Hydrogen energy levels Eₙ = −13.6 - n² eV: Worked examples

The scenario matrix

Before solving anything, let us list the complete set of case-classes. Every exam question is one of these cells (or a combination). Each worked example below is tagged with the cell(s) it covers.

Cell	Case class	What makes it tricky	Example
A	Single level, "how deep is shelf $n$ ?"	just divide by $n^{2}$	Ex 1
B	Ionization from an excited state (not ground)	must go up to $E = 0$ , not to $n = 1$	Ex 2
C	Emission (electron drops) → photon out	difference of two levels, wavelength	Ex 3
D	Absorption (electron climbs) → photon in	same $Δ E$ , but sign/direction reversed	Ex 4
E	Limiting case $n \to \infty$ (series limit)	the $1/ n^{2} \to 0$ boundary	Ex 5
F	Degenerate / "impossible" input	$n = 0$ , non-integer, or downward "emission" that can't happen	Ex 6
G	One-electron ion, hidden $Z^{2}$	the formula is not the pure hydrogen one	Ex 7
H	Real-world word problem	translate physics from a story	Ex 8
I	Exam twist: given a wavelength, find the transition	run the machine backwards	Ex 9

Two constants we reuse everywhere:

The figure below is the map we will keep pointing back to. Pin it in your mind.

Figure — Hydrogen energy levels Eₙ = −13.6 - n² eV

The worked examples

Cell A — one level

Cell B — ionization from an excited state

Worked example Ex 2 · Ionizing hydrogen already sitting in

$n = 2$ An atom is in the first excited state ( $n = 2$ ). How much energy must a photon supply to knock the electron completely free? Compare to ionizing from the ground state (that famous $13.6$ eV — see Ionization energy).

Forecast: Should it need more or less than $13.6$ eV? The electron is already higher up the well, so it should need less.

Free means $E = 0$ . Energy needed $= 0 - E_{2}$ . Why this step? Ionization = climb all the way to the top of the well, where the electron is at rest and infinitely far ( $E = 0$ ). We built this reference in the parent note.
Find $E_{2}$ . $E_{2} = - \frac{13.6}{2 ^{2}} = - 3.40 eV$ .
Subtract. $Δ E = 0 - (- 3.40) = 3.40 eV$ .

Verify: $3.40 < 13.6$ — needs less than from the ground state, as forecast. This is exactly the depth of the $n = 2$ shelf, which is the point. ✓

Cell C — emission, drop, photon out

Worked example Ex 3 · The

$n = 4 \to n = 2$ Balmer line The electron falls from $n = 4$ to $n = 2$ . Find the emitted photon's energy and wavelength. Is it visible?

Forecast: Ending at $n = 2$ is a Balmer transition, and Balmer lines are visible. Guess: a few eV, a few hundred nm.

Transition energy. $Δ E = 13.6 (\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}}) = 13.6 (\frac{1}{4} - \frac{1}{16})$ . Why this step? A photon carries the difference of two shelves, not one shelf's value. Here $n_{f} = 2$ (lower, so its $1/ n^{2}$ is the bigger term ⇒ $Δ E > 0$ , energy released).
Arithmetic. $\frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16}$ , so $Δ E = 13.6 \times \frac{3}{16} = 2.55 eV$ .
Wavelength. $λ = \frac{h c}{Δ E} = \frac{1240}{2.55} \approx 486 nm$ . Why this step? Energy released becomes light; $h c = 1240$ eV·nm converts eV to nm in one division.

Verify: $486$ nm is blue-green — inside the visible band (400–700 nm), a real Balmer line ( $H β$ ). Matches forecast. ✓

Cell D — absorption, climb, photon in

Worked example Ex 4 · Absorbing your way up:

$n = 1 \to n = 3$ A ground-state atom absorbs a photon and jumps to $n = 3$ . What photon energy and wavelength does it need? Which spectral series?

