Bohr Model Calculations: Energy Levels, Radii and Transitions

The Bohr model is no longer a complete description of atoms, but for hydrogen and other one-electron species, its equations give exactly the right energies. That makes it ideal for practising calculations with energy levels, photons and spectra. This article collects the key equations and works through the main types of problem.

Constants you’ll need

Constant	Symbol	Value
Planck constant	h	6.626 × 10⁻³⁴ J s
Speed of light	c	2.998 × 10⁸ m s⁻¹
Avogadro constant	N_A	6.022 × 10²³ mol⁻¹
Electronvolt	eV	1.602 × 10⁻¹⁹ J
Rydberg energy (hydrogen)	R_H	2.18 × 10⁻¹⁸ J (13.6 eV)
Bohr radius	a₀	5.29 × 10⁻¹¹ m (52.9 pm)

The key equations

Energy of level n (hydrogen): Eₙ = −R_H ÷ n² = −2.18 × 10⁻¹⁸ J ÷ n²

Energy of level n (one-electron ion with nuclear charge Z): Eₙ = −R_H × Z² ÷ n²

Energy of a transition from nᵢ (initial) to n_f (final): ΔE = R_H × (1/n_f² − 1/nᵢ²) (Positive for emission when nᵢ > n_f; this is the energy of the emitted photon.)

Photon relationships: E = hf and c = fλ, so E = hc ÷ λ

Radius of orbit n: rₙ = a₀ × n² ÷ Z

Worked example 1: energies of the first four levels

Eₙ = −2.18 × 10⁻¹⁸ ÷ n²

n	Eₙ (J)	Eₙ (eV)
1	−2.18 × 10⁻¹⁸	−13.6
2	−5.45 × 10⁻¹⁹	−3.40
3	−2.42 × 10⁻¹⁹	−1.51
4	−1.36 × 10⁻¹⁹	−0.850
∞	0	0

The levels get closer together as n increases, converging at zero (ionisation).

Worked example 2: a Balmer transition (n = 3 → n = 2)

Energy released: ΔE = 2.18 × 10⁻¹⁸ × (1/2² − 1/3²) = 2.18 × 10⁻¹⁸ × (0.2500 − 0.1111) = 2.18 × 10⁻¹⁸ × 0.1389 = 3.03 × 10⁻¹⁹ J

Frequency: f = E ÷ h = 3.03 × 10⁻¹⁹ ÷ 6.626 × 10⁻³⁴ = 4.57 × 10¹⁴ Hz

Wavelength: λ = c ÷ f = 2.998 × 10⁸ ÷ 4.57 × 10¹⁴ = 6.56 × 10⁻⁷ m = 656 nm

This is the red H-alpha line of the Balmer series, one of the most important lines in astronomy. See the hydrogen emission spectrum.

Worked example 3: a Lyman transition (n = 2 → n = 1)

ΔE = 2.18 × 10⁻¹⁸ × (1/1² − 1/2²) = 2.18 × 10⁻¹⁸ × 0.750 = 1.635 × 10⁻¹⁸ J

λ = hc ÷ E = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ 1.635 × 10⁻¹⁸ = 1.986 × 10⁻²⁵ ÷ 1.635 × 10⁻¹⁸ = 1.215 × 10⁻⁷ m = 122 nm

This is in the ultraviolet, invisible to the eye. Transitions ending at n = 1 release more energy than those ending at n = 2, because the gap to the ground state is so large.

Worked example 4: absorption

What wavelength of light must a ground-state hydrogen atom absorb to reach n = 4?

ΔE = 2.18 × 10⁻¹⁸ × (1/1² − 1/4²) = 2.18 × 10⁻¹⁸ × 0.9375 = 2.044 × 10⁻¹⁸ J

λ = 1.986 × 10⁻²⁵ ÷ 2.044 × 10⁻¹⁸ = 9.72 × 10⁻⁸ m = 97.2 nm (ultraviolet)

Absorption and emission between the same two levels involve exactly the same photon energy and wavelength. See emission vs absorption spectra.

Worked example 5: ionisation energy of hydrogen

Ionisation means removing the electron completely: n = 1 → n = ∞.

Per atom: ΔE = 2.18 × 10⁻¹⁸ × (1/1² − 0) = 2.18 × 10⁻¹⁸ J = 13.6 eV

Per mole: 2.18 × 10⁻¹⁸ × 6.022 × 10²³ = 1.313 × 10⁶ J/mol = 1,313 kJ/mol

The measured first ionisation energy of hydrogen is 1,312 kJ/mol, an excellent match. See ionization energy trend.

Minimum wavelength to ionise hydrogen: λ = 1.986 × 10⁻²⁵ ÷ 2.18 × 10⁻¹⁸ = 9.11 × 10⁻⁸ m = 91.1 nm

Worked example 6: from an excited state

How much energy is needed to ionise a hydrogen atom already in the n = 2 state?

