Hydrogen atom — solving in spherical coordinates: Worked examples

This page is a drill. The parent note built the machinery — quantum numbers $n, ℓ, m_{ℓ}$ , the energy ladder $E_{n} = - 13.6/ n^{2}$ eV, degeneracy $n^{2}$ . Here we hit every kind of question that machinery can throw at you: the smallest case, the biggest allowed case, the illegal case (so you learn to reject it), the limit $n \to \infty$ , a real photon in a lab, and an exam-style trap.

Before we start, four tiny reminders so no symbol is unearned:

Definition The three counters and two constants (from the parent)

$== n ==$ = principal quantum number = $1, 2, 3, \dots$ It sets the energy and the size of the atom.
$== ℓ ==$ = orbital angular-momentum quantum number = $0, 1, \dots, n - 1$ . It sets the shape and how much the electron "orbits".
$== m_{ℓ} ==$ = magnetic quantum number = $- ℓ, - ℓ + 1, \dots, + ℓ$ . It sets the tilt (which way the orbit points).
$ℏ$ = the reduced Planck constant = $h /2 π$ , the fundamental "chunk size" of angular momentum ( $\approx 1.055 \times 1 0^{- 34}$ J·s).
$μ$ = the reduced mass of the electron–proton pair, $μ = \frac{m _{e} m _{p}}{m _{e} + m _{p}} \approx m_{e}$ . It is the effective mass that turns the two-body atom into a one-body problem (see Reduced mass and two-body problem).

Read the arrows as a waterfall: $n$ decides how far $ℓ$ may climb; $ℓ$ decides how far $m_{ℓ}$ may spread. Nothing skips a level.

The scenario matrix

Every hydrogen-atom problem is one of these cells. The worked examples below are tagged with the cell they cover.

Cell	Case class	What makes it tricky	Example
A	Smallest input ( $n = 1$ )	Only one state exists — the "degenerate/trivial" end	Ex 1
B	A middling $n$ , count all states	Must sum $2 ℓ + 1$ over all $ℓ$	Ex 2
C	Illegal quantum numbers	You must reject, not compute	Ex 3
D	Energy difference (emission)	Sign of $Δ E$ , which is the photon	Ex 4
E	Energy difference (absorption)	Opposite sign, atom climbs up	Ex 5
F	Limit $n \to \infty$ (ionisation)	The ladder's top rung; series limit	Ex 6
G	Real-world word problem (photon → wavelength)	Convert eV to nanometres	Ex 7
H	Exam twist: max $ℓ$ , then angular momentum magnitude	Combine $ℓ_{m a x} = n - 1$ with $ℓ (ℓ + 1) ℏ$	Ex 8
I	Centrifugal barrier: degenerate $ℓ = 0$ (zero barrier) vs large $ℓ$	Division-by-zero edge case	Ex 9

We reference two figures: the energy ladder (Ex 4–6) and the centrifugal barrier (Ex 9).

Figure — Hydrogen atom — solving in spherical coordinates

Cell A — The smallest input

Cell B — Count all states at a middling $n$

Worked example Ex 2 — Degeneracy of

$n = 4$ How many orbital states (ignore electron spin) share the energy level $n = 4$ ?

Forecast: Odd numbers are about to appear. Guess the total before summing.

List allowed $ℓ$ : $0, 1, 2, 3$ . Why this step? $ℓ$ runs $0$ to $n - 1 = 3$ .
Count $m_{ℓ}$ for each $ℓ$ : each $ℓ$ gives $2 ℓ + 1$ values.
- $ℓ = 0 : 1$ , $ℓ = 1 : 3$ , $ℓ = 2 : 5$ , $ℓ = 3 : 7$ . Why this step? $m_{ℓ}$ spans $- ℓ$ to $+ ℓ$ , which is $2 ℓ + 1$ integers.
Add them: $1 + 3 + 5 + 7 = 16$ . Why this step? Every distinct $(ℓ, m_{ℓ})$ at this $n$ is a separate state with the same energy.

