The spectrum of hydrogen

A fingerprint in light

Pass the light of a hot solid through a prism. You see a smooth band of colours, a continuous spectrum. Do the same with a glowing gas, and you see only a few bright lines on a dark background. That is an emission line spectrum.

Now shine the light of a hot solid through a cool gas, and look at it through a prism. The band is crossed by dark lines. That is an absorption spectrum. The dark lines are at the same wavelengths as the bright lines of the same gas. Each atom takes in the colours it can also give out.

Every element has its own set of lines, so a spectrum is a fingerprint. Dark lines in sunlight showed that the Sun holds hydrogen, sodium and many other elements. In 1868 a new line in the solar spectrum led to the discovery of helium, on the Sun, before anyone found it on Earth.

The formula for hydrogen

In 1885 Johann Balmer found that four lines of hydrogen in the visible region obey a simple rule. Later Johannes Rydberg put every line into one formula:

Here $R = 1.097 \times 10^{7}\ \text{m}^{-1}$ is the Rydberg constant. For each $n_f$ we get a series of lines, from the jumps $n_i \to n_f$.

Where Bohr's model gives the formula

A jump gives a photon of energy $E_i - E_f = 13.6\ \text{eV}\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$. Divide by $hc$ to find $\dfrac{1}{\lambda}$. So the constant is

That is the value that Rydberg measured. It was the first great success of Bohr's model.

The five series

Each series has a name, a lowest level $n_f$ and a region of the spectrum:

• Lyman, $n_f = 1$: ultraviolet, from 91 nm to 122 nm.
• Balmer, $n_f = 2$: mostly visible, from 365 nm to 656 nm.
• Paschen, $n_f = 3$: infrared, from 820 nm to 1875 nm.
• Brackett, $n_f = 4$: infrared, from 1459 nm to 4051 nm.
• Pfund, $n_f = 5$: infrared, from 2279 nm to 7458 nm.

In every series the lines crowd together as $n_i$ grows. They stop at the series limit, the shortest wavelength of the series. It comes from a jump from $n_i = \infty$:

The Balmer series is the only one with lines in the visible region. Its four visible lines are red at 656 nm, blue-green at 486 nm, violet-blue at 434 nm and violet at 410 nm.

Choose $n_f = 3$. Is the first line shorter or longer than for $n_f = 2$? Now set $Z = 2$. Where do the lines move, and by what factor?

Ions with one electron

For an ion with one electron, $R$ becomes $RZ^2$:

For $\text{He}^+$, $Z = 2$ and $Z^2 = 4$. Take the jump $n_i = 4$ to $n_f = 2$:

That is exactly the value of the first Lyman line of hydrogen, $R\left(1 - \dfrac{1}{4}\right)$. The wavelength is $121.5\ \text{nm}$. In general, a jump of $\text{He}^+$ between the levels $2n_i$ and $2n_f$ matches a jump of hydrogen between $n_i$ and $n_f$. The other lines of $\text{He}^+$ are new.

Practise with new numbers

Try it in Python

The program prints the first lines and the limit of all five series of hydrogen.

Common mistakes

• Turning the formula upside down. The formula gives $\dfrac{1}{\lambda}$, not $\lambda$.
• Using $\dfrac{1}{n}$ in place of $\dfrac{1}{n^2}$. The levels are squared.
• Taking the first line as the shortest. The first line has the longest wavelength. The shortest is the limit.
• Forgetting $Z^2$ for an ion. The lines of $\text{He}^+$ have wave numbers four times larger than the matching jump of hydrogen.

Remember these

• Emission line spectrum from a glowing gas. Absorption spectrum: dark lines at the same wavelengths.
• $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$, $R = 1.097 \times 10^7\ \text{m}^{-1}$. For an ion, replace $R$ by $RZ^2$.
• Series by lowest level: Lyman 1, Balmer 2, Paschen 3, Brackett 4, Pfund 5.
• Lyman: ultraviolet. Balmer: visible. The others: infrared.
• Series limit: $\lambda = \dfrac{n_f^2}{R}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why does each element have its own line spectrum?

2. Which series of hydrogen lies partly in the visible region, and on which level do its jumps end?

3. Which jump gives the limit of the Balmer series?

4. A jump of $\text{He}^+$ from $n = 8$ to $n = 4$ gives a wavelength that is also found in hydrogen. Which jump of hydrogen is it?