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Lyman alpha and Balmer alpha wavelengths, and the ground state of He+

Hydrogen emits light at a fixed list of wavelengths. One formula produces the whole list.

$$\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$

$n_i$ is the level the electron falls from, $n_f$ is the level it lands on, and $R$ is the Rydberg constant. The formula returns $1/\lambda$, not $\lambda$. Forgetting the reciprocal at the end is the most common way to lose this question.

The series are named by where the electron lands. $n_f = 1$ is the Lyman series, in the ultraviolet. $n_f = 2$ is the Balmer series, which is where the visible hydrogen lines live. Within a series the alpha line is the shortest drop, so Lyman alpha is $2 \to 1$ and Balmer alpha is $3 \to 2$.

Lyman alpha and Balmer alpha

Use $R_\infty = 1.0973732\times10^{7}$ m$^{-1}$. That is the CODATA 2022 value, known to about one part in $10^{12}$, and it is the number a textbook problem means by “the Rydberg constant”.

Lyman alpha is $n_i = 2$ falling to $n_f = 1$.

$$\frac{1}{\lambda} = R_\infty\left(\frac{1}{1} - \frac{1}{4}\right) = 1.0973732\times10^{7} \times 0.75 = 8.230299\times10^{6}\ \text{m}^{-1}$$

$$\lambda = \frac{1}{8.230299\times10^{6}} = 1.2150\times10^{-7}\ \text{m} = 121.50\ \text{nm}$$

That is deep ultraviolet. Air absorbs it, so Lyman alpha astronomy is done from orbit and Lyman alpha lab work is done in vacuum.

Balmer alpha is $n_i = 3$ falling to $n_f = 2$. Keep the bracket as the exact fraction $1/4 - 1/9 = 5/36$ rather than rounding it to 0.139.

$$\frac{1}{\lambda} = R_\infty \times \frac{5}{36} = 1.524129\times10^{6}\ \text{m}^{-1}$$

$$\lambda = 6.5611\times10^{-7}\ \text{m} = 656.11\ \text{nm}$$

That is red, and it is the colour of every hydrogen discharge tube and every emission nebula you have seen a photograph of.

Take $n_i$ to infinity and the second term vanishes. The Lyman series limit is $\lambda = 1/R_\infty = 91.13$ nm, the shortest wavelength hydrogen can emit and the wavelength of a photon that just barely ionises the atom from its ground state.

Why the measured lines are not quite those numbers

Look up Lyman alpha and you get 121.567 nm, not 121.50. Look up Balmer alpha and you get 656.45 nm in vacuum, not 656.11. The calculation is not sloppy. The constant is.

$R_\infty$ assumes an infinitely heavy nucleus that never moves. A real proton is only about 1836 times the electron mass, so both particles orbit their common centre of mass and the electron behaves as if it were slightly lighter. Correcting for that gives the Rydberg constant for hydrogen.

$$R_H = \frac{R_\infty}{1 + m_e/m_p}$$

With the CODATA masses $m_e = 9.1093837139\times10^{-31}$ kg and $m_p = 1.67262192595\times10^{-27}$ kg, the ratio $m_e/m_p$ is $5.44617\times10^{-4}$, so $R_H = 1.0967758\times10^{7}$ m$^{-1}$. That is 0.054 percent smaller, which is invisible at three significant figures and obvious at five.

Redo both lines with $R_H$ and you get 121.57 nm and 656.47 nm. The NIST Atomic Spectra Database lists the measured vacuum wavelengths as 121.567 nm and 656.452 nm. The finite-mass calculation lands within 0.02 nm of both, and what is left over is fine structure, which the Bohr model does not describe at all.

So there are two defensible answers to “what is the Lyman alpha wavelength”, and they differ in the fourth figure. Use $R_\infty$ when a problem hands you the Rydberg constant, which is what the simulations linked below do. Use $R_H$ when you are comparing against a spectrum somebody measured.

Energy levels, and why Z is squared

The same physics written in energy instead of wavelength gives the level diagram directly.

$$E_n = -\frac{Z^2 E_0}{n^2}$$

$E_0$ is the Rydberg energy, $13.605693$ eV from CODATA, rounded to 13.6 eV in most textbooks. $Z$ is the atomic number, the number of protons in the nucleus. For hydrogen $Z = 1$, so the ground state is $E_1 = -13.6$ eV and the ionisation energy is $+13.6$ eV.

The two pictures have to agree. A $2 \to 1$ transition in hydrogen releases $E_2 - E_1 = -3.4014 - (-13.6057) = 10.204$ eV, and with $hc = 1239.84$ eV nm the photon wavelength is $1239.84/10.204 = 121.50$ nm. That is the $R_\infty$ answer again, because $E_0$ carries the same infinite-mass assumption that $R_\infty$ does.

The $Z^2$ is the interesting part. Doubling the nuclear charge doubles the Coulomb attraction, and it also halves the orbit radius, and energy is force times distance. Both factors pull the same way, so the binding energy goes as the square rather than linearly.

Singly ionised helium is the clean test. He$^+$ has two protons and one electron, so it is hydrogen-like with $Z = 2$.

$$E_1 = -\frac{2^2 \times 13.6057}{1^2} = -4 \times 13.6057 = -54.4\ \text{eV}$$

Four times hydrogen’s binding energy, from one extra proton. NIST lists the measured ionisation energy of He$^+$ as 54.4178 eV, and the leftover 0.005 eV is the same finite-nuclear-mass correction as before, smaller here because a helium nucleus is four times heavier than a proton.

The scaling runs up the periodic table. Every hydrogen-like ion, He$^+$, Li$^{2+}$, and on, has hydrogen’s spectrum with every wavelength divided by $Z^2$. The $3 \to 2$ line in He$^+$ sits near $656.11/4 = 164.0$ nm, in the ultraviolet rather than the red.

Each of these simulations builds the level diagram and the emission spectrum from $Z$ and the two quantum numbers, so the arithmetic above is a check on it and not its source.

References