Verify: The sum of the first $n$ odd numbers equals $n^{2}$ . Here $n = 4 \Rightarrow n^{2} = 16$ . ✓ The tidy match confirms energy ignores $ℓ$ and $m_{ℓ}$ — the special "accidental" degeneracy of the pure $1/ r$ potential (see Zeeman effect, where a magnetic field breaks this and splits the 16).

Cell C — Reject the illegal case

Worked example Ex 3 — Which of these are impossible?

For each proposed triple, say if it is a legal hydrogen state, and if not, which rule it breaks. (i) $(n, ℓ, m_{ℓ}) = (2, 2, 0)$ (ii) $(3, 2, - 2)$ (iii) $(2, 1, - 2)$ (iv) $(3, 0, 0)$

Forecast: Before computing, expect at least two to be forbidden — the question is which rule.

(i) Check $ℓ \leq n - 1$ : need $ℓ \leq 1$ , but $ℓ = 2$ . Illegal. Why this step? The radial solution only terminates for $ℓ \leq n - 1$ ; $ℓ = n$ blows up.
(ii) Check both rules: $ℓ = 2 \leq n - 1 = 2$ ✓, and $∣ m_{ℓ} ∣ = 2 \leq ℓ = 2$ ✓. Legal. Why this step? Both waterfall steps satisfied.
(iii) Check $∣ m_{ℓ} ∣ \leq ℓ$ : here $∣ - 2 ∣ = 2 > ℓ = 1$ . Illegal. Why this step? $m_{ℓ}$ cannot exceed $ℓ$ in magnitude — a full turn's single-valued condition caps it.
(iv) Check both: $ℓ = 0 \leq 2$ ✓, $m_{ℓ} = 0$ (only choice) ✓. Legal.

Verify: Legal ones are (ii) and (iv). Sanity: for (ii) the state is $3 d$ ; $d$ means $ℓ = 2$ , and $n = 3$ allows up to $ℓ = 2$ , so a $3 d$ orbital exists — matches reality. ✓ For (iii), a $2 p$ shell only has $m_{ℓ} \in {- 1, 0, + 1}$ , so $m_{ℓ} = - 2$ genuinely can't exist. ✓

Cell D — Emission (energy comes out)

Worked example Ex 4 — The

$H α$ line, $n = 3 \to n = 2$ Find the energy released when the electron drops from $n = 3$ to $n = 2$ . Which way does energy flow?

Forecast: Falling to a lower rung — does the atom gain or lose energy? Guess the sign.

Compute both levels. $E_{3} = - 13.6/9 = - 1.511$ eV, $E_{2} = - 13.6/4 = - 3.400$ eV. Why this step? Each rung's height is $- 13.6/ n^{2}$ ; we need both to take a difference.
Take $Δ E = E_{final} - E_{initial} = E_{2} - E_{3}$ . $Δ E = - 3.400 - (- 1.511) = - 1.889 eV .$ Why this step? Final minus initial tells us how the atom's energy changed. Negative ⇒ atom lost energy.
Photon energy = $∣Δ E ∣ = 1.889$ eV. Why this step? Energy is conserved: what the atom loses, the photon carries away.

Verify: On the ladder figure (s01), the arrow from $n = 3$ to $n = 2$ points down and is short. $∣Δ E ∣ = 1.89$ eV lands in the red visible band — this is the famous red $H α$ line. ✓ Positive photon energy, negative $Δ E$ for the atom — consistent. ✓

Cell E — Absorption (energy goes in)

Worked example Ex 5 — Climbing

$n = 1 \to n = 4$ A photon is absorbed, kicking the electron from ground state to $n = 4$ . What photon energy is required?

Forecast: Climbing up the ladder — sign of $Δ E$ for the atom this time?

Compute both levels. $E_{1} = - 13.6$ eV, $E_{4} = - 13.6/16 = - 0.850$ eV. Why this step? Same ladder formula, different rungs.
$Δ E = E_{4} - E_{1} = - 0.850 - (- 13.6) = 12.750$ eV. Why this step? Final minus initial. Positive now — the atom gained energy.
Photon must supply exactly $12.75$ eV. Why this step? Absorption only happens if the incoming photon matches the gap exactly (that's why absorption spectra are sharp lines).

