Atomic spectra of hydrogen

The colours of a flame

Put a pinch of common salt in a flame, and the flame turns yellow. Put in a copper salt, and it turns green. A firework uses the same idea. Each element gives out its own colours, and this was one of the first ways to tell what a substance contains.

Continuous spectra and line spectra

When white light passes through a prism, it spreads into a band of colours from red to violet, with no gaps. This is a continuous spectrum. A hot solid, such as the filament of a lamp, gives such a spectrum.

A gas of free atoms is different. If we heat a gas, or pass an electric discharge through it at low pressure, it gives out light of only a few wavelengths. Through a prism, this is a set of bright thin lines on a dark background, an emission spectrum.

If white light goes through a cool gas, the gas takes in the same wavelengths that it would give out. We see dark lines in a bright band, an absorption spectrum.

Each element has its own set of lines, a fingerprint. In $1868$, astronomers saw a yellow line in the light of the Sun that did not belong to any known element. They named the new element helium, from the Greek word for the Sun. It was found on Earth almost thirty years later.

The spectrum of hydrogen

Hydrogen has the simplest spectrum. In the visible part there are four lines, and they come closer together toward the violet end. In $1885$, Johann Balmer found a simple formula for them. Johannes Rydberg then wrote one formula for all the lines of hydrogen:

Here $n_1$ and $n_2$ are whole numbers, and $R_H = 1.097 \times 10^{7}\ \text{m}^{-1}$ is the Rydberg constant (in $\text{cm}^{-1}$ it is $109\,677$). Each pair of numbers gives one line, and each value of $n_1$ gives one series.

Using the formula

The first line of the Balmer series has $n_1 = 2$ and $n_2 = 3$.

We want the wavelength, so $\lambda = \dfrac{1}{\tilde{\nu}} = 6.563 \times 10^{-7}\ \text{m} = 656.3\ \text{nm}$, which is a red line.

The limit of a series is the shortest wavelength, when $n_2 \to \infty$. Then $\dfrac{1}{n_2^2} = 0$. For the Balmer series, $\tilde{\nu} = \dfrac{R_H}{4}$, so $\lambda = \dfrac{4}{R_H} = 364.6\ \text{nm}$. For the Lyman series, $\lambda = \dfrac{1}{R_H} = 91.2\ \text{nm}$, and the first Lyman line has $\lambda = \dfrac{1}{R_H(1 - \frac{1}{4})} = 121.5\ \text{nm}$.

How many lines? If an electron is in level $n$ and falls to level $1$ in all possible ways (directly, or in steps), the number of different lines is $\dfrac{n(n-1)}{2}$. For $n = 4$, this gives $6$ lines.

You should find that every Lyman line lies below $122\ \text{nm}$, in the ultraviolet, and every Paschen line beyond $820\ \text{nm}$, in the infrared.

The puzzle

The formula agrees with the measurements very well. But why would an atom give out only certain wavelengths? The old physics had no answer. Why should a hydrogen atom have energies that are arranged in such a pattern? Niels Bohr gave the first explanation, and we meet it in the next lesson.

Practise with new numbers

Try it in Python

This program uses the Rydberg formula to find the wavelengths of the first lines of three series.

Remember these

• An element gives out a line spectrum, its fingerprint. A hot solid gives a continuous spectrum.
• $\tilde{\nu} = R_H \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$, with $R_H = 1.097 \times 10^{7}\ \text{m}^{-1}$ and $n_2 > n_1$.
• Lyman: $n_1 = 1$, ultraviolet. Balmer: $n_1 = 2$, visible. Paschen: $n_1 = 3$, infrared.
• A series limit is the case $n_2 \to \infty$.

Check yourself

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

1. Which series of hydrogen has its lines in the infrared?

2. What is the wavenumber of the limit of the Balmer series?

3. Why is a line spectrum called the fingerprint of an element?

4. A line of hydrogen has a wavelength of $1282\ \text{nm}$. Which series is it in?