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v0.1.1065
NotesPhysics HLTopic 5.1Bohr model and quantized energy levels (HL)
Back to Physics HL Topics
5.1.63 min read

Bohr model and quantized energy levels (HL)

IB Physics • Unit 5

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Contents

  • The Bohr energy ladder
  • Reading the levels for hydrogen
  • Photons from jumps between levels
  • The ground state and ionisation
  • In the exam
The big idea: In the Bohr model the electron in a hydrogen atom can only sit on a fixed set of energy levels — like rungs on a ladder, not a smooth ramp. Each level is numbered by a whole number n (the principal quantum number), n = 1, 2, 3, …

The energy of level n is quantized (it can only take these discrete values):
Given for hydrogen in the data booklet. n = 1, 2, 3, … numbers the level.
energy of level n (eV)
principal quantum number (1, 2, 3, …)
the hydrogen constant (eV)
Two things to notice: (1) The energies are negative. The zero is set when the electron has just escaped the atom (n → ∞), so a bound electron has less energy than that — hence the minus sign.

(2) The levels bunch together as n rises: because of the 1/n², the gap from n = 1 to n = 2 is huge, but the gaps higher up shrink toward zero.

To find any level, just put its n into En = −13.6/n². The ground state (n = 1) is the lowest (most negative) — the most tightly bound. The numbers below are the ones you keep meeting in hydrogen problems.

Level nn²En = −13.6/n²Meaning
11−13.6 eVground state (lowest, most bound)
24−3.40 eVfirst excited state
39−1.51 eVsecond excited state
416−0.85 eVthird excited state
∞∞0 eVelectron just free (ionised)

Worked example — the n = 2 level

Use the Bohr formula to find the energy of the n = 2 level of hydrogen.

Solution

  1. Write the given formula first:
  2. Put in n = 2 (so n² = 4):
  3. Work it out — keep the unit and the sign:

Final answer

E₂ = −3.40 eV. (Negative because the electron is still bound to the atom.)

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How a photon is made: An electron can jump down from a higher level to a lower one. It loses energy, and that exact energy is carried off as a single photon of light. Because the levels are discrete, the photon energies are discrete too — that is why each gas emits a line spectrum (sharp coloured lines), not a continuous rainbow.

A jump up works in reverse: the atom absorbs a photon of exactly the right energy.
Given in the data booklet. The photon energy equals the size of the energy drop.
Planck constant (J s)
frequency of the photon (Hz)
energy of the initial (higher) level (J or eV)
energy of the final (lower) level (J or eV)

Worked example — the n = 3 → n = 2 photon

An electron in a hydrogen atom drops from n = 3 to n = 2. Find the energy of the emitted photon, in eV and in joules. (E₃ = −1.51 eV, E₂ = −3.40 eV; 1 eV = 1.60×10⁻¹⁹ J.)

Solution

  1. Photon energy = the drop in energy (given relation):
  2. Substitute the two levels (mind the signs):
  3. Add them — this is the photon energy in eV:
  4. Convert to joules (× 1.60×10⁻¹⁹):

Final answer

Photon energy = 1.89 eV = 3.0×10⁻¹⁹ J. (This is the red line of the hydrogen spectrum.)

Watch the signs: Use higher minus lower, Ei − Ef, so the photon energy comes out positive. Subtracting two negatives is where marks are lost: (−1.51) − (−3.40) = +1.89 eV, not −1.89 eV.
Pulling the electron right off: The ground state of hydrogen is n = 1 with E₁ = −13.6 eV. To ionise the atom you must lift the electron from this level all the way to E = 0 (n → ∞), where it is just free.

So the ionisation energy of hydrogen is the full 13.6 eV gap from the ground state up to zero.
The ground state from the given formula with n = 1.
ground-state energy of hydrogen (eV)
principal quantum number (here n = 1)

Worked example — ionising hydrogen

How much energy is needed to remove the electron from a hydrogen atom that starts in its ground state? Give the answer in eV.

Solution

  1. Energy needed = lift from the ground state (E₁) up to free (E = 0):
  2. Put in the ground-state energy E₁ = −13.6 eV:
  3. Work it out — keep the unit:

Final answer

Ionisation energy = 13.6 eV. (You add energy, so a positive amount.)

Emission (jump down)

  • Electron falls to a lower level
  • Atom gives out a photon
  • Photon energy = Ei − Ef
  • Makes the bright lines of a spectrum

Absorption / ionisation (jump up)

  • Electron rises to a higher level (or escapes)
  • Atom takes in a photon
  • Needs exactly the right energy gap
  • Full escape from n = 1 needs 13.6 eV

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Where it shows up: The Bohr model is HL only (E.1):

- Paper 1A — 'which transition gives the highest-energy photon?', 'what is the ionisation energy?', or read a level off En = −13.6/n². - Paper 2 — determine a photon's energy (eV → J), then often its frequency or wavelength, from a stated transition.
Three easy marks: (1) Photon energy = higher minus lower level, so it comes out positive. (2) Convert eV → J with × 1.60×10⁻¹⁹ before using hf or hc/λ. (3) The biggest energy drop (e.g. n = 2 → 1) gives the highest-frequency photon.

IB-style question — biggest jump, biggest photon

In hydrogen the levels are E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV. An electron can fall by 3→2, 2→1, or 3→1. Determine which transition emits the highest-energy photon, and state that energy.

Solution

  1. Photon energy = the size of the drop (Ei − Ef). Compute each:
  2. And the other two drops:
  3. The largest drop is 3 → 1:

Final answer

The 3 → 1 transition emits the highest-energy photon, 12.1 eV (the largest energy gap).

IB Exam Questions on Bohr model and quantized energy levels (HL)

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Related Physics HL Topics

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5.1.1Nuclear model and atomic structure
5.1.2Energy levels and atomic spectra
5.1.3The electronvolt
5.1.4Quantisation of charge
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