E = 2.18 × 10⁻¹⁸ × (1/2² − 0) = 5.45 × 10⁻¹⁹ J (3.40 eV)

Only a quarter of the ground-state ionisation energy. Excited electrons are much easier to remove.

Worked example 7: one-electron ions (He⁺)

He⁺ has Z = 2 and one electron. Energies scale with Z²:

E₁(He⁺) = −2.18 × 10⁻¹⁸ × 2² ÷ 1² = −8.72 × 10⁻¹⁸ J (−54.4 eV)

So removing the electron from He⁺ takes four times as much energy as from hydrogen. The measured second ionisation energy of helium is 5,250 kJ/mol, compared with 4 × 1,312 = 5,248 kJ/mol predicted: again excellent.

Transition n = 3 → n = 2 in He⁺: ΔE = 2.18 × 10⁻¹⁸ × 4 × 0.1389 = 1.211 × 10⁻¹⁸ J → λ = 1.986 × 10⁻²⁵ ÷ 1.211 × 10⁻¹⁸ = 164 nm

Every wavelength is divided by Z² = 4 compared with hydrogen (656 ÷ 4 = 164 nm).

For Li²⁺ (Z = 3), energies are 9 times hydrogen’s. The Bohr model works for any one-electron species, but not for neutral helium or anything else with two or more electrons.

Worked example 8: orbit radii

rₙ = a₀ × n² ÷ Z

• Hydrogen, n = 1: r = 52.9 pm (the Bohr radius)
• Hydrogen, n = 2: r = 52.9 × 4 = 212 pm
• Hydrogen, n = 3: r = 52.9 × 9 = 476 pm
• He⁺, n = 1: r = 52.9 ÷ 2 = 26.5 pm

The radius grows with n², so excited atoms are much larger. The modern quantum model doesn’t have fixed orbits, but the Bohr radius remains the most probable distance of the 1s electron from the nucleus in hydrogen.

A strategy for any Bohr problem

1. Identify the species and Z. Hydrogen is Z = 1; He⁺ is Z = 2; Li²⁺ is Z = 3. If the species has more than one electron, the Bohr equations don’t apply.
2. Identify the levels. For emission, the electron starts high and ends low; for absorption, the reverse. For ionisation, the final level is n = ∞.
3. Calculate ΔE using R_H × Z² × (1/n_low² − 1/n_high²).
4. Convert to whatever the question asks for: frequency (f = E/h), wavelength (λ = hc/E), electronvolts (÷ 1.602 × 10⁻¹⁹) or kJ/mol (× N_A ÷ 1000).
5. Sense-check the region. Transitions to n = 1 in hydrogen are always ultraviolet; to n = 2, mostly visible; to n = 3 and above, infrared. If your Lyman-series answer comes out as 5,000 nm, something has gone wrong.

Working through these steps in the same order every time makes even multi-part questions routine.

Common mistakes

1. Getting the sign wrong. For emission, the photon energy is always positive; use the magnitude of ΔE.
2. Mixing up nᵢ and n_f in the bracket. Put the smaller n first in the bracket (1/n_small² − 1/n_large²) to get a positive photon energy.
3. Forgetting to convert nm to m (1 nm = 10⁻⁹ m) or eV to J.
4. Forgetting Avogadro’s constant when converting per-atom energy to kJ/mol.
5. Applying the formula to multi-electron atoms. It works only for H, He⁺, Li²⁺ and other one-electron species.

Practice questions

1. Calculate the wavelength emitted when hydrogen’s electron falls from n = 4 to n = 2.
2. Calculate the energy (J) needed to excite hydrogen from n = 1 to n = 3.
3. Calculate the first ionisation energy of Li²⁺ in kJ/mol.

Answers:

1. ΔE = 2.18 × 10⁻¹⁸ × (1/4 − 1/16) = 4.09 × 10⁻¹⁹ J; λ = 1.986 × 10⁻²⁵ ÷ 4.09 × 10⁻¹⁹ = 486 nm (blue-green H-beta line)
2. 2.18 × 10⁻¹⁸ × (1 − 1/9) = 1.94 × 10⁻¹⁸ J
3. 9 × 1,313 = 11,800 kJ/mol (measured third ionisation energy of lithium: 11,815 kJ/mol)

For the same calculations expressed with the Rydberg constant in wavelength form, see the Rydberg equation. For general photon conversions, see wavelength, frequency and photon energy calculations.

Key takeaways

• Eₙ = −2.18 × 10⁻¹⁸ J × Z² ÷ n² for hydrogen and one-electron ions.
• Transition energy ΔE = 2.18 × 10⁻¹⁸ J × Z² × (1/n_f² − 1/nᵢ²); convert to wavelength with λ = hc ÷ E.
• Hydrogen’s ionisation energy is 2.18 × 10⁻¹⁸ J per atom, or 1,312 kJ/mol.
• Energies scale with Z² and radii with n² ÷ Z.
• The formulas are exact for one-electron species but fail for multi-electron atoms.