Hydrogen Energy Levels Calculator
Enter initial and final quantum numbers to compute hydrogen energy levels, transition energy and photon wavelength via the Bohr model.
Input Data
Results
At a glance:The Bohr model gives hydrogen energy levels E_n = −13.6 eV/n² = −(m_e·e⁴)/(8ε₀²h²)·(1/n²), where n is the principal quantum number. A transition n1→n2 changes energy by ΔE = E_n2 − E_n1; for an electron dropping (n2<n1) a photon of energy |ΔE| is emitted with wavelength λ = hc/|ΔE| (hc ≈ 1240 eV·nm). The Rydberg formula 1/λ = R_H·(1/n1² − 1/n2²) is equivalent, with R_H ≈ 1.097×10⁷ m⁻¹. The Lyman series (to n=1) is UV, Balmer (to n=2) is visible, Paschen (to n=3) is IR. This tool returns E_n1, E_n2, ΔE and λ for any transition.
Formula
E_n = −13.6 eV/n²
ΔE = E_n2 − E_n1
λ = hc/|ΔE|
Rydberg: 1/λ = R_H(1/n1² − 1/n2²)
$$E_n = -\frac{13.6\,\text{eV}}{n^2}, \quad \Delta E = E_{n_2} - E_{n_1} = 13.6\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\,\text{eV}, \quad \lambda = \frac{hc}{|\Delta E|}$$How to Use
- Enter the initial quantum number n1 and final n2.
- The calculator returns E_n1, E_n2, ΔE and the photon wavelength λ.
Case Studies
Balmer red line
n1 = 3 → n2 = 2.
E_3 = −1.51 eV, E_2 = −3.40 eV, ΔE = −1.89 eV.
λ = 1240/1.89 ≈ 656 nm (red Hα).
FAQ
Why are energy levels negative?
The zero of energy is set at a free electron (infinity); bound states have less energy, hence negative. The ground state n=1 is −13.6 eV.
Which transitions are visible?
Balmer transitions ending at n=2 fall in the visible range (e.g. 3→2 is the red Hα line at 656 nm). Lyman (to n=1) is UV; Paschen (to n=3) is IR.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.