Forecast: Jumps involving $n = 1$ are the biggest gaps → high energy → short wavelength → ultraviolet (Lyman).

Same $Δ E$ machine. The atom must gain exactly the shelf difference: $Δ E = E_{3} - E_{1} = 13.6 (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) .$ Why this step? Absorption is emission run backwards. The photon must deliver precisely the gap $E_{3} - E_{1}$ ; anything else and the atom simply won't absorb it. Discrete shelves ⇒ picky about colour.
Arithmetic. $1 - \frac{1}{9} = \frac{8}{9}$ , so $Δ E = 13.6 \times \frac{8}{9} = 12.09 eV$ .
Wavelength. $λ = \frac{1240}{12.09} \approx 102.6 nm$ .

Verify: $102.6$ nm is deep UV — a Lyman line, ends/starts at $n = 1$ , huge energy. Matches forecast. Note $12.09 < 13.6$ : not enough to ionize, exactly enough to reach $n = 3$ . ✓

Cell E — the limiting case $n \to \infty$

Worked example Ex 5 · The series limit (where the lines pile up)

As $n_{i} \to \infty$ for transitions ending at $n_{f} = 2$ , what wavelength do the Balmer lines approach? This "series limit" is where the discrete lines crowd into a continuum.

Forecast: Higher and higher starting shelves sit ever closer to $E = 0$ . So $Δ E$ approaches a fixed ceiling — the full depth of the $n_{f} = 2$ shelf — giving a shortest Balmer wavelength.

Take the limit. As $n_{i} \to \infty$ , $\frac{1}{n _{i}^{2}} \to 0$ , so $Δ E \to 13.6 (\frac{1}{2 ^{2}} - 0) = \frac{13.6}{4} = 3.40 eV .$ Why this step? The starting shelf melts into the top of the well ( $E = 0$ ), so the photon simply carries away the whole depth of the $n = 2$ shelf — which is $3.40$ eV (compare Ex 2!).
Wavelength. $λ_{min} = \frac{1240}{3.40} \approx 365 nm$ .

Verify: $365$ nm is the known Balmer series limit (just past violet, into near-UV). Because it uses the largest possible $Δ E$ for this series, it is the shortest wavelength — the edge of the line pile-up. Consistency check: same $3.40$ eV as the $n = 2$ ionization energy, as it must be. ✓

Cell F — degenerate / impossible inputs

Worked example Ex 6 · When the formula refuses to answer

(a) What is $E_{0}$ ? (b) Can the electron emit a photon by going from $n = 2$ to $n = 3$ ?

Forecast: Both feel "off." $n = 0$ isn't a real shelf; and $2 \to 3$ is a climb, not a fall — you can't emit by going up.

(a) Try $n = 0$ . $E_{0} = - \frac{13.6}{0 ^{2}}$ — division by zero, undefined. Why this step? The Bohr quantization was $m v r = n ℏ$ with $n = 1, 2, 3, \dots$ . There is no $n = 0$ orbit (zero angular momentum would mean the electron falling straight into the proton). The formula's domain simply starts at $n = 1$ .
(b) Compute $Δ E$ for $2 \to 3$ . $Δ E = E_{2} - E_{3} = - 3.40 - (- 1.51) = - 1.89 eV$ . Why this step? A negative $Δ E$ means the atom would have to gain energy, not release it. So $2 \to 3$ is absorption, impossible as spontaneous emission. To emit, you need $n_{i} > n_{f}$ (a fall), giving $Δ E > 0$ .

Verify: (a) $1/0$ genuinely undefined ⇒ $n = 0$ excluded, correct. (b) $∣ Δ E ∣ = 1.89$ eV is the same size as the $3 \to 2$ emission (just opposite sign) — the ladder is symmetric in energy, only the direction flips. ✓

Cell G — one-electron ion, the hidden $Z^{2}$

Cell H — real-world word problem

Worked example Ex 8 · A hydrogen lamp and a solar cell

A hydrogen discharge lamp emits its $H α$ red line at $656$ nm. A student wants to check whether these photons can free an electron from a metal whose "work function" (minimum energy to eject an electron) is $2.5$ eV. Will the red photons work?

