The Hydrogen Spectrum and Its Series – Electron Configurations of Atoms and Ions (Senior 4 Chemistry) | Ganzaa

Electron Configurations of Atoms and Ions

Lessons

Senior 4 · Chemistry

Electron Configurations of Atoms and Ions

5 lessons

Lesson 2 of 5

The Hydrogen Spectrum and Its Series

25 min

Four lines and a schoolteacher's formula

Look at a hydrogen discharge tube through a spectroscope and you see four lines: red, blue-green, blue and violet. In 1885 Johann Balmer, a Swiss schoolteacher, found that their wavelengths fitted a simple formula, although nobody then knew why. Later, more lines were found in the ultraviolet and infrared, and every one of them fitted a general form of his formula. Bohr's energy levels explained it: each line is one jump of the electron, and each family of lines ends on the same lower level. In this lesson you will calculate the wavelength of any hydrogen line and find the ionisation energy of hydrogen from its spectrum.

Five series of lines

A series is a family of lines produced by electrons falling to the same lower level from different higher levels. The series is named after its discoverer and identified by its lower level, $n_{1}$ .

Table 2.1: The series of the hydrogen spectrum

Series	Electron falls to	Region of spectrum	Reported
Lyman	$n_{1} = 1$	ultraviolet	1906
Balmer	$n_{1} = 2$	visible (and near ultraviolet)	1885
Paschen	$n_{1} = 3$	infrared	1908
Brackett	$n_{1} = 4$	infrared	1922
Pfund	$n_{1} = 5$	infrared	1924

The Lyman lines carry the most energy because the drop to $n = 1$ is the biggest gap in the atom. Figure 2.1 shows the series as arrows on an energy-level diagram.

Definition

Rydberg equation: the wavelength $\lambda$ of the line emitted when a hydrogen electron falls from level $n_{2}$ to level $n_{1}$ ( $n_{2} > n_{1}$ ) is given by

\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right)

where $R_H = 1.097 \times 10^{7}\,\mathrm{m^{-1}}$ is the Rydberg constant. It follows from Bohr's $E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$ and $\Delta E = \frac{hc}{\lambda}$ , since $\frac{2.18 \times 10^{-18}}{hc} = 1.097 \times 10^{7}\,\mathrm{m^{-1}}$ .

Worked example 1: the first line of the Lyman series

Calculate the wavelength of the first (lowest-energy) line of the Lyman series, and say why you cannot see it.

Solution. The first line is the smallest jump into $n_{1} = 1$ , which is from $n_{2} = 2$ :

\frac{1}{\lambda} = 1.097 \times 10^{7}\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 1.097 \times 10^{7} \times 0.750 = 8.23 \times 10^{6}\,\mathrm{m^{-1}}

\lambda = \frac{1}{8.23 \times 10^{6}} = 1.22 \times 10^{-7}\,\mathrm{m} = 122\,\mathrm{nm}

Answer: 122 nm. It is in the ultraviolet (shorter than 400 nm), so the eye cannot detect it.

Worked example 2: the Balmer series converges

Use the Rydberg equation to find the wavelengths of the Balmer lines from $n_{2} = 3, 4, 5, 6$ and the series limit ( $n_2 = \infty$ ). What happens to the spacing?

Solution. With $n_{1} = 2$ : $\lambda = \frac{1}{R_H\left(\frac{1}{4} - \frac{1}{n_2^2}\right)}$ .

$n_{2}$	3	4	5	6	∞
$\lambda$ / nm	656.3	486.2	434.1	410.2	364.6
colour	red	blue-green	blue	violet	ultraviolet

Answer: the gaps between lines shrink (170, 52, 24 nm, …) and the lines crowd together towards 364.6 nm, the series limit, at the short-wavelength, high-energy end.

The convergence limit and the ionisation energy of hydrogen

Lines converge because the energy levels themselves crowd together as $n$ increases. The convergence limit of a series is the line for a jump from $n = \infty$ , where the electron is just free of the atom. For the Lyman series, which ends on the ground state, the convergence limit corresponds to the energy needed to remove the electron completely from a hydrogen atom in its ground state: the ionisation energy of hydrogen.

The convergence limit can be calculated, or found from measured lines. In the graphical method, the frequency difference $\Delta f$ between neighbouring lines is plotted against frequency, and the curve is extended to $\Delta f = 0$ , where the lines would merge (Figure 2.2).

Worked example 3: the ionisation energy of hydrogen

Figure 2.2 gives a Lyman convergence frequency of $3.29 \times 10^{15}\,\mathrm{Hz}$ . Calculate the first ionisation energy of hydrogen in $\mathrm{kJ\,mol^{-1}}$ . ( $h = 6.626 \times 10^{-34}\,\mathrm{J\,s}$ , $N_A = 6.022 \times 10^{23}\,\mathrm{mol^{-1}}$ )

Solution. Energy to ionise one atom:

E = hf = 6.626 \times 10^{-34} \times 3.29 \times 10^{15} = 2.18 \times 10^{-18}\,\mathrm{J}

For one mole of atoms:

2.18 \times 10^{-18} \times 6.022 \times 10^{23} = 1.31 \times 10^{6}\,\mathrm{J\,mol^{-1}} = 1310\,\mathrm{kJ\,mol^{-1}}

Answer: about $1310\,\mathrm{kJ\,mol^{-1}}$ (data books give 1312). This matches $- E_{1}$ in Bohr's formula, which is strong evidence for his model.

