Look at an old sodium street lamp through the shiny side of a CD, held at a slant. Beside the lamp you see a copy of it in one colour only, deep yellow. In the physics laboratory, a thin tube of glowing hydrogen seen the same way gives four copies: red, blue-green, and two violet ones close together.
From energies to wavelengths
A jump from level \(n_i\) down to level \(n_f\) gives a photon of energy \(\Delta E = E_{n_i} - E_{n_f}\). Its wavelength is \(\lambda = hc/\Delta E\). Putting in Bohr's energies turns this into one formula for every line of every hydrogen-like atom.
\[ \Delta E = E_{n_i} - E_{n_f} = 13.6\,Z^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)\ \text{eV} \]
Why. \(E_n = -13.6\,Z^2/n^2\) eV. The minus signs turn the difference round.
\[ \dfrac{1}{\lambda} = \dfrac{\Delta E}{hc} \]
Why. the photon's energy is \(hc/\lambda\).
\[ \dfrac{1}{\lambda} = \dfrac{13.6\,\text{eV}}{hc}\,Z^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) = RZ^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) \]
Read it. the constant \(R = 13.6\,\text{eV}/hc\) collects everything that does not depend on the jump.
The constant is the Rydberg constant, \(R = 13.6/1240\,\text{nm}^{-1} = 1.097 \times 10^7\,\text{m}^{-1}\). Its inverse is a handy length: \(1/R = 1240/13.6 = 91.2\,\text{nm}\). Before Bohr, spectroscopists had found this formula by fitting measured lines, and they had measured \(R\). Bohr's model gave its value from \(e\), \(m\), \(h\) and \(c\).
Fig 5.1 shows the formula as a graph. Every line of hydrogen, whatever its jump, falls on one straight line through the origin. Its slope is \(R\): for the jump from 2 to 1, \(1/\lambda = 8.23 \times 10^6\,\text{m}^{-1}\) at a bracket of \(0.75\), and \(8.23 \times 10^6/0.75 = 1.097 \times 10^7\,\text{m}^{-1}\).
Find the wavelength of the line from \(n = 3\) to \(n = 2\) in hydrogen, (a) from the energies and (b) from the Rydberg formula. Then find the same line for \(\text{He}^+\).
(a) \(\Delta E = 13.6(1/4 - 1/9) = 13.6 \times 5/36 = 1.89\,\text{eV}\), so \(\lambda = 1240/1.889 = 656.5\,\text{nm}\), red.
(b) \(1/\lambda = R(1/4 - 1/9) = 5R/36\), so \(\lambda = 36/5R = 7.2 \times 91.18 = 656.5\,\text{nm}\).
(c) For \(\text{He}^+\), \(Z^2 = 4\), so every \(1/\lambda\) is four times larger: \(\lambda = 656.5/4 = 164.1\,\text{nm}\), in the ultraviolet.
Values used: \(hc = 1240\,\text{eV nm}\), ground state \(-13.6\,\text{eV}\), \(1/R = 1240/13.6 = 91.18\,\text{nm}\).
Different books give this line as \(656.3\) or \(656.5\,\text{nm}\). With \(R = 1.097 \times 10^7\,\text{m}^{-1}\) exactly you get \(656.3\,\text{nm}\); with \(1240/13.6\) you get \(656.5\,\text{nm}\). The difference is only the rounding of the constants, about three parts in ten thousand. Either answer is right; do not mix the two routes in one calculation.
The line from \(n = 4\) to \(n = 2\) in hydrogen has a wavelength of \(486.3\,\text{nm}\). What is the wavelength, in nm, of the same jump in \(\text{He}^+\) (\(Z = 2\))?
Families of lines
The lines of hydrogen fall into families. All the lines that end on the same lower level form a spectral series, named after its discoverer. The series ending on \(n_f = 1\) is the Lyman series. The one ending on \(n_f = 2\) is the Balmer series, and the ones ending on \(3\), \(4\) and \(5\) are the Paschen, Brackett and Pfund series.
The size of the last step decides the colour. Falls to \(n = 1\) cross the big first gap of \(10.2\,\text{eV}\) or more, so the Lyman lines are all ultraviolet. Falls to \(n = 2\) release between \(1.89\) and \(3.40\,\text{eV}\), which puts the Balmer lines in the visible, the higher ones near its violet edge. Falls to \(n = 3\) and higher release less than \(1.51\,\text{eV}\), so those series are all infrared.
