A full lesson: a cartoon movie to start, then the energy ladder of hydrogen, ground and excited states, ionisation, emission and absorption of photons, the Rydberg formula, the Lyman, Balmer, Paschen, Brackett and Pfund series, counting spectral lines, NEET shortcuts, problems to solve yourself, and 12 MCQs.
Each colour of a firework is light from electrons jumping down between fixed energy levels. One jump gives one photon of one exact colour, so every element shows a barcode of lines. Hydrogen's barcode is the simplest of all.
On Diwali night Dadi asks a simple question: why does one anar burst red and another green? The answer is inside the atoms. Firework makers add strontium salts for red, barium salts for green and copper salts for blue. The heat pushes electrons in these atoms up to higher energy levels. When they fall back down, each fall releases one packet of light, a photon, whose energy exactly equals the drop. Each element has its own fixed drops, so it gives its own fixed colours.
Try at home. Hold an old CD or DVD near your eye and look at a white LED torch or a filament bulb in its reflection: you see a smooth rainbow (a continuous spectrum). Now look at a street lamp, a CFL or a tube light the same way: you see separate bright stripes with dark gaps. That is a line spectrum, the barcode of the gas inside. Hydrogen, the simplest atom, has the simplest barcode, and Bohr's model explains every line of it.
"Remember one sentence for the whole lesson: one jump, one photon, one colour. The photon's energy is exactly the gap between the two levels."
"If the levels were not fixed, and an electron could have any energy at all, what would the spectrum of a hot gas look like?"
It would be a smooth rainbow with every colour, like a hot filament. We see sharp lines with gaps, so the electron's energy can take only certain values. Those allowed values are the energy levels.
In Bohr's model the electron of hydrogen can only have the energies En = −13.6 / n² eV, n = 1, 2, 3, … The lowest rung, n = 1, is the ground state. Higher rungs are excited states. The top, n → ∞, is E = 0: the electron is free. Tap or drag on the ladder (or use the slider) to put the electron on any level. Press Ionise to set it free.
| Level | Energy Eₙ | To ionise from here | To reach it from n = 1 |
|---|---|---|---|
| n = 1 (ground) | −13.6 eV | 13.6 eV | 0 |
| n = 2 (1st excited) | −3.4 eV | 3.4 eV | 10.2 eV |
| n = 3 (2nd excited) | −1.51 eV | 1.51 eV | 12.09 eV |
| n = 4 | −0.85 eV | 0.85 eV | 12.75 eV |
| n → ∞ | 0 | 0 | 13.6 eV |
The minimum energy to free the electron from the ground state. Ionisation potential = 13.6 V.
Ground to n = 2 is 10.2 eV: the first excitation potential is 10.2 V.
Switch to true scale: the levels crowd together just below 0 eV.
"The minus sign means the electron is bound. 13.6 eV is the ionisation energy only from the ground state. From n = 2 you need just 3.4 eV, the size of that level's energy."
"At room temperature, almost all hydrogen atoms are in which level? Why?"
The ground state, n = 1. The first excitation needs 10.2 eV, while the typical thermal energy at room temperature is only about 0.025 eV. That is why cold hydrogen absorbs only the Lyman lines (Stage 5).
When the electron drops from level nᵢ to a lower level nf, the atom emits one photon with hν = Enᵢ − Enf. In absorption, an atom takes in a photon only if its energy exactly equals the gap between two levels (or is enough to ionise it). Choose a mode. In Absorb, set the photon energy and fire it at a ground-state atom. In Emit, set the starting level and watch the electron fall.
Electron falls to a lower level. Bright lines on a dark background: the emission spectrum of a hot gas.
White light through a cool gas: dark lines at the same frequencies. The atom cannot take half a photon.
Any photon above the ionisation energy is absorbed: it frees the electron and the extra becomes kinetic energy.
"An 11 eV photon is more than 10.2 eV, but less than 12.09 eV. It matches no gap, so a ground-state hydrogen atom simply ignores it. Bohr's explanation of these lines won him the 1922 Nobel Prize."
"Cool hydrogen gas absorbs a 12.09 eV photon. The atom then de-excites. Which photon energies can come out?"
The electron is now on n = 3. It can fall 3 → 1 (12.09 eV) or 3 → 2 (1.89 eV, red) and then 2 → 1 (10.2 eV). Three possible photon energies, and they are emitted in all directions.
Put the photon energy hν = hc/λ into Bohr's level formula and the wavelength of every hydrogen line falls out. Choose the lower level nf and the upper level nᵢ. The arrow on the ladder and the marker on the wavelength scale move together.
Both energies are negative; the upper one is less negative, so the difference is positive.
The constant 13.6 eV / hc works out to 1.097 × 10⁷ m⁻¹.
This is the formula Balmer, Rydberg and others found from experiment. Bohr's model predicts R from e, m, h and ε₀, and it matches.
For 3 → 2: ΔE = 1.89 eV, so λ = 1240 / 1.89 ≈ 656 nm, the red H-alpha line.
"Bigger jump, bigger energy, shorter wavelength. And because the upper levels crowd together, the lines of a series crowd together too, towards a series limit."