Verify: Emission from $4 \to 1$ would release the same $12.75$ eV — reversing an arrow flips the sign but not the magnitude. ✓ Bigger jump than Ex 4's $1.89$ eV, as expected for a longer climb. ✓

Cell F — The limit $n \to \infty$ (ionisation)

Worked example Ex 6 — Ionisation energy from the ground state

How much energy frees the electron completely, starting from $n = 1$ ?

Forecast: "Completely free" means which final rung? Guess the number the ladder approaches.

Identify the top of the ladder. As $n \to \infty$ , $E_{n} = - 13.6/ n^{2} \to 0$ . Why this step? A free (unbound) electron has zero binding energy — that's the ceiling the rungs crowd up toward.
Take the difference to that limit. $Δ E = E_{\infty} - E_{1} = 0 - (- 13.6) = 13.6$ eV. Why this step? Ionisation = lift the electron from its rung all the way to $E = 0$ .
State result: the ionisation energy of ground-state hydrogen is $13.6$ eV. Why this step? This single number is exactly $∣ E_{1} ∣$ — the depth of the deepest well.

Verify: On the ladder (s01), the rungs bunch up and pile against the $E = 0$ line — the gaps shrink like $1/ n^{2}$ , confirming they converge. ✓ The measured ionisation energy of hydrogen is $13.6$ eV. ✓ Note this equals the Bohr model of hydrogen prediction — the full Schrödinger solution reproduces Bohr's number.

Cell G — Real-world word problem (photon → wavelength)

Worked example Ex 7 — What colour is the

$H α$ line? The $n = 3 \to 2$ transition emits a $1.889$ eV photon (Ex 4). Find its wavelength in nanometres.

Forecast: Visible light runs roughly $400$ – $700$ nm. Guess before computing whether $H α$ is blue or red.

Recall the photon relation. A photon's energy and wavelength obey $E = \frac{h c}{λ}$ , so $λ = \frac{h c}{E}$ . Why this step? Energy and wavelength are inversely linked — more energy, shorter wave. We invert to get $λ$ .
Use a handy constant. $h c = 1240$ eV·nm. Why this step? This packaged value lets us divide eV directly and get nm — no unit gymnastics.
Divide: $λ = \frac{1240 eV\cdotnm}{1.889 eV} = 656.4 nm$ . Why this step? Plug the photon energy from Ex 4 straight in.

Verify: Units: $\frac{eV\cdotnm}{eV} = nm$ . ✓ $656$ nm sits in the red part of the visible spectrum — matching the observed deep-red hydrogen line astronomers use to map glowing gas. ✓

Cell H — Exam twist: max $ℓ$ and angular momentum magnitude

Worked example Ex 8 — Largest orbital angular momentum at

$n = 3$ Within $n = 3$ , what is the biggest allowed $ℓ$ , and what is the actual magnitude $∣ L ∣$ of the angular momentum for that state?

Forecast: A common trap says $∣ L ∣ = ℓ ℏ$ . Guess whether that's right before step 3.

Max $ℓ$ . $ℓ_{m a x} = n - 1 = 2$ . Why this step? The waterfall rule caps $ℓ$ one below $n$ .
Recall the magnitude formula (from the parent's $\hat{L}^{2}$ eigenvalue). $\hat{L}^{2} Y = ℓ (ℓ + 1) ℏ^{2} Y$ , so the length is $∣ L ∣ = ℓ (ℓ + 1) ℏ$ . (Here $ℏ$ is the reduced Planck constant, the chunk size of angular momentum.) Why this step? The measurable magnitude is the square root of the $\hat{L}^{2}$ eigenvalue — not $ℓ ℏ$ .
Plug $ℓ = 2$ : $∣ L ∣ = 2 \cdot 3 ℏ = 6 ℏ \approx 2.449 ℏ$ . Why this step? Direct substitution into the earned formula.