The first line and the limit of a series
Each series has a longest wavelength and a shortest one. The longest comes from the smallest jump into \(n_f\), from \(n_i = n_f + 1\); it is the first or α line of the series. The shortest comes from the largest jump, from \(n_i = \infty\). This is the series limit: the bracket becomes \(1/n_f^2\), so \(\lambda_{\text{limit}} = n_f^2/R\). Between them the lines crowd towards the limit, because the upper levels crowd together.
With \(hc = 1240\,\text{eV nm}\) and \(13.6\,\text{eV}\), the series of hydrogen are:
- Lyman, \(n_f = 1\): \(121.6\), \(102.6\), \(97.3\,\text{nm}\), …, limit \(91.2\,\text{nm}\). Ultraviolet.
- Balmer, \(n_f = 2\): \(656.5\) (red), \(486.3\) (blue-green), \(434.2\) (violet), \(410.3\,\text{nm}\) (violet), …, limit \(364.7\,\text{nm}\). Visible, with the limit in the near ultraviolet.
- Paschen, \(n_f = 3\): first line \(1876\,\text{nm}\), limit \(821\,\text{nm}\). Infrared.
- Brackett, \(n_f = 4\): first line \(4052\,\text{nm}\), limit \(1459\,\text{nm}\). Infrared.
- Pfund, \(n_f = 5\): first line \(7460\,\text{nm}\), limit \(2279\,\text{nm}\). Infrared.
Find the longest and the shortest wavelengths of the Paschen series of hydrogen. In which region of the spectrum do they lie?
The Paschen series ends on \(n_f = 3\). The longest line is the jump from 4; the shortest is the jump from \(\infty\).
Longest: \(1/\lambda = R(1/9 - 1/16) = 7R/144\). Shortest: \(1/\lambda = R/9\).
Longest: \(\lambda = 144/7R = 20.57 \times 91.18 = 1876\,\text{nm}\). Shortest: \(\lambda = 9/R = 9 \times 91.18 = 821\,\text{nm}\).
The Paschen series runs from \(1876\,\text{nm}\) down to \(821\,\text{nm}\). Both are longer than red light, about \(750\,\text{nm}\), so the whole series is infrared.
The electron of a hydrogen atom falls from \(n = 4\) to \(n = 2\). To which series does the line belong, and where does it lie?
Find the ratio of the longest wavelength of the Lyman series to the longest wavelength of the Balmer series of hydrogen. Give it as a decimal.
Which jump gives the shortest wavelength of the Lyman series, \(91.2\,\text{nm}\)?
Johann Balmer taught mathematics at a girls' school in Basel and loved numerical puzzles. A friend gave him the four visible wavelengths of hydrogen, measured by Anders Ångström. In 1885, at the age of sixty, Balmer found that all four follow one simple rule with the whole numbers 3, 4, 5 and 6. He predicted a fifth line, and it had already been seen. Twenty-eight years later his formula led Bohr to the atom.
How many lines?
Suppose a gas of hydrogen atoms is excited to level \(n\). The atoms do not all come down the same way. An electron on level \(n\) can fall to any of the \(n - 1\) levels below it, and from there to any level below that. Every pair of levels gives one line. So the count of lines is the number of pairs among the \(n\) levels.
Count from the top, as in Fig 5.3. Level \(n\) offers \(n - 1\) jumps, level \(n - 1\) offers \(n - 2\), and so on down to level 2, which offers one. The sum is \((n - 1) + (n - 2) + \ldots + 1 = n(n - 1)/2\). A single atom, however, gives at most \(n - 1\) photons, one per step, if it comes down one level at a time.
Hydrogen atoms are excited to the level \(n = 5\). What is the largest number of different spectral lines they can give out?
Hydrogen atoms excited to \(n = 4\) give six lines. How many of these belong to the Balmer series?
A single hydrogen atom is in the level \(n = 4\). What is the largest number of photons it can give out on its way to the ground state?
Now the two lamps through the CD. Hydrogen's visible lines are its Balmer lines, the falls to \(n = 2\) from \(3\), \(4\), \(5\) and \(6\): \(656.5\) red, \(486.3\) blue-green, \(434.2\) and \(410.3\,\text{nm}\) violet. The upper levels crowd together, so the two violet lines sit close. The sodium lamp shows one yellow line because one gap of sodium, about \(2.1\,\text{eV}\), outshines all its others.
- Naming a series by the upper level.The lower level names the series: every Balmer line ends on \(n = 2\), whatever level it starts from.
- Taking the first line of a series as its shortest wavelength.The first line is the smallest jump, so it is the longest wavelength. The series limit is the shortest.
- Using \(n(n - 1)/2\) for a single atom.That counts lines from many atoms. One atom gives at most \(n - 1\) photons.