All lines that end on the same lower level form a series. Pick a series and watch its lines appear one by one, from the first line (longest wavelength, nᵢ = nf + 1) to the series limit (shortest wavelength, nᵢ → ∞). Slide to add more lines.
| Series | Ends on | Region | First line | Series limit |
|---|---|---|---|---|
| Lyman | n_f = 1 | ultraviolet | 121.5 nm | 91.2 nm |
| Balmer | n_f = 2 | visible | 656.3 nm | 364.6 nm |
| Paschen | n_f = 3 | infrared | 1875 nm | 820.4 nm |
| Brackett | n_f = 4 | infrared | 4051 nm | 1458 nm |
| Pfund | n_f = 5 | infrared | 7458 nm | 2279 nm |
"Remember the order with 'Lazy Boys Play Badminton Pretty': Lyman 1, Balmer 2, Paschen 3, Brackett 4, Pfund 5. Only Balmer is visible. Balmer found his formula first, in 1885, precisely because those lines can be seen."
"The Balmer series is called visible, but its series limit is 364.6 nm. Is every Balmer line visible?"
No. The first four lines (656, 486, 434, 410 nm) are the bright visible ones. The next lines (397, 389, 384 nm …) sit at the violet edge where the eye barely responds, and then the series runs into the near ultraviolet, ending at 364.6 nm.
A discharge tube holds billions of atoms. If they are excited up to level n, different atoms fall by different routes, and together they show every possible line. A single atom takes only one route. Choose n, then compare a whole gas with one atom.
Every pair of levels gives one line. From n = 4: 4 × 3 / 2 = 6 lines.
Falling one step at a time, n → n−1 → … → 1, gives the most photons from one atom.
Glowing gas clouds in space look red because of the 3 → 2 Balmer line. Dark lines in sunlight (absorption) tell astronomers which elements are in the Sun.
"Read the question carefully. 'A sample of hydrogen atoms' means count all pairs, n(n − 1)/2. 'A single hydrogen atom' means at most n − 1 photons."
Learn the first five. Excitation from n = 1: 10.2, 12.09, 12.75, 13.06 eV.
10.2 eV → 121.6 nm, 1.89 eV → 656 nm. In ångström use 12400.
Lyman 91.2 nm, Balmer 4 × 91.2 = 364.6 nm, Paschen 9 × 91.2 = 820.4 nm.
Lyman 4/(3R) = 121.5 nm, Balmer 36/(5R) = 656.3 nm, Paschen 144/(7R) = 1875 nm.
In general λ_max / λ_min = (n+1)² / (2n + 1) for n = n_f.
From level n₂ down to n₁ only: N = (n₂ − n₁)(n₂ − n₁ + 1)/2.
In the ground state: KE = 13.6 eV, PE = −27.2 eV, total −13.6 eV.
Ground-state atoms can only jump up from n = 1, so the absorption spectrum of cold hydrogen shows only the Lyman series.
"In NEET, never use the full Rydberg formula when a ratio will do. For a ratio of two wavelengths, R cancels: just compare the brackets (1/n_f² − 1/nᵢ²) and flip them."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute. Use R = 1.097 × 10⁷ m⁻¹ and hc = 1240 eV nm.
The electron of a hydrogen atom is in the n = 3 level. Find (a) its total energy, (b) the energy needed to ionise the atom from this level, (c) the energy that was needed to excite it from the ground state.
Eₙ = −13.6/n² eV. Ionising means reaching 0 eV.
Find the wavelength of the photon emitted when the electron in hydrogen falls from n = 4 to n = 2. Which series and region is it in?
1/λ = R(1/2² − 1/4²).
Find (a) the shortest wavelength of the Paschen series and (b) the longest wavelength of the Lyman series.
Shortest: nᵢ → ∞. Longest: nᵢ = n_f + 1.
Hydrogen gas in the ground state is exposed to photons of energy 12.75 eV. (a) Up to which level are the atoms excited? (b) How many different lines can be emitted afterwards? (c) How many of them are visible?
Find n with 13.6(1 − 1/n²) = 12.75. Then list every pair of levels.
Find the ratio of the longest wavelength of the Lyman series to the longest wavelength of the Balmer series.
λ ∝ 1 / (1/n_f² − 1/nᵢ²). R cancels in a ratio.
Take Eₙ = −13.6/n² eV, R = 1.097 × 10⁷ m⁻¹ and hc = 1240 eV nm. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "ionisation energy from level n" | 13.6 / n² eV | Answering 13.6 eV for an excited state |
| "excitation energy" or "first excitation potential" | 13.6 (1 − 1/n²) eV; 10.2 V | Giving the level energy (−3.4 eV) instead of the gap |
| "first line" or "longest wavelength" of a series | nᵢ = n_f + 1 | Putting nᵢ = ∞ (that is the shortest) |
| "series limit" or "shortest wavelength" | nᵢ → ∞, λ = n_f² / R | Thinking the limit is the longest line |
| "visible lines of hydrogen" | Balmer (n_f = 2) | Picking Lyman because it is "the first series" |
| "number of spectral lines" from level n | n(n − 1)/2 (a sample) | Using n − 1, which is for one atom only |
| "photon of E eV falls on ground-state hydrogen" | E must equal a gap, or be ≥ 13.6 eV | Letting the atom absorb part of the photon |
| "absorption spectrum of cold hydrogen" | only Lyman lines | Expecting the visible Balmer lines |
| "ratio of wavelengths" | ratio of the brackets, flipped | Forgetting λ ∝ 1/bracket and not flipping |
Lesson 5 asks why only these levels are allowed. De Broglie's answer: the electron is a wave that must fit around its orbit, 2πr = nλ. That one picture gives Bohr's rule L = nh/2π, and also shows where the Bohr model stops working.
Values used: Eₙ = −13.6/n² eV, R = 1.097 × 10⁷ m⁻¹, hc = 1240 eV nm, visible light ≈ 380–750 nm. Back to the chapter hub.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.