Atomic Models

Lessons

Senior 5 · Physics

Atomic Models

4 lessons

Lesson 3 of 4

Energy Levels and Spectral Lines

25 min

An element found in the Sun first

In 1868, astronomers studying the light of the Sun during an eclipse found a bright yellow spectral line that matched no element known on Earth. The new element was named helium, after the Greek word for the Sun, helios. It was not found on Earth until 1895. Every element gives out and absorbs light only at its own set of wavelengths, as individual as a fingerprint, so a spectrum tells you what something is made of, even if it is 150 million kilometres away. Bohr's energy levels explain why.

Three kinds of spectrum

A continuous spectrum contains all wavelengths over a range. It comes from hot solids and dense gases, such as a lamp filament or the surface layers of the Sun.
An emission line spectrum is a set of bright lines on a dark background. It comes from a hot gas at low pressure, such as the hydrogen in a discharge tube.
An absorption line spectrum is a continuous spectrum crossed by dark lines. It appears when white light passes through a cooler gas: the gas absorbs exactly the wavelengths it would emit. The dark lines in sunlight, discovered by Joseph von Fraunhofer, are made by gases in the Sun's outer layers.

Why lines?

In a discharge tube, collisions with fast electrons lift hydrogen atoms into excited states. Each atom soon drops back to lower levels, and every downward jump gives out one photon whose energy equals the gap:

hf = \frac{hc}{\lambda} = E_{\text{upper}} - E_{\text{lower}} \qquad\text{so}\qquad \lambda = \frac{hc}{\Delta E}

Since only certain energies are allowed, only certain gaps exist, and so only certain wavelengths appear: a line spectrum. A useful shortcut: with $\Delta E$ in electronvolts, $\lambda$ in nanometres $= \frac{1240}{\Delta E}$ .

The hydrogen series and the Rydberg formula

Using $E_n = -\frac{13.6\,\mathrm{eV}}{n^2}$ for a jump from $n_{i}$ down to $n_{f}$ :

\frac{1}{\lambda} = \frac{E_{n_i} - E_{n_f}}{hc} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right), \qquad R_H = \frac{me^4}{8\varepsilon_0^2h^3c} = 1.097 \times 10^7\,\mathrm{m^{-1}}

$R_{H}$ is the Rydberg constant. Johannes Rydberg had found this formula by fitting measured wavelengths in 1888, before anyone knew why it worked; Bohr derived it, with the right value of the constant.

Series	Lower level $n_{f}$	Wavelength range	Region
Lyman	1	91.2 nm to 121.6 nm	ultraviolet
Balmer	2	364.6 nm to 656.3 nm	visible (and near ultraviolet)
Paschen	3	820 nm to 1875 nm	infrared

In each series the lines crowd together towards a series limit, the wavelength of a photon emitted when a free electron ( $n_i = \infty$ ) falls straight to $n_{f}$ .

Definition

A spectral series is the set of lines produced by electron transitions that all end on the same lower level. For hydrogen, $\frac{1}{\lambda} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ , with $R_H = 1.097 \times 10^7\,\mathrm{m^{-1}}$ . Example: the Balmer line from $n_{i} = 3$ has $\frac{1}{\lambda} = 1.097 \times 10^7 \times \left(\frac{1}{4} - \frac{1}{9}\right)$ , so $\lambda = 656\,\mathrm{nm}$ , red.

Worked example 1: the red line of hydrogen

Find the energy of the photon emitted when a hydrogen electron falls from $n = 3$ to $n = 2$ , and its wavelength and frequency.

Solution

$\Delta E = E_3 - E_2 = -1.51 - (-3.40) = 1.89\,\mathrm{eV} = 1.89 \times 1.60 \times 10^{-19} = 3.02 \times 10^{-19}\,\mathrm{J}$ .

$\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.02 \times 10^{-19}} = 6.58 \times 10^{-7}\,\mathrm{m} \approx 656\,\mathrm{nm}$ (the small difference comes from rounding). $f = \frac{c}{\lambda} = 4.56 \times 10^{14}\,\mathrm{Hz}$ . This is the red H-alpha line.

Worked example 2: how many lines?

A sample of hydrogen atoms is excited to $n = 4$ . (a) How many different spectral lines can it emit as the atoms return to the ground state? (b) Which has the shortest wavelength, and what is it? (c) Which lines are visible?

Solution

(a) The possible jumps are 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1: six lines (in general $\frac{n(n - 1)}{2}$ ).

(b) The largest gap, 4→1: $\Delta E = 13.6 - 0.85 = 12.75\,\mathrm{eV}$ , $\lambda = \frac{1240}{12.75} = 97\,\mathrm{nm}$ (ultraviolet).

(c) Only the Balmer jumps 4→2 (486 nm) and 3→2 (656 nm) give visible light; 4→3 (1875 nm) is infrared and the three jumps to $n = 1$ are ultraviolet.

Activity 6.1: looking at spectra

You need: a hand spectroscope or a diffraction grating (a piece of an old CD or DVD will also work), a filament lamp, a fluorescent tube or compact fluorescent lamp, and, if your school has them, gas discharge tubes run by your teacher.

