Line spectra of hydrogen
Jumps make lines
In a discharge tube, collisions knock hydrogen atoms into excited states. The electron then drops back to lower levels, emitting one photon in each jump.
By Bohr’s third postulate, a jump from an initial level n i to a final level n f gives a photon of energy h ν equal to E n i minus E n f.
Using E n equals minus m e⁴ over 8 n² ε₀² h², the frequency is ν equals m e⁴ over 8 ε₀² h³, times 1 over n f² minus 1 over n i².
Divide by c to get 1 over λ. This is the Rydberg formula: 1 over λ equals R times, 1 over n f² minus 1 over n i², with R equal to m e⁴ over 8 ε₀² h³ c.
Put in the values of m, e, ε₀, h and c, and R comes out at very nearly 1.097 × 10⁷ per metre. That is exactly the constant in Balmer’s formula, found from experiment years before. This was Bohr’s great success.
Now we can see where each series comes from. All jumps that end on n equal to 1 form the Lyman series. They are large jumps, so the photons are energetic: ultraviolet.
Jumps ending on n equal to 2 form the Balmer series, in the visible. Jumps ending on n equal to 3 give the Paschen series, in the infrared. Ending on 4 and 5: the Brackett and Pfund series, further in the infrared.
Within each series, a jump from n i equal to infinity gives the shortest wavelength, the series limit. That is why the lines crowd together at one end.
Working out the lines
Let us find some wavelengths. A handy shortcut: h c is about 1240 electron volt nanometre, so λ in nanometres is 1240 divided by the photon energy in eV.
Jump from n equal to 3 to n equal to 2. The energy given out is minus 1.51 minus minus 3.40, which is 1.89 eV. λ equals 1240 over 1.89, which is 656 nanometre. The red line of hydrogen, the first Balmer line.
Jump from n equal to 2 to n equal to 1. Energy: 13.6 minus 3.4, which is 10.2 eV. λ equals 1240 over 10.2, about 122 nanometre. This is ultraviolet, the first Lyman line.
From infinity down to n equal to 1, the photon carries 13.6 eV. λ equals 1240 over 13.6, which is 91.2 nanometre: the Lyman series limit.
And a jump from 4 to 3 releases only 1.51 minus 0.85, about 0.66 eV. Its wavelength is about 1875 nanometre, in the infrared: the first Paschen line.
An atom excited to n equal to 4 can come down in several ways: straight to 1, or in steps. Count the possible jumps: 4 to 3, 4 to 2, 4 to 1, 3 to 2, 3 to 1, and 2 to 1. Six lines in all.
In general, from level n the number of possible lines is n times n minus 1, over 2. A gas has billions of atoms, so all these lines appear together.
Absorption works the other way. Hydrogen at room temperature has its electrons in n equal to 1. It can absorb only photons that lift the electron up from n equal to 1: the Lyman wavelengths.
Try it: make the lines
- The electron starts in n = 3 and drops to n = 2: a red photon of 656 nm. See where it falls on the spectrum.
- Make every jump that ends on n = 2. They are the Balmer lines, all in or near the visible band.
- Jumps ending on n = 1 (Lyman) land in the ultraviolet; jumps ending on n = 3 (Paschen) in the infrared.
- Send the electron to n = 4, then find all six different ways down. Each one is a separate line.
- Press ∞ and then 1: that photon marks the Lyman series limit, 91.2 nm.
From n = 4 to n = 2
An electron in a hydrogen atom drops from n = 4 to n = 2. What is the photon’s energy, and where is it in the spectrum?
- 12.75 eV, ultraviolet
- 2.55 eV, visible (486 nm)
- 0.66 eV, infrared
- 10.2 eV, ultraviolet
Answer
B. E₄ = −13.6/16 = −0.85 eV and E₂ = −13.6/4 = −3.40 eV. The photon carries −0.85 − (−3.40) = 2.55 eV, so λ = 1240/2.55 ≈ 486 nm, the blue-green Balmer line. 12.75 eV is the jump 4 → 1, 0.66 eV is 4 → 3, and 10.2 eV is 2 → 1.
How many lines?
Hydrogen atoms are excited to the n = 4 level. How many different spectral lines can they emit on returning to the ground state?
- 3
- 4
- 6
- 10
Answer
C. Each pair of levels gives one line: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1. That is n(n − 1)/2 = 4 × 3/2 = 6 lines. 3 counts only the jumps from n = 4; 10 would be for n = 5.
Summary
- Photon from a jump:
hν = E_i − E_f - Rydberg formula:
1/λ = R (1/n_f² − 1/n_i²), R = me⁴/(8ε₀²h³c) = 1.097 × 10⁷ m⁻¹ - Handy form:
λ (nm) ≈ 1240 / ΔE (eV) - Number of lines from level n:
n(n − 1)/2
- Bohr’s model derives the Rydberg constant from m, e, ε₀, h and c, and it matches the experimental value.
- Series by final level: n_f = 1 Lyman (UV), 2 Balmer (visible), 3 Paschen, 4 Brackett, 5 Pfund (IR).
- Examples: 3→2 gives 656 nm (red), 2→1 gives 122 nm (UV), ∞→1 gives the Lyman limit 91.2 nm.
- Cold hydrogen absorbs only lines that start from n = 1, the Lyman series.