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Hydrogen spectrum (Bohr model): interactive simulation

In Bohr’s model the electron in hydrogen can only have certain energies, Eₙ = −13.6/n² eV. When it drops to a lower level, the energy difference leaves as one photon.

Useful for: JEE Main, NEET

Try this

Pick a starting and a final level. The photon’s wavelength comes from λ = hc/ΔE, so you can see why only drops to n = 2 give visible light.

  1. 1

    Drop the electron from n = 3 to n = 2.

    What you should see

    ΔE ≈ 1.89 eV and λ ≈ 656 nm — the red H-alpha line of the Balmer series.

  2. 2

    Try 4 → 2 and 6 → 2.

    What you should see

    About 486 nm (blue-green) and 410 nm (violet). Higher starting levels give shorter wavelengths.

  3. 3

    Try 2 → 1.

    What you should see

    ΔE = 10.2 eV and λ ≈ 122 nm, the first Lyman line. It is ultraviolet, so you cannot see it.

  4. 4

    Try 4 → 3.

    What you should see

    λ ≈ 1875 nm, in the infrared Paschen series.

Formulas behind it

Energy of level n

En=−13.6n2 eVE_n = -\frac{13.6}{n^2}\ \text{eV}

Photon wavelength

λ=hcΔE≈1240ΔE (eV) nm\lambda = \frac{hc}{\Delta E} \approx \frac{1240}{\Delta E\,(\text{eV})}\ \text{nm}

Rydberg formula

1λ=RH(1n12−1n22)\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

Common mistakes

  • Writing the energy as positive. Bound states have negative energy; −13.6 eV means 13.6 eV is needed to free the electron.
  • Swapping n₁ and n₂ in the Rydberg formula (n₁ is the lower level).
  • Thinking the Balmer series is entirely visible. Its lines crowd towards the ultraviolet limit near 365 nm.
Practise 2 questionsCheck your answer instantly, with a worked solution.Modern Physics chapter guideFormula sheet, solved examples and PYQ patterns.Stuck on a hydrogen spectrum (bohr model) question?Snap it and get a whiteboard video solution.

FAQ

How many spectral lines can come from level n?

n(n − 1)/2. From n = 4 there are 6 possible lines.

What is the ionisation energy of hydrogen?

13.6 eV — the energy to move the electron from n = 1 to n = ∞.

Does the Bohr model work for other atoms?

Only for one-electron species like He⁺ and Li²⁺, where Eₙ = −13.6 Z²/n² eV.

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