C. Structure and Properties of Matter · Structure and Properties of MatterLight, Line Spectra and the Bohr Model
C3.1 C2 A1.12About 75 minutes
Learning goals- I can use c = λν and E = hν to relate wavelength, frequency and photon energy.
- I can explain the difference between continuous, emission and absorption spectra.
- I can explain how Bohr's model accounts for the line spectrum of hydrogen.
- I can calculate the energy and wavelength of photons for transitions in hydrogen.
Words to know
wavelength (λ)
The distance between two neighbouring crests of a wave.
frequency (ν)
The number of waves passing a point per second, in hertz (s⁻¹).
photon
A single packet (quantum) of electromagnetic energy, with E = hν.
emission spectrum
The pattern of specific wavelengths given off by excited atoms of an element.
energy level
One of the allowed energies an electron can have in an atom.
ground state / excited state
The lowest-energy arrangement of an atom's electrons / any higher-energy arrangement.
Why do fireworks have colours?
Strontium compounds make fireworks red, sodium makes them yellow-orange, and copper compounds make blue-green. Each element gives off its own set of colours when its atoms are heated. That set is like a fingerprint, and explaining it was the key that unlocked the inside of the atom.
To understand it, you first need to know a little about light.
Light as a wave
Light is an electromagnetic wave. Like any wave, it has:
- Wavelength (λ, lambda): the distance between two crests, often measured in nanometres (1 nm = 10⁻⁹ m). Visible light runs from about 400 nm (violet) to 700 nm (red).
- Frequency (ν, nu): the number of waves passing a point each second, in hertz (Hz = s⁻¹).
- Speed: in a vacuum all electromagnetic waves travel at c = 3.00 × 10⁸ m/s.
These are linked by c = λν. Longer wavelength means lower frequency.
Light as particles: photons
In 1900 Max Planck explained the glow of hot objects by proposing that energy is given off in small packets called quanta. In 1905 Albert Einstein used the same idea to explain the photoelectric effect (light knocking electrons out of a metal): light itself comes in packets, now called photons. The energy of one photon is
E = hν = hc/λ
where h = 6.626 × 10⁻³⁴ J·s is Planck's constant. Red light, however bright you make it, cannot knock electrons out of some metals, while even weak ultraviolet light can. Each red photon simply does not carry enough energy. Light therefore behaves as both a wave and a stream of particles.
Continuous and line spectra
A glowing solid such as a stove element or an incandescent bulb filament gives a continuous spectrum: all colours blend into a rainbow. A gas of one element, excited by heat or electricity, gives an emission (line) spectrum: only certain wavelengths, as bright lines on a dark background. If white light shines through a cool gas, the gas absorbs those same wavelengths, giving dark lines on a rainbow: an absorption spectrum.
Bohr's model of hydrogen (1913)
Niels Bohr combined Rutherford's nucleus with Planck's quanta. His model of the hydrogen atom has three main ideas:
- The electron can only move in certain allowed orbits, called energy levels (n = 1, 2, 3, ...). In these orbits it does not radiate energy, so the atom is stable.
- The ground state is the lowest level (n = 1). When the atom absorbs exactly the right amount of energy, the electron jumps to a higher level, an excited state.
- When the electron drops to a lower level, it gives off one photon whose energy equals the difference between the two levels: E(photon) = E(higher) − E(lower).
Because only certain levels exist, only certain energy differences exist, so only certain wavelengths are given off. That is exactly a line spectrum. Energy is quantized: like standing on a staircase, not a ramp.
For hydrogen, the energy of level n is
Eₙ = −2.18 × 10⁻¹⁸ J ÷ n²
The energies are negative because the electron is bound: you must add energy to remove it, and a free electron (n = ∞) has energy 0.
What Bohr explained, and where he failed
Bohr's formula reproduced the wavelengths of hydrogen's lines extremely well, including the ultraviolet Lyman series (drops to n = 1), the visible Balmer series (drops to n = 2), and the infrared Paschen series (drops to n = 3). It also gave hydrogen's ionization energy correctly.
But the model failed for every atom with more than one electron, even helium. It could not explain why some spectral lines are brighter than others, or why lines split into several closely spaced lines in a magnetic field. And it gave no reason why only certain orbits are allowed. Bohr's lasting idea is the quantized energy level; his circular orbits were replaced by the quantum mechanical model (next lesson).
Worked exampleExample 1: Wavelength and frequency
The red line of hydrogen has a wavelength of 656 nm. Find its frequency and the energy of one photon.
- Convert: 656 nm = 656 × 10⁻⁹ m = 6.56 × 10⁻⁷ m.
- ν = c/λ = (3.00 × 10⁸ m/s) ÷ (6.56 × 10⁻⁷ m) = 4.57 × 10¹⁴ Hz.