Verify: $6 ℏ \approx 2.449ℏ$ is larger than the naive $ℓ ℏ = 2ℏ$ — the trap answer under-counts. ✓ See Angular momentum operators and Spherical harmonics for why $ℓ (ℓ + 1)$ , not $ℓ$ , is the true length. Also $ℓ = 2$ is a $d$ -orbital, matching "3d exists" from Ex 3. ✓

Cell I — Centrifugal barrier: the $ℓ = 0$ zero and division-by-zero edge case

Here is the effective potential the parent note built. The extra piece beyond Coulomb attraction is the centrifugal barrier: $V_{cf} (r) = \frac{ℏ ^{2} ℓ ( ℓ + 1 )}{2 μ r ^{2}},$ where $ℏ$ is the reduced Planck constant and $μ$ is the reduced mass of the electron–proton pair (both recalled in the opening definition box). Notice the barrier's only dependence on the state is through the factor $ℓ (ℓ + 1)$ — the constants $ℏ, μ$ and the radius $r$ are common to every state, so they cancel in any ratio.

Worked example Ex 9 — The

$ℓ = 0$ zero, and why a naive ratio breaks At a fixed radius $r$ , compare the centrifugal barrier for $ℓ = 0$ , $ℓ = 1$ , and $ℓ = 2$ . The task asks for the ratio "going from $ℓ = 0$ to $ℓ = 2$ " — handle it carefully.

Forecast: For $ℓ = 0$ the barrier is... something special. Guess it before step 1, then guess what the requested ratio does.

Evaluate the state-factor at $ℓ = 0$ . $ℓ (ℓ + 1) = 0 \cdot 1 = 0$ , so $V_{cf} (ℓ = 0) = 0$ exactly, at every radius. Why this step? An $s$ -state ( $ℓ = 0$ ) has no angular motion, hence no centrifugal barrier — the electron can sit right at the nucleus. This is the degenerate/zero case.
Evaluate for $ℓ = 1$ and $ℓ = 2$ . $ℓ (ℓ + 1) = 1 \cdot 2 = 2$ and $2 \cdot 3 = 6$ . Why this step? Higher $ℓ$ carries more angular motion, so a larger barrier factor.
Form the requested ratio $\frac{V _{cf} ( ℓ = 2 )}{V _{cf} ( ℓ = 0 )}$ — and stop. The denominator is $0$ , so this ratio is undefined (it diverges to $+ \infty$ ). You cannot report a finite number here. Why this step? Division by zero is not "a big number" — it is a genuine edge case. Since $ℓ = 0$ gives literally zero barrier, "how many times bigger" has no answer: any positive barrier is infinitely larger than none. The honest statement is: $ℓ = 2$ has a barrier where $ℓ = 0$ has exactly none.
Give a well-defined comparison instead. To compare two nonzero barriers, ratio $ℓ = 2$ against $ℓ = 1$ : $\frac{V _{cf} ( ℓ = 2 )}{V _{cf} ( ℓ = 1 )} = \frac{6}{2} = 3$ . Why this step? Both denominators are nonzero, so the constants and $r$ cancel cleanly and we get a finite, meaningful factor: the $ℓ = 2$ wall is $3 \times$ taller than the $ℓ = 1$ wall at every radius.

Verify: On the barrier figure (s02), the $ℓ = 0$ curve is pure attraction (no bump — the " $0$ " case), while $ℓ = 1$ and $ℓ = 2$ rise steeply near the origin, with $ℓ = 2$ three times higher than $ℓ = 1$ at each $r$ . ✓ The $ℓ = 2$ -to- $ℓ = 0$ ratio has no finite value because you'd divide by that flat-zero curve — exactly what step 3 warned. ✓ Bigger $ℓ$ ⇒ stronger barrier ⇒ wavefunction pushed outward, as the parent's skater analogy said. ✓

Recall Self-test (try to answer, then reveal)

Q — Number of states at $n = 4$ ? A — $n^{2} = 16$ .

Q — Photon energy for $3 \to 2$ ? A — $1.89$ eV (red $H α$ , $656$ nm).

Q — Ionisation energy from ground state? A — $13.6$ eV.

Q — Magnitude of $L$ for $ℓ = 2$ ? A — $6 ℏ \approx 2.45ℏ$ .

Q — Centrifugal barrier for an $s$ -state, and the $ℓ = 2$ -to- $ℓ = 0$ ratio? A — Zero ( $ℓ = 0$ ); the ratio is undefined (divides by zero).