In Physics 2 you used hydrogen's energy levels, $-13.6\ \text{eV}/n^2$, to find the colors of its light. Where does that formula come from? In 1913 Niels Bohr combined Coulomb's law and circular orbits with one new rule: the electron's angular momentum comes in whole multiples of $\hbar = h/2\pi$. The model is not the final word (the Schrödinger equation is), but it gets hydrogen's energies exactly right.
🔑 Key idea
For hydrogen (and for one-electron ions of nuclear charge $Z$): $$r_n = \frac{n^2}{Z}a_0,\qquad E_n = -\frac{13.6\ \text{eV}\cdot Z^2}{n^2},\qquad n = 1, 2, 3, \ldots$$ with $a_0 = 0.0529$ nm (the Bohr radius). The ground state $n = 1$ has the lowest (most negative) energy, and $+13.6$ eV is needed to ionize it.
💡 Why it works
Two conditions fix the orbit. Newton's second law with the Coulomb force: $\frac{ke^2}{r^2} = \frac{m_ev^2}{r}$. Bohr's rule: $m_evr = n\hbar$. Square the second, $m_e^2v^2r^2 = n^2\hbar^2$, and use the first, $m_ev^2 = \frac{ke^2}{r}$: $m_eke^2r = n^2\hbar^2$, so $$r_n = n^2\,\frac{\hbar^2}{m_eke^2} = n^2a_0.$$ The energy: $K = \frac{1}{2}m_ev^2 = \frac{ke^2}{2r}$ and $U = -\frac{ke^2}{r}$, so $E = -\frac{ke^2}{2r_n} = -\frac{13.6\ \text{eV}}{n^2}$. Bohr's rule has a wave meaning, too: with de Broglie's $\lambda = h/m_ev$, $m_evr = n\hbar$ says $2\pi r = n\lambda$, a whole number of wavelengths around the orbit.
Bohr orbits, with radii growing like $n^2$, beside the energy ladder. An electron dropping from $n = 3$ to $n = 2$ gives off the red 656 nm photon.
Worked examples
✏️ Worked example: The n = 3 level
Find the energy and orbit radius of hydrogen's $n = 3$ level.
$E_3 = -\frac{13.6}{9} = -1.51$ eV.
$r_3 = 9a_0 = 9 \times 0.0529 = 0.476$ nm.
Answer: $E_3 = -1.51$ eV, $r_3 = 0.476$ nm
✏️ Worked example: Ionizing an excited atom
How much energy does it take to ionize a hydrogen atom that is already in its $n = 2$ state?
$E_2 = -\frac{13.6}{4} = -3.40$ eV.
Ionized means $n \to \infty$, where $E = 0$. Needed: $0 - (-3.40) = 3.40$ eV.
Answer: $3.40$ eV (a photon of 365 nm or shorter can do it)
✏️ Worked example: A heavier nucleus
Find the ground-state energy and radius of He$^+$ (one electron, $Z = 2$).
$E_1 = -13.6 \times \frac{2^2}{1} = -54.4$ eV.
$r_1 = \frac{a_0}{2} = 0.0265$ nm: twice the charge pulls the electron in closer.
Answer: $-54.4$ eV and $0.0265$ nm
Bohr's model has limits. It cannot handle atoms with two or more electrons, it says nothing about how bright each spectral line is, and its picture of definite orbits is wrong in detail. What it gets right, the energy levels, survives in the full quantum theory you will meet later in this unit, where orbits are replaced by clouds of probability.
⚠️ Common mistake
Thinking the $n = 1$ level has the highest energy because it comes first. The energies are negative, and $-13.6$ eV is the lowest: it takes the most energy to pull the electron out from there. As $n$ grows the levels rise toward 0 and crowd together. Check: an atom gives off light when it drops to lower $n$, so lower $n$ must mean lower energy.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.