sqrt{2}$. Answer: $\boxed{|\vec{L}|=\sqrt{2}\hslash \approx 1.414\hslash }$.

Magnitude: $|\vec{L}|=\sqrt{2(3)}\hslash =\sqrt{6}\hslash \approx 2.449\hslash$. Max projection: biggest $m_l=+2$, so $L_z=2\hslash$. Why $\cos \theta =L_z/|\vec{L}|$: $L_z$ is the shadow of the vector on $z$; the shadow over the hypotenuse (full length) is the cosine of the tilt angle — the same right-triangle idea as orbital geometry. $\cos \theta =\frac{2}{\sqrt{6}}=0.8165\Rightarrow \theta =\arccos (0.8165)\approx 35.26^{\circ }.$ What it means: even at "most aligned," $\vec{L}$ tilts $\approx 35.3^{\circ }$ off $z$ — it can never point straight up because $|\vec{L}|>L_z^{\max }$. See the cone below.

The rule that governs $l$: $l\le n-1$. This comes from the radial equation only giving finite, normalizable solutions when angular motion doesn't overspend the shell's budget. Test: $n=2\Rightarrow l_{\max }=n-1=1$. A "d" needs $l=2>1$. Answer: $\boxed{\text{"2d" is forbidden}}$ — the $n=2$ shell holds only $s(l=0)$ and $p(l=1)$.

WHY it splits: $m_l$ sets the orbital's orientation; a field picks a special direction, so each $m_l$ now has a slightly different energy. Count: $l=1$ gives $m_l=-1,0,+1$ → three values. Answer: $\boxed{3}$ levels. This is the meaning of "magnetic quantum number."

Fixed for all: $n=2$, $l=1$ (that's "2p"). Hund's rule: singly occupy each $m_l$ with parallel spins before pairing.

• Electron 1: $(2,1,+1,+\frac{1}{2})$
• Electron 2: $(2,1,0,+\frac{1}{2})$
• Electron 3: $(2,1,-1,+\frac{1}{2})$

Why not pair up first? Parallel spins in separate orbitals minimise electron–electron repulsion, lowering total energy. All three share $n,l,m_s$ but each has a distinct $m_l$, so no two share all four numbers — Pauli is satisfied.

WHY this is $2n^2$: total states $=2\sum _{l=0}^{n-1}(2l+1)$. For $n=3$: $l=0,1,2$ give $(2l+1)=1,3,5$. $2(1+3+5)=2(9)=18.$ Answer: $\boxed{18}$ distinct addresses $=2n^2=2(3{)}^2$. Each is one electron slot in the $n=3$ shell.

Set up: slots $=2(2l+1)=10\Rightarrow 2l+1=5\Rightarrow l=2$. Decode: $l=2$ is the d subshell; orbitals $=2l+1=5$. Answer: $\boxed{l=2,\text{d},5\text{orbitals}}$.

Subshells ($l=0\to 3$):

Sum orbitals: $1+3+5+7=16=n^2$. Sum capacity: $2+6+10+14=32=2n^2=2(4{)}^2$. ✓ Answer: $\boxed{32}$ electrons total.

Fixed magnitude: $|\vec{L}|=\sqrt{1(2)}\hslash =\sqrt{2}\hslash \approx 1.414\hslash$. Use $\cos \theta =\frac{m_l}{\sqrt{l(l+1)}}=\frac{m_l}{\sqrt{2}}$.

• $m_l=+1$: $\cos \theta =\frac{1}{\sqrt{2}}=0.7071\Rightarrow \theta =45^{\circ }$ (tilts up toward $+z$).
• $m_l=0$: $\cos \theta =0\Rightarrow \theta =90^{\circ }$ (lies in the $xy$-plane).
• $m_l=-1$: $\cos \theta =-\frac{1}{\sqrt{2}}=-0.7071\Rightarrow \theta =135^{\circ }$ (tilts down toward $-z$).

All cases covered: the vector points above ($45^{\circ }$), flat ($90^{\circ }$), and below ($135^{\circ }$) — never along $z$. This mirrors the cone picture at every allowed setting.

No field (degeneracy): in hydrogen, energy depends only on $n$, so all orbital states of $n=3$ share one energy. Count them: ${\sum }_{l=0}^2(2l+1)=1+3+5=9$ orbital states. In a field: each state now labelled by its $m_l$. Across the whole shell $m_l$ ranges over $\{-2,-1,0,+1,+2\}$ (the widest comes from $l=2$). Distinct $m_l$ energies $=5$. Why the numbers differ: $9$ counts orbitals (all degenerate without a field); $5$ counts distinct energies after the field, since orbitals with the same $m_l$ (e.g. the $m_l=0$ of 3s, 3p, 3d) share one Zeeman shift. Answers: $\boxed{9}$ degenerate orbital states; $\boxed{5}$ distinct $m_l$ energies.

Cover the answers. Capacity of $n=5$ shell? ::: $2n^2=2(25)=50$ $|\vec{L}|$ for an f electron ($l=3$)? ::: $\sqrt{3\cdot 4}\hslash =\sqrt{12}\hslash =2\sqrt{3}\hslash$ Angle of a d electron's most-aligned $\vec{L}$ from $z$? ::: $\arccos (2/\sqrt{6})\approx 35.3^{\circ }$ Distinct $m_l$ energies for the whole $n=4$ shell in a field? ::: $7$ (from $m_l=-3\cdots +3$)