Hydrogen's bright-line spectrum has a few sharp lines, not a rainbow. Photons and the Photoelectric Effect explains why that means its energy is fixed in steps: each line is one photon, and one photon size means one energy jump. Niels Bohr found the energies of hydrogen's levels in 1913.
Key idea
Hydrogen's electron can have only the energies $$E_n = -\frac{2.18 \times 10^{-18}\ \text{J}}{n^2}, \qquad n = 1, 2, 3, \ldots$$ A jump from level $n_i$ to $n_f$ changes the atom's energy by $\Delta E = E_f - E_i$. Falling ($\Delta E < 0$) emits a photon; rising ($\Delta E > 0$) needs one to be absorbed. Either way the photon's energy is $|\Delta E|$ and its wavelength is $\lambda = hc/|\Delta E|$.
Why it works
Energy zero is the electron just free of the atom. A bound electron has less energy than that, so $E_n$ is negative, and most negative at $n = 1$, the ground state. Because of the $1/n^2$, the levels crowd together as $n$ grows: $E_1 = -21.8$, $E_2 = -5.45$, $E_3 = -2.42$, $E_4 = -1.36 \times 10^{-19}$ J, approaching zero. The drop from 2 to 1 is bigger than all the higher gaps put together, so it gives ultraviolet light, while drops to $n = 2$ give visible light. Energy is conserved: the atom loses exactly what the photon carries away.
Worked example: The red line
Find the wavelength of the photon emitted when hydrogen's electron falls from $n = 3$ to $n = 2$.
- $$\Delta E = E_2 - E_3 = -2.18 \times 10^{-18}\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right) = -2.18 \times 10^{-18}(0.1389) = -3.03 \times 10^{-19}\ \text{J}$$
- Negative: the atom loses energy and emits a photon of $3.03 \times 10^{-19}\ \text{J}$.
- $$\lambda = \dfrac{hc}{|\Delta E|} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{3.03 \times 10^{-19}} = 6.57 \times 10^{-7}\ \text{m} = 657\ \text{nm}$$ red light. (The measured line is 656 nm; the small difference is rounding in the constants.)
Answer: About $657\ \text{nm}$.
Worked example: Absorbing ultraviolet light
What wavelength must hydrogen absorb to go from $n = 1$ to $n = 4$?
- $$\Delta E = -2.18 \times 10^{-18}\left(\dfrac{1}{16} - \dfrac{1}{1}\right) = +2.04 \times 10^{-18}\ \text{J}$$ positive because the atom gains energy.
- $$\lambda = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{2.04 \times 10^{-18}} = 9.73 \times 10^{-8}\ \text{m} = 97.3\ \text{nm}$$ ultraviolet.
Answer: $97.3\ \text{nm}$
Worked example: Removing the electron
How much energy ionizes a hydrogen atom whose electron is in $n = 2$?
- Ionizing means going to $n = \infty$, where $E = 0$.
- $\Delta E = 0 - E_2 = \dfrac{2.18 \times 10^{-18}}{4} = 5.45 \times 10^{-19}\ \text{J}$.
Answer: $5.45 \times 10^{-19}\ \text{J}$
Common mistake
Two traps. First, picturing the levels as evenly spaced or spreading apart; they crowd together near zero, so high jumps are small. Second, a negative wavelength: $\Delta E$ is negative for emission, but a wavelength is a length. Use $|\Delta E|$, and let the sign tell you whether light was emitted or absorbed. Check: any drop to $n = 1$ is ultraviolet (under 122 nm), and drops to $n = 2$ fall between 365 and 656 nm.
Drops to $n = 1$ form the Lyman series (ultraviolet), drops to $n = 2$ the Balmer series (visible), and drops to $n = 3$ the Paschen series (infrared). Bohr's formula works only for one-electron species; other atoms need the full quantum theory of the next topics.