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beginner · Physics · Quantum Numbers & the Hydrogen Atom (Intro)

Hydrogen Energy Levels

The Bohr energy formula

The most important quantitative result in atomic physics is the hydrogen energy spectrum. Solving the time-independent Schrödinger equation for an electron in a 1/r1/r Coulomb potential yields a discrete set of allowed energies labelled by the principal quantum number n=1,2,3,…n = 1, 2, 3, \ldots:

En=−mee42(4πϵ0)2ℏ2⋅1n2=E1n2,E_n = -\frac{m_e e^4}{2(4\pi\epsilon_0)^2 \hbar^2} \cdot \frac{1}{n^2} = \frac{E_1}{n^2},

where the ground-state energy is

E1=−mee42(4πϵ0)2ℏ2≈−13.6 eV.E_1 = -\frac{m_e e^4}{2(4\pi\epsilon_0)^2 \hbar^2} \approx -13.6\,\text{eV}.

The negative sign means the electron is bound — energy must be added to remove it from the atom. As n→∞n \to \infty, En→0E_n \to 0, which corresponds to a free (unbound) electron.

Deriving the −13.6 eV-13.6\,\text{eV} ground state

The combination of constants fixes the energy scale. Using SI values:

The full expression (keeping track of the 4πϵ04\pi\epsilon_0 in SI) is

E1=−mee42(4πϵ0)2ℏ2≈−2.18×10−18 J=−13.6 eV.E_1 = -\frac{m_e e^4}{2(4\pi\epsilon_0)^2 \hbar^2} \approx -2.18 \times 10^{-18}\,\text{J} = -13.6\,\text{eV}.

Converting to electron-volts: 1 eV=1.602×10−19 J1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J}, so 2.18×10−18 J÷1.602×10−19 J/eV≈13.6 eV2.18 \times 10^{-18}\,\text{J} \div 1.602 \times 10^{-19}\,\text{J/eV} \approx 13.6\,\text{eV}.

Working out the excited states

Because En=E1/n2E_n = E_1/n^2, every excited-state energy follows from a single division:

| nn | EnE_n (eV) | |-----|------------| | 1 | −13.6-13.6 | | 2 | −3.40-3.40 | | 3 | −1.51-1.51 | | 4 | −0.850-0.850 |

For example, E2=−13.6/4=−3.40 eVE_2 = -13.6/4 = -3.40\,\text{eV} and E3=−13.6/9≈−1.51 eVE_3 = -13.6/9 \approx -1.51\,\text{eV}.

Photon emission and the Rydberg formula

When an electron falls from level nin_i to a lower level nfn_f, the energy difference is carried away by a single photon:

Eγ=Eni−Enf=13.6 eV ⁣(1nf2−1ni2).E_\gamma = E_{n_i} - E_{n_f} = 13.6\,\text{eV}\!\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right).

This is the Rydberg formula. Setting nf=2n_f = 2 gives the Balmer series — the visible hydrogen lines measured long before quantum mechanics was invented. The formula reproduces each wavelength exactly, which was one of the first triumphs of the quantum theory of the atom.

Try it

This is a numerical exercise — return a number. Using E1=−13.6 eVE_1 = -13.6\,\text{eV} and the formula En=E1/n2E_n = E_1/n^2, compute the energy of hydrogen in the n=3n = 3 state.

Run your code to see the quantum state.

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