Forecast: Red light is fairly low-energy. The work function $2.5$ eV is a decent barrier. Guess: probably not enough.

Photon energy from wavelength. $E_{photon} = \frac{h c}{λ} = \frac{1240}{656} \approx 1.89 eV$ . Why this step? We are told a colour ( $λ$ ); the physics lever between colour and energy is $E = h c / λ$ . This is the same $1.89$ eV that produced the line ( $3 \to 2$ ) in the first place.
Compare to the barrier. $1.89 eV < 2.5 eV$ . Why this step? An electron is ejected only if a single photon carries at least the work function's worth of energy (photoelectric threshold — one photon, one electron).

Verify: $1.89 < 2.5$ , so no electron is ejected — matches forecast. Even a brighter lamp (more photons) won't help; you'd need bluer (higher-energy) photons, e.g. an $n = 4 \to 2$ line at $2.55$ eV from Ex 3, which would just clear it. ✓

Cell I — exam twist: given $λ$ , find the transition

Worked example Ex 9 · Reverse-engineering a spectral line

A hydrogen line is measured at $λ = 1216 \overset{˚}{A} = 121.6 nm$ . Identify the transition ( $n_{i} \to n_{f}$ ).

Forecast: $121.6$ nm is deep UV — very high energy — so it must involve $n_{f} = 1$ (Lyman). Probably the smallest such jump, $2 \to 1$ .

Wavelength → photon energy. $E_{photon} = \frac{1240}{121.6} \approx 10.2 eV$ . Why this step? Run the photon rule backwards: from the colour we recover the energy the transition released.
Match to a level gap. We need $13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) = 10.2$ . Try $n_{f} = 1, n_{i} = 2$ : $13.6 (1 - \frac{1}{4}) = 13.6 \times \frac{3}{4} = 10.2 eV . ✓$ Why this step? High energy ⇒ ends at $n = 1$ ; test the smallest Lyman jump first. It lands exactly.
Conclusion. The transition is $n_{i} = 2 \to n_{f} = 1$ — the Lyman- $α$ line.

Verify: $10.2$ eV back to wavelength: $1240/10.2 \approx 121.6$ nm — round-trips perfectly. UV, Lyman, as forecast. ✓

Recall Rebuild the matrix from memory

Cover the table. For each cell, name the one idea: A single level ::: divide $13.6$ by $n^{2}$ . B ionization from level $n$ ::: cost is $13.6/ n^{2}$ , i.e. climb to $E = 0$ , not to $n = 1$ . C emission ::: fall, $n_{i} > n_{f}$ , $Δ E > 0$ , photon out. D absorption ::: climb, $n_{i} < n_{f}$ , atom gains exactly the gap. E limit $n_{i} \to \infty$ ::: $1/ n_{i}^{2} \to 0$ , gives the series-limit (shortest) wavelength. F degenerate ::: $n = 0$ undefined; upward "emission" impossible ( $Δ E < 0$ ). G one-electron ion ::: multiply by $Z^{2}$ . I reverse ::: from $λ$ get $E = 1240/ λ$ , then match a level gap.

Connections

2.3.14 Hydrogen energy levels Eₙ = −13.6 - n² eV (Hinglish) — parent topic
Photon energy E = hf = hc/λ — the $λ = h c /Δ E$ tool used in Ex 3, 4, 5, 8, 9
Hydrogen spectral series — Lyman/Balmer naming used throughout
Ionization energy — Cells B and G
Rydberg constant and formula — the $Z^{2}$ generalisation in Ex 7
Bohr model of the atom — why $n$ starts at $1$ (Ex 6)
Coulomb's law — origin of the $Z^{2}$ factor
Quantum numbers — meaning of $n$