Warning

Common mistakes. (1) A series is named by its lower level: all Balmer lines end on $n = 2$ . (2) Lines converge at the short-wavelength (high-frequency) end, not the red end. (3) Only the Lyman convergence limit gives the ionisation energy, because only Lyman lines start from the ground state. (4) In the Rydberg equation $n_{1} < n_{2}$ ; if you swap them, $\frac{1}{\lambda}$ comes out negative.

Tip

To count the possible lines when electrons fall from level $n$ to the ground state, count the pairs of levels: $\frac{n(n-1)}{2}$ . From $n = 4$ : $\frac{4 \times 3}{2} = 6$ lines (4→3, 4→2, 4→1, 3→2, 3→1, 2→1).

Cohesive Chemistry: The hydrogen emission spectrum (9 min). Watch how each series is linked to one lower energy level, and why the lines in a series get closer together.

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Practice

Name the series and region for a hydrogen line produced by a fall from $n = 5$ to $n = 3$ .
Calculate the wavelength of the line for $n = 5 \rightarrow n = 2$ . Check it against Worked example 2.
Calculate the shortest wavelength in the Lyman series.
Use the shortest Lyman wavelength to calculate the ionisation energy of hydrogen in $\mathrm{kJ\,mol^{-1}}$ .
How many different lines can be emitted by a sample of hydrogen atoms whose electrons are excited to $n = 5$ ?
Explain why the lines in any series converge.
Explain why the ionisation energy of hydrogen is found from the Lyman series and not from the Balmer series. What would the Balmer convergence limit measure?

Show answersHide answers

Paschen series (ends on $n = 3$ ); infrared.
$\frac{1}{\lambda} = 1.097 \times 10^{7}\left(\frac{1}{4} - \frac{1}{25}\right) = 2.304 \times 10^{6}\,\mathrm{m^{-1}}$ ; $\lambda = 434\,\mathrm{nm}$ , as in the table.
Series limit, $n_2 = \infty$ : $\frac{1}{\lambda} = R_H$ , so $\lambda = \frac{1}{1.097 \times 10^{7}} = 9.12 \times 10^{-8}\,\mathrm{m} = 91.2\,\mathrm{nm}$ .
$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{9.12 \times 10^{-8}} = 2.18 \times 10^{-18}\,\mathrm{J}$ ; $\times 6.022 \times 10^{23} = 1310\,\mathrm{kJ\,mol^{-1}}$ .
$\frac{5 \times 4}{2} = 10$ lines.
The energy levels get closer together as $n$ increases, so the energy differences between successive upper levels and the lower level differ less and less, and the lines crowd together, approaching the jump from $n = \infty$ .
Ionisation energy is measured from the ground state, $n = 1$ , and only Lyman lines start from or end on $n = 1$ . The Balmer limit gives the energy needed to remove an electron from $n = 2$ , an excited atom ( $328\,\mathrm{kJ\,mol^{-1}}$ ).

Quick check

All the lines of one series end on n = 2. Which series is it?

Key points

A series is all the lines ending on one lower level: Lyman $n = 1$ (UV), Balmer $n = 2$ (visible), Paschen $n = 3$ , Brackett $n = 4$ , Pfund $n = 5$ (infrared).
$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ , with $R_H = 1.097 \times 10^{7}\,\mathrm{m^{-1}}$ .
Lines in a series converge at high energy because the levels crowd together.
The Lyman convergence limit (91.2 nm, $3.29 \times 10^{15}\,\mathrm{Hz}$ ) gives the ionisation energy of hydrogen, $1312\,\mathrm{kJ\,mol^{-1}}$ .

Sources and credits

Image: Hydrogen dischrge tube.jpg by Mohit Jhanjhotia, via Wikimedia Commons, licensed under CC BY 4.0.
Image: Hydrogen emission spectrum.png by Umop503, via Wikimedia Commons, dedicated to the public domain under CC0 1.0.
The Rydberg constant and other physical constants are CODATA values published by NIST; hydrogen wavelengths are calculated by Ganzaa from the Rydberg equation.
Video: S1.3.1 - The hydrogen emission spectrum, Cohesive Chemistry, embedded from YouTube.
The text, diagrams, tables, worked examples and Practice items were written and drawn by Ganzaa.

1Four lines and a schoolteacher's formula

2Five series of lines

3The convergence limit and the ionisation energy of hydrogen

Four lines and a schoolteacher's formula

Five series of lines

The convergence limit and the ionisation energy of hydrogen