- E = hν = (6.626 × 10⁻³⁴ J·s)(4.573 × 10¹⁴ s⁻¹) = 3.03 × 10⁻¹⁹ J.
ν = 4.57 × 10¹⁴ Hz; E = 3.03 × 10⁻¹⁹ J per photon.
Worked exampleExample 2: A Bohr transition
Calculate the wavelength of the photon emitted when a hydrogen electron drops from n = 4 to n = 2. Which line is it?
- E₄ = −2.18 × 10⁻¹⁸ ÷ 16 = −1.3625 × 10⁻¹⁹ J. E₂ = −2.18 × 10⁻¹⁸ ÷ 4 = −5.45 × 10⁻¹⁹ J.
- ΔE = E₄ − E₂ = −1.3625 × 10⁻¹⁹ − (−5.45 × 10⁻¹⁹) = 4.0875 × 10⁻¹⁹ J. (A shortcut: ΔE = 2.18 × 10⁻¹⁸ × (1/4 − 1/16) = 2.18 × 10⁻¹⁸ × 0.1875.)
- λ = hc/ΔE = (6.626 × 10⁻³⁴)(3.00 × 10⁸) ÷ 4.0875 × 10⁻¹⁹ = 1.9878 × 10⁻²⁵ ÷ 4.0875 × 10⁻¹⁹ = 4.86 × 10⁻⁷ m.
λ = 486 nm, the blue-green line in the Balmer series.
Worked exampleExample 3: Ionization energy of hydrogen
How much energy is needed to remove the electron from a ground-state hydrogen atom? Express it in kilojoules per mole.
- Removing the electron means going from n = 1 (−2.18 × 10⁻¹⁸ J) to n = ∞ (0 J). ΔE = 2.18 × 10⁻¹⁸ J per atom.
- Per mole: 2.18 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹ = 1.313 × 10⁶ J/mol.
- Convert: 1.313 × 10⁶ J/mol = 1313 kJ/mol.
About 1.31 × 10³ kJ/mol, which matches the measured first ionization energy of hydrogen (about 1312 kJ/mol).
Avoiding common slips
- Always convert nm to m before using c = λν or E = hc/λ.
- The photon energy is always positive; it is the size of the gap between levels.
- Bigger jumps give more energetic photons and shorter wavelengths: n = 6 → 2 (410 nm, violet) is a bigger drop than n = 3 → 2 (656 nm, red).
Just for funHelium was discovered in the Sun before it was found on Earth. In 1868 astronomers studying sunlight saw a yellow spectral line that matched no known element, and the new element was named after Helios, the Greek word for the Sun. It is probably the only element first detected about 150 million kilometres away.
Hands-on activitySee line spectra at home
You need: an old CD or DVD (the shiny side works as a diffraction grating), a dark room, and several light sources: an incandescent or halogen bulb if you have one, an LED or compact fluorescent bulb, and a phone or computer screen showing a white page.
Safety: never look at the Sun, at a laser, or directly into a very bright bulb; look at the light reflected in the disc, not at the bulb itself. Compact fluorescent bulbs contain a little mercury; do not use a broken one.
- In a dark room, hold the disc about 30 cm from your eye and tilt it until you see a rainbow band reflected from the light source.
- Compare the sources. Sketch what you see for each: a smooth rainbow or separate coloured bands or lines?
- If you can visit a place with neon signs or a street lamp, look at them with the disc too.
What to look for: an incandescent bulb gives a smooth, continuous spectrum (a hot solid). A compact fluorescent bulb shows bright separate lines (green, blue-violet, orange) because the mercury vapour inside gives a line spectrum. White LEDs usually show a broad blue band and a broad yellow-to-red band. If you can, look at a sodium street lamp: it is almost pure yellow-orange, one strong line. Write a sentence explaining which sources contain a hot solid and which contain an excited gas.
Your turn
1Find the frequency of violet light with λ = 410 nm.
ν = c/λ = 3.00 × 10⁸ ÷ 4.10 × 10⁻⁷ = 7.32 × 10¹⁴ Hz.
2Find the energy of one photon of the 410 nm light.
E = hc/λ = 1.9878 × 10⁻²⁵ ÷ 4.10 × 10⁻⁷ = 4.85 × 10⁻¹⁹ J.
3Which has more energy per photon, a microwave photon or an X-ray photon? Explain.
The X-ray photon: it has a much shorter wavelength and higher frequency, and E = hν.
4Explain the difference between a continuous spectrum and a line spectrum, with one source of each.
A continuous spectrum contains all wavelengths (a hot solid, such as an incandescent filament). A line spectrum contains only certain wavelengths (an excited gas of one element, such as hydrogen or neon in a discharge tube).
5Use Bohr's model to explain why hydrogen gives off only certain wavelengths.
The electron can only have certain energy levels. When it drops between levels it emits a photon with energy equal to the difference. Only certain differences exist, so only certain photon energies, and therefore wavelengths, are produced.