Look at the filament lamp through the grating: you see a continuous spectrum.
Look at a fluorescent lamp: you see bright lines (from mercury vapour inside) on top of a continuous band (from the white coating that converts mercury's ultraviolet into visible light).
If available, look at hydrogen, helium and neon tubes. Sketch the lines you see for each and compare them with published spectra. Each gas has its own pattern.

Warning

Safety. Never look at the Sun through a grating, spectroscope or any optical instrument: it causes permanent eye damage. Gas discharge tubes run at several thousand volts: only your teacher should connect and switch them on, and nobody should touch the tube or its connections while it is on. Do not stare at bright lamps for long; some discharge tubes also give out ultraviolet.

Warning

Common mistakes. (1) Mixing up the order in the Rydberg formula: $n_{f}$ is the lower level, and $\frac{1}{n_f^2} - \frac{1}{n_i^2}$ must be positive. (2) Thinking the longest arrow on the diagram gives the longest wavelength: the largest energy gap gives the shortest wavelength. (3) Forgetting to convert eV to J. (4) Saying absorption lines appear at different wavelengths from emission lines: for the same gas they are at the same wavelengths.

Tip

Keep $\lambda\,(\text{nm}) = \frac{1240}{\Delta E\,(\text{eV})}$ in mind. Visible light runs from about 1.8 eV (red) to 3.1 eV (violet): any gap in that range gives a visible line.

Physics 104 Lab: Balmer Series (Flexible Physics, University of Wisconsin–Madison). Watch how a hydrogen lamp and a diffraction grating are used to measure the Balmer wavelengths.

✦

Practice

Distinguish between continuous, emission and absorption spectra, giving a source of each.
Explain why a hot gas at low pressure gives a line spectrum and not a continuous one.
Calculate the wavelength of the first line of the Lyman series ( $n = 2$ to $n = 1$ ) and state the region of the spectrum.
Use the Rydberg formula to find the wavelength of the 5→2 transition.
Find the series limit of the Paschen series.
Hydrogen atoms are excited to $n = 5$ . How many different lines may be emitted? How many are in the Balmer series?
Sodium street lamps glow yellow at 589 nm. Find the energy gap, in eV, between the levels involved.
A student sees dark lines in the spectrum of a distant star at 656 nm and 486 nm. Evaluate the conclusion that "the star contains hydrogen".

Show answersHide answers

Continuous: all wavelengths, from a hot solid (filament). Emission: bright lines on dark, from a hot low-pressure gas (discharge tube). Absorption: dark lines on a continuous band, from white light through a cool gas (sunlight).
The atoms are far apart, so each behaves as an isolated atom with sharp energy levels; photons come only from jumps between levels, so only certain wavelengths appear.
$\Delta E = 10.2\,\mathrm{eV}$ ; $\lambda = \frac{1240}{10.2} = 122\,\mathrm{nm}$ (121.6 nm), ultraviolet.
$\frac{1}{\lambda} = 1.097 \times 10^7 \times \left(\frac{1}{4} - \frac{1}{25}\right) = 2.30 \times 10^6\,\mathrm{m^{-1}}$ ; $\lambda = 434\,\mathrm{nm}$ (violet).
$\frac{1}{\lambda} = \frac{1.097 \times 10^7}{9}$ , $\lambda = 820\,\mathrm{nm}$ .
$\frac{5 \times 4}{2} = 10$ lines; three Balmer lines (5→2, 4→2, 3→2).
$\Delta E = \frac{1240}{589} = 2.11\,\mathrm{eV}$ .
Reasonable: these are the Balmer wavelengths, at which hydrogen absorbs, and two matching lines are good evidence. A careful astronomer would check more lines and allow for a Doppler shift if the star is moving, which shifts all lines by the same fraction.

Key points

Hot solid → continuous spectrum; hot low-pressure gas → emission lines; cool gas in front of a hot source → absorption lines at the same wavelengths.
Each line: $\frac{hc}{\lambda} = E_{\text{upper}} - E_{\text{lower}}$ ; $\lambda(\mathrm{nm}) = \frac{1240}{\Delta E(\mathrm{eV})}$ .
Hydrogen: $\frac{1}{\lambda} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ , $R_H = 1.097 \times 10^7\,\mathrm{m^{-1}}$ .
Lyman (to $n = 1$ , UV), Balmer (to $n = 2$ , visible), Paschen (to $n = 3$ , IR).

Sources and credits

6.4 Bohr's Model of the Hydrogen Atom, from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Hydrogen Lamp, UCL Mathematical & Physical Sciences, Wikimedia Commons, licensed under CC BY 2.0.
Photo: Spektrum einer Wasserstoff-Gasentladungsröhre, Stiftland-Gymnasium Tirschenreuth, Wikimedia Commons, CC0.
Video: Physics 104, Lab LC-5: Balmer Series, Flexible Physics UW Madison, embedded from YouTube.

The activity, the three diagrams, the worked examples and the practice exercises are written by Ganzaa.

1An element found in the Sun first

2Three kinds of spectrum

3Why lines?

4The hydrogen series and the Rydberg formula

5Activity 6.1: looking at spectra

An element found in the Sun first

Three kinds of spectrum

Why lines?

The hydrogen series and the Rydberg formula

Activity 6.1: looking at spectra