The colours of hydrogen
Tonight's idea
Quantum theory explains the sharp colours. The electron in a hydrogen atom cannot have just any energy. Only certain levels are allowed, numbered \(n = 1, 2, 3, \dots\), and level \(n\) has energy about \(-\frac{13.6}{n^2}\) electronvolts (eV, a tiny unit of energy). When the atom drops from a higher level to a lower one, it gives out the difference as one particle of light, a photon. Fixed levels mean fixed drops, and each drop has its own colour. A photon's energy is proportional to \(\frac{1}{\lambda}\), where \(\lambda\) is the wavelength of the light. Bigger drop, shorter wavelength.
Drops down to level 2 give the Balmer lines. From level \(n = 3, 4, 5, \dots\) the wavelength satisfies \[\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{n^2}\right), \qquad R \approx 1.097 \times 10^{7} \text{ per metre.}\]
The \(\frac{1}{n^2}\) pattern belongs to hydrogen, the atom with one electron. Other atoms glow in sharp lines too, but not by this formula.
Watch one: the drop from level 3 to level 2
Energy. Level 2 is at \(-\frac{13.6}{4} = -3.40\) eV and level 3 is at \(-\frac{13.6}{9} \approx -1.51\) eV. The photon carries off the difference, \(-1.51 - (-3.40) = 1.89\) eV.
Wavelength. Put \(n = 3\) in the formula: \(\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}\). So \(\frac{1}{\lambda} = \frac{5R}{36}\), and \[\lambda = \frac{36}{5R} = \frac{36}{5 \times 1.097 \times 10^{7}} \approx 6.56 \times 10^{-7} \text{ m}.\] In nanometres (1 nm is \(10^{-9}\) m) that is about 656 nm. Eyes see from roughly 380 nm (violet) to roughly 750 nm (red), so this is red: the red line of hydrogen, measured in air at 656.3 nm.
Your turn
- The drop from level 4 to level 2. (a) Work out \(\frac{1}{4} - \frac{1}{16}\) as a fraction, then find \(\lambda\) in nm. What colour is it? (b) Find the energy of level 4 and the energy the photon carries off. (c) For this line and for the red line, multiply the photon's energy in eV by its wavelength in nm. What do you notice?
- Without using the value of \(R\), find the exact ratio of the red wavelength to the wavelength in exercise 1. Then compare it with the wavelengths measured in air, 656.3 nm and 486.1 nm.
- Find the wavelengths for the drops from level 5 and from level 6 to level 2. Then list the gaps, in nm, between neighbouring lines from red downwards. What is happening to the gaps?
- Now let \(n\) get very large. (a) What does \(\frac{1}{n^2}\) get close to, and what wavelength do the lines close in on? Can you see light of that wavelength? (b) Which Balmer line has the longest wavelength? (c) True or false: hydrogen has a Balmer line at 550 nm, in the green.
Stretch: the other families
Drops down to level 1 are called the Lyman lines, and drops down to level 3 the Paschen lines. The formula changes in one place: \(\frac{1}{4}\) becomes \(\frac{1}{1}\) for Lyman and \(\frac{1}{9}\) for Paschen. (a) For each family, find the longest wavelength (the drop from the next level up) and the wavelength the lines close in on as \(n\) grows. (b) Can you see any Lyman or Paschen line? (c) What about drops down to level 4 or higher?
Answers
- (a) \(\frac{1}{4} - \frac{1}{16} = \frac{3}{16}\), so \(\lambda = \frac{16}{3R} \approx 4.86 \times 10^{-7}\) m, about 486 nm. That is blue-green. (b) Level 4 is at \(-\frac{13.6}{16} = -0.85\) eV, so the photon carries \(-0.85 - (-3.40) = 2.55\) eV. A bigger drop than the red line's 1.89 eV, and a shorter wavelength. (c) \(1.89 \times 656 \approx 1240\) and \(2.55 \times 486 \approx 1239\). The products agree, up to rounding. This is what "energy is proportional to \(\frac{1}{\lambda}\)" means: energy times wavelength is the same for every photon, about 1240 when you use eV and nm.
- \(\frac{36}{5R} \div \frac{16}{3R} = \frac{36}{5} \times \frac{3}{16} = \frac{27}{20} = 1.35\). The \(R\) cancels, so the ratio is a pure number. Measured: \(656.3 \div 486.1 \approx 1.350\). The match is to four figures.
- From level 5: \(\frac{1}{4} - \frac{1}{25} = \frac{21}{100}\), so \(\lambda = \frac{100}{21R} \approx 434\) nm. From level 6: \(\frac{1}{4} - \frac{1}{36} = \frac{2}{9}\), so \(\lambda = \frac{9}{2R} \approx 410\) nm. Both are violet. The lines are at about 656, 486, 434 and 410 nm, so the gaps are 170, 52 and 24 nm. The gaps shrink quickly: the lines crowd together.
- (a) \(\frac{1}{n^2}\) gets close to 0, so \(\frac{1}{\lambda}\) gets close to \(\frac{R}{4}\) and \(\lambda\) closes in on \(\frac{4}{R} \approx 365\) nm. That is shorter than about 380 nm, so it is ultraviolet and you cannot see it. The lines pile up just above 365 nm and never pass it. (b) The red one, from level 3. A bigger \(n\) makes \(\frac{1}{4} - \frac{1}{n^2}\) bigger, so \(\lambda\) is smaller. (c) False. Level 3 gives 656 nm and level 4 gives 486 nm, and there is no level between 3 and 4, so nothing lands between those two lines. Every Balmer line lies between about 365 and 656 nm. Johann Balmer, a Swiss schoolteacher, found this formula in 1885 by fitting the measured lines. Niels Bohr explained it with energy levels in 1913.
Stretch. (a) Lyman: the longest comes from level 2, with \(1 - \frac{1}{4} = \frac{3}{4}\), so \(\lambda = \frac{4}{3R} \approx 122\) nm. As \(n\) grows the lines close in on \(\frac{1}{R} \approx 91\) nm. Paschen: the longest comes from level 4, with \(\frac{1}{9} - \frac{1}{16} = \frac{7}{144}\), so \(\lambda = \frac{144}{7R} \approx 1875\) nm. The lines close in on \(\frac{9}{R} \approx 820\) nm. (b) No. Every Lyman line is between about 91 and 122 nm, far into the ultraviolet. Every Paschen line is between about 820 and 1875 nm, which is infrared, past the red end of what eyes see. (c) Drops to level \(m\) close in on \(\frac{m^2}{R}\), and all their lines are longer than that. For \(m = 4\) it is already about 1459 nm, and it grows with \(m\). So these are all infrared. Level 2 is the only landing level that gives visible light, taking visible to mean roughly 380 to 750 nm.