6Calculate the energy of the n = 3 level of hydrogen.
E₃ = −2.18 × 10⁻¹⁸ ÷ 9 = −2.42 × 10⁻¹⁹ J.
7Calculate the wavelength of the photon emitted in the n = 3 → n = 2 transition.
ΔE = 2.18 × 10⁻¹⁸(1/4 − 1/9) = 2.18 × 10⁻¹⁸ × 0.13889 = 3.028 × 10⁻¹⁹ J. λ = 1.9878 × 10⁻²⁵ ÷ 3.028 × 10⁻¹⁹ = 6.56 × 10⁻⁷ m = 656 nm (red).
More practice: warm-up
1What is the speed of light in a vacuum?
2Convert 500 nm to metres.
A packet (quantum) of light energy.
4What is the lowest energy level of an atom called?
5Does an electron absorb or emit energy when it moves from n = 2 to n = 5?
It absorbs energy (it moves to a higher level).
More practice: on your own
1A radio station broadcasts at 99.9 MHz. Find the wavelength.
λ = c/ν = 3.00 × 10⁸ ÷ 9.99 × 10⁷ = 3.00 m.
2A photon has an energy of 3.98 × 10⁻¹⁹ J. Find its wavelength and colour.
λ = hc/E = 1.9878 × 10⁻²⁵ ÷ 3.98 × 10⁻¹⁹ = 4.99 × 10⁻⁷ m = 499 nm, in the blue-green part of the visible spectrum.
3Calculate the wavelength of the n = 2 → n = 1 transition. In which region of the spectrum is it?
ΔE = 2.18 × 10⁻¹⁸ (1 − 1/4) = 1.635 × 10⁻¹⁸ J. λ = 1.9878 × 10⁻²⁵ ÷ 1.635 × 10⁻¹⁸ = 1.216 × 10⁻⁷ m = 122 nm, ultraviolet (Lyman series).
4Why is dim ultraviolet light able to knock electrons out of a metal surface when bright red light cannot?
Whether an electron is ejected depends on the energy of each photon (E = hν), not on the number of photons. UV photons have enough energy each; red photons do not, however many there are.
5State two things the Bohr model could not explain.
Any of: spectra of atoms with more than one electron; the different brightness of lines; the splitting of lines in a magnetic field; why only certain orbits are allowed.
Challenge
1A hydrogen electron emits a 1875 nm infrared photon when it drops to n = 3. From which level did it start?
ΔE = hc/λ = 1.9878 × 10⁻²⁵ ÷ 1.875 × 10⁻⁶ = 1.060 × 10⁻¹⁹ J. Then 1/9 − 1/nᵢ² = 1.060 × 10⁻¹⁹ ÷ 2.18 × 10⁻¹⁸ = 0.04863. 1/nᵢ² = 0.11111 − 0.04863 = 0.06248, so nᵢ² = 16.0 and nᵢ = 4 (Paschen series).
2How many different photon energies could be emitted by a collection of hydrogen atoms whose electrons all start at n = 4 and end at n = 1, if they can drop in one step or several?
Count the pairs of levels from 1 to 4: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1 = 6 different energies (6 lines).
3Show that the energy needed to ionize a hydrogen atom from n = 2 is one quarter of that needed from n = 1, and calculate it in kJ/mol.
From n: ΔE = 2.18 × 10⁻¹⁸/n². For n = 2: 2.18 × 10⁻¹⁸ ÷ 4 = 5.45 × 10⁻¹⁹ J, exactly ¼ of the n = 1 value. Per mole: 5.45 × 10⁻¹⁹ × 6.022 × 10²³ = 3.28 × 10⁵ J = 328 kJ/mol.
Quiz
1. Which colour of visible light has photons with the most energy?
2. What does a line spectrum tell us about the electrons in an atom?
3. In the Bohr model, what happens when an electron drops from n = 3 to n = 2?
4. What is the frequency of light with a wavelength of 600 nm?
5. Who first proposed that energy is emitted in small packets called quanta?
6. What was the main weakness of the Bohr model?
Big idea (C3.1): light comes in packets (photons), and atoms can only absorb or emit photons of particular energies because their electrons have quantized energy levels. This explains line spectra.
How to help: the CD activity is a genuinely good way to see the difference between a continuous spectrum and a line spectrum. Make sure your teen looks at reflected light only, never at the Sun. When checking calculations, look first at unit conversions (nm to m) and powers of ten.
Common difficulties:
- Forgetting to convert nanometres to metres.
- Thinking a brighter light has more energetic photons (brightness is the number of photons; energy per photon depends on frequency).
- Sign confusion with Eₙ. The levels are negative; the photon energy is the positive difference.
- Mixing up "absorbs" (moves up) and "emits" (moves down).