Section 4.8: The Bohr Atom and Atomic Spectra



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In 1913 Bohr constructed a model to explain the spectrum of the hydrogen atom. He started with a classical planetary model where the electron orbits the nucleus. Elliptical orbits can also be considered, but for simplicity we will use circular orbits.7  For such a circular orbit,

|F| = ma = mv2/r = e2/4πε0r2, (4.9)

since the electron experiences a Coulomb attraction to the nucleus. Given this result we immediately know that the electrons are accelerating toward the nucleus.   Accelerating charged particles radiate classically with the power radiated, P is proportional to q22a2/c, where a is the magnitude of the charged particle's acceleration. In fact, they radiate so fast that atoms would exist for only 10−10 seconds. Restart.

To counter this problem, Bohr postulated that electrons could be found in stable orbits, so-called stationary states. These stationary states were characterized by the electron's angular momentum being quantized, L = nh/2π = (n = 1, 2, 3, ...).  In addition he postulated that atomic radiation occurred when electrons transitioned from one stationary state to another.

For the hydrogen atom, the total energy is the kinetic plus the potential energy,

E = p2/2me2/4πε0r = L2/2mr2e2/4πε0r = n2ħ2/2mr2e2/4πε0r ,

where we have used Bohr's relationship for L.  Re-examining the equation for the force the electron experiences yields a relationship for the radius,

r = e2/4πε0mv2 = me2r2/4πε0L2, (4.10)

which can be written, using angular momentum quantization, as

rn = n2 [4πε0ħ2/me2], (4.11)

or n2 times the Bohr radius (0.53 × 10−10 m).  Using rn in the energy equation yields

E = −me2/8πε0n2ħ2 = −R/n2, (4.12)

where R = −13.6 eV is the Rydberg constant.

Using these results, Bohr could also explain the well-known Balmer lines, the visible emission spectra from the hydrogen atom as shown in this animation. You can click in the animation to measure the wavelength of a given spectral line. Recall that Bohr postulated that atomic radiation occurs when electrons transition from one stationary state to another. Therefore, consider the differences in the energies of two stationary states,

Eγ = ( 1/n2 − 1/m2 ) R , (4.13)

where n and m are integers with m > n. The Balmer lines, as shown in the animation, obey this equation with n = 2.  For these emission lines, we are seeing light emitted from the atom as it undergoes a transition down to n = 2.

Despite this success, there are problems with this model. This simple approach, often called old quantum theory, fails to explain the spectra of atoms with more than one electron. It also fails to predict transition rates and spectral line intensities, it gets the electron's angular momentum wrong (for example, in the ground state the electron has zero angular momentum and not ħ), and its use of exact circular orbits with definite angular momentum violates the uncertainly principle (which, to be fair to Bohr, was developed later by Heisenberg).8


7In fact, elliptical orbits are necessary for the large n limit to agree with the classical prediction of the frequency of the radiation that results from transitions. For large n, transitions require Δn = 2, 3, 4,....  In the case of circular orbits, the large n limit agrees with the classical prediction for the radiation frequency only for transitions with Δn = 1.
8For a nice description of all of the deficiencies of the Bohr model, including the fortuitous two errors that he made which canceled, see pages 198-199 of K. Krane, Modern Physics, 2nd ed. (John Wiley and Sons, New York, 1997).