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Enter transition details

We assume an ideal hydrogen-like ion: a single electron bound to a nucleus with charge +Z.

Use positive integers (1, 2, 3, …). For emission, ninitial > nfinal; for absorption, ninitial < nfinal.

Options:

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How this calculator works

  • We treat the system as a hydrogen-like ion with a single electron and nuclear charge +Z.
  • Energy levels follow En = −(Z²·RH) / n², where RH ≈ 2.18×10⁻¹⁸ J.
  • The photon energy is |ΔE| = |Efinal − Einitial|. We convert between energy, frequency, and wavelength via ΔE = h·ν = h·c/λ.
  • We also identify the series (e.g., Lyman, Balmer, Paschen) from the lower n, and classify the photon as UV, visible, or IR from λ.

Formula & Equations Used

Bohr energy levels (J): En = −(Z²·RH) / n²

Energy difference: ΔE = Efinal − Einitial

Photon relations: |ΔE| = h·ν = h·c/λ

Wavelength: λ = h·c / |ΔE|

Frequency: ν = |ΔE| / h

Energy in eV: E (eV) = E (J) / (1.602×10⁻¹⁹ J·eV⁻¹)

Example Problems & Step-by-Step Solutions

Example 1 — H Balmer-α (3 → 2)

For hydrogen (Z = 1): En = −RH/n². So E3 ≈ −2.42×10⁻¹⁹ J and E2 ≈ −5.45×10⁻¹⁹ J. ΔE = E2 − E3 ≈ −3.03×10⁻¹⁹ J, so |ΔE| ≈ 3.03×10⁻¹⁹ J. Using λ = h·c/|ΔE| gives λ ≈ 656 nm (red, Balmer-α).

Example 2 — H Lyman-α (2 → 1)

E2 ≈ −5.45×10⁻¹⁹ J, E1 ≈ −2.18×10⁻¹⁸ J. ΔE = E1 − E2 ≈ −1.64×10⁻¹⁸ J, so |ΔE| ≈ 1.64×10⁻¹⁸ J. Then λ = h·c/|ΔE| ≈ 121.6 nm (ultraviolet, Lyman-α).

Frequently Asked Questions

Q: Which ions does this work for?

Any hydrogen-like ion with one electron (H, He⁺, Li²⁺, Be³⁺, …). It does not apply to multi-electron atoms.

Q: How do I know if it’s emission or absorption?

If ninitial > nfinal, the electron drops and emits a photon. If ninitial < nfinal, the electron is excited and absorbs a photon.

Q: Why doesn’t this match real hydrogen exactly?

The Bohr model is an idealized one-electron model. Real spectra can shift slightly due to reduced mass, fine structure, and other quantum effects.