Hydrogen spectrum: spectral series, emission and absorption
An electron falling between levels emits hν = Eₙ − Eₘ. Falls ending on levels 1, 2 and 3 form the Lyman (UV), Balmer (visible) and Paschen (infrared) series.
Emission and absorption spectra
Heat a gas, or pass an electric discharge through it at low pressure, and it glows. Seen through a spectroscope, its light is a set of bright lines at fixed wavelengths. This is an emission line spectrum. Each element has its own set of lines, like a fingerprint. Now pass white light through the cool gas. Dark lines appear at the same wavelengths, because the gas takes in just those photons. This is an absorption spectrum.
How Bohr's model explains the lines
- 1Heat or an electric discharge gives energy to the atoms and lifts electrons to higher energy levels (excited states).
- 2An excited atom stays in that state only for a very short time.
- 3The electron falls to a lower level m and gives out one photon: hν = Eₙ − Eₘ.
- 4The levels are fixed, so only certain energy gaps exist. So only certain wavelengths appear: a line spectrum.
- 5In absorption, an electron takes in a photon only if its energy exactly matches a gap, which gives dark lines at the same wavelengths.
The spectral series of hydrogen
| Series | Falls to | Falls from | Region |
|---|---|---|---|
| Lyman | m = 1 | n = 2, 3, 4, ... | Ultraviolet |
| Balmer | m = 2 | n = 3, 4, 5, ... | Visible |
| Paschen | m = 3 | n = 4, 5, 6, ... | Near infrared |
| Brackett | m = 4 | n = 5, 6, 7, ... | Mid infrared |
| Pfund | m = 5 | n = 6, 7, 8, ... | Far infrared |
Worked example: a blue-green line of hydrogen
In a hydrogen atom the electron falls from n = 4 to n = 2. Find the wavelength of the light given out. Is it visible, and which series is it in? (R = 1.097 × 10⁷ m⁻¹)
- Formula: 1/λ = R(1/m² − 1/n²), with m = 2 and n = 4.
- 1/2² − 1/4² = 1/4 − 1/16 = 3/16.
- 1/λ = 1.097 × 10⁷ × 3/16 = 2.057 × 10⁶ m⁻¹.
- λ = 1 ÷ 2.057 × 10⁶ = 4.86 × 10⁻⁷ m = 486 nm.
- 486 nm lies between 400 nm and 700 nm, so it is visible (blue-green). It ends on m = 2, so it is in the Balmer series.
λ ≈ 486 nm (4860 Å): visible blue-green light in the Balmer series
Worked example: limits of the Balmer series
Find the longest and shortest wavelengths of the Balmer series of hydrogen. (R = 1.097 × 10⁷ m⁻¹)
- Longest wavelength comes from the smallest gap: n = 3 to m = 2.
- 1/λ = R(1/4 − 1/9) = 5R/36, so λ = 36 ÷ (5 × 1.097 × 10⁷) = 6.56 × 10⁻⁷ m = 656 nm.
- Shortest wavelength comes from the largest gap: n = ∞ to m = 2.
- 1/λ = R(1/4 − 0) = R/4, so λ = 4 ÷ (1.097 × 10⁷) = 3.65 × 10⁻⁷ m = 365 nm.
Longest ≈ 656 nm (red); shortest ≈ 365 nm (just into the ultraviolet)
How to draw the transitions diagram
- Draw the levels n = 1 to 5 and ∞ as horizontal lines that crowd together near the top.
- Draw downward arrows ending on n = 1 and label the group Lyman series (ultraviolet).
- Draw arrows ending on n = 2 and label them Balmer series (visible); mark the 3 to 2 arrow red line, 656 nm.
- Draw arrows ending on n = 3, 4 and 5 and label them Paschen, Brackett and Pfund (infrared).
- In each group, the longest arrow gives the shortest wavelength.
The dark Fraunhofer lines in sunlight are an absorption spectrum: cooler gases in the Sun's outer layer absorb certain wavelengths. Matching them with lines seen on Earth showed hydrogen, sodium, iron and calcium in the Sun.
3 marks: “An electron in a hydrogen atom falls from n = 5 to n = 2. Find the wavelength of the light emitted and name its series. (R = 1.097 × 10⁷ m⁻¹)” 1/λ = R(1/4 − 1/25) = 0.21R = 2.30 × 10⁶ m⁻¹. λ ≈ 4.34 × 10⁻⁷ m = 434 nm, violet light, in the Balmer series.
2 marks: “Distinguish between an emission spectrum and an absorption spectrum.” Emission: bright lines on a dark background, given out by an excited gas. Absorption: dark lines in a continuous spectrum when white light passes through a cool gas. For one element, both have the same wavelengths.
1 marks: “How many spectral lines can hydrogen atoms excited to n = 4 emit?” n(n − 1)/2 = 4 × 3 ÷ 2 = 6.
Lazy Boys Prefer Bed Pillows: Lyman, Balmer, Paschen, Brackett and Pfund end on levels 1, 2, 3, 4 and 5.