38.7.2 The Emission Spectrum

According to the third assumption of Bohr’s model of atomic quantization, the frequency of the photon emitted in an $n \to m$ transition is

f = \frac{Δ E_{atom}}{h} = \frac{E_{n} - E_{m}}{h}

(38.37)

We can use Equation 38.33 for the energies $E_{n}$ and $E_{m}$ to predict that the emitted photon has frequency

\begin{array}{l} f & = & \frac{1}{h} {[- \frac{1}{n^{2}} (\frac{1}{4 π ϵ_{0}} \frac{e^{2}}{2 a_{B}})] - [\frac{1}{m^{2}} (\frac{1}{4 π ϵ_{0}} \frac{e^{2}}{2 a_{B}})]} \\ = & \frac{1}{4 π ϵ_{0}} \frac{e^{2}}{2 h a_{B}} (\frac{1}{m^{2}} - \frac{1}{n^{2}}) \end{array}

(38.38)

The frequency is a positive number because $m < n$ and thus $1 / m^{2} > 1 / n^{2}$ .

We are more interested in wavelength than frequency, because wavelengths are the quantity measured by experiment. The wavelength of the photon emitted in an $n \to m$ transition is

λ_{n \to m} = \frac{c}{f} = \frac{8 π ϵ_{0} h c a_{B} / e^{2}}{(\frac{1}{m^{2}} - \frac{1}{n^{2}})}

(38.39)

This looks rather gruesome, but notice that the numerator is simply a collection of various constants. The value of the numerator, which we can call $λ_{0}$ , is

λ_{0} = \frac{8 π ϵ_{0} h c a_{B}}{e^{2}} = 9.112 \times 10^{- 8} m = 91.12 nm

With this definition, our prediction for the wavelengths in the hydrogen emission spectrum is

\begin{array}{l} λ_{n \to m} & = & \frac{λ_{0}}{(\frac{1}{m^{2}} - \frac{1}{n^{2}})} & \begin{array}{l} m = 1, 2, 3, \dots \\ n = any integer greater than m \end{array} \end{array}

(38.40)

This should look familiar. It is the Balmer formula from Chapter 37! However, there is one slight difference: Bohr’s analysis of the hydrogen atom has predicted $λ_{0} = 91.12 nm,$ whereas Balmer found, from experiment, that $λ_{0} = 91.18 nm$ . Could Bohr have come this close but then fail to predict the Balmer formula correctly?

The problem, it turns out, is in our assumption that the proton remains at rest while the electron orbits it. In fact, both particles rotate about their common center of mass, rather like a dumbbell with a big end and a small end. The center of mass is very close to the proton, which is far more massive than the electron, but the proton is not entirely motionless. The good news is that a more advanced analysis can account for the proton’s motion. It changes the energies of the stationary states ever so slightly—about 1 part in 2000—but that is precisely what is needed to give a revised value:

λ_{0} = 91.18 nm when corrected for the nuclear motion

It works! Unlike all previous atomic models, the Bohr hydrogen atom correctly predicts the discrete spectrum of the hydrogen atom. FIGURE 38.22 shows the Balmer series and the Lyman series transitions on an energy-level diagram. Only the Balmer series, consisting of transitions ending on the $m = 2$ state, gives visible wavelengths, and this is the series that Balmer initially analyzed. The Lyman series, ending on the $m = 1$ ground state, is in the ultraviolet region of the spectrum and was not measured until later. These series, as well as others in the infrared, are observed in a discharge tube where collisions with electrons excite the atoms upward from the ground state to state $n$ . They then decay downward by emitting photons. Only the Lyman series is observed in the absorption spectrum because, as noted previously, essentially all the atoms in a quiescent gas are in the ground state.

The figure illustrates transitions producing the Lyman series and the Balmer series of lines in the hydrogen spectrum.

FIGURE 38.22 Full Alternative Text

EXAMPLE 38.12 Hydrogen absorption

Whenever astronomers look at distant galaxies, they find that the light has been strongly absorbed at the wavelength of the $1 \to 2$ transition in the Lyman series of hydrogen. This absorption tells us that interstellar space is filled with vast clouds of hydrogen left over from the Big Bang. What is the wavelength of the $1 \to 2$ absorption in hydrogen?

SOLVE Equation 38.40 predicts the absorption spectrum of hydrogen if we let $m = 1$ . The absorption seen by astronomers is from the ground state of hydrogen $(m = 1)$ to its first excited state $(n = 2)$ . The wavelength is

λ_{1 \to 2} = \frac{91.18 nm}{(\frac{1}{1^{2}} - \frac{1}{2^{2}})} = 121.6 nm

REVIEW This wavelength is far into the ultraviolet. Ground-based astronomy cannot observe this region of the spectrum because the wavelengths are strongly absorbed by the atmosphere, but with space-based telescopes, first widely used in the 1970s, astronomers see 121.6 nm absorption in nearly every direction they look.