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Deuterium is an isotope of hydrogen, in which the nucleus contains a proton and a single neutron.  It was originally discovered because deuterium exhibits a slightly different atomic spectrum from hydrogen.  The Bohr theory can be extended to explain why and how the spectrum of deuterium should be slightly different from that of hydrogen.

Previous:  Correspondence Principle


In the first few articles, we have seen how the simple Bohr theory can be used to produce remarkably accurate estimates of the energy levels of the hydrogen atom.  However, one of the major failings of the Bohr theory is that it cannot reproduce all of the observed characteristics of the energy levels.  For example, it cannot explain why the spectral lines of hydrogen split when the hydrogen atoms are subject to a magnetic field (the Zeeman effect).  To explain such phenomena, we need to introduce more advanced concepts such as electron spin and angular momentum.

However, we can make a further modification to our simple Bohr theory, and that is to consider the motion of the nucleus.  In the simple theory presented in the first couple of articles, it was assumed that the hydrogen nucleus (a proton) was fixed, and that the electron in the hydrogen atom rotated around the fixed nucleus.

In reality, both the electron and the nucleus will rotate around the centre of mass of the system, as shown in the following diagram:



The motion around the centre of mass can be taken into account in the Bohr theory by using a standard trick of classical mechanics, in which the rotation of the electron-nucleus system can be replaced by an equivalent system, in which the electron rotates about the position of the nucleus.  However, for this equivalence to hold, the electron must be assigned an effective mass, which is referred to as the “reduced mass”.

To compute the reduced mass, consider the situation in the following diagram.  (The following is a more detailed treatment of that given by Beiser in Concepts of Modern Physics, 1984)



The top half of the diagram shows the original system in which both the electron (mass m) and the nucleus (mass M) rotate about the centre of mass.  By definition, we must have


The lower half of the diagram shows the equivalent system with which we wish to replace the original rotating system.  There are several ways to deduce the required reduced mass of the electron.  One is to note that the total angular momenta of the original and equivalent systems must be equal.  This leads to the following condition:


From these two equations it is easy to show that the reduced mass is given by


The modified equations for the radii and energies of the electron orbits are:



It can be seen that the correction to the energy levels by assuming rotation about the centre of mass is equal to


Noting that a proton has a mass about 1,836 times greater than that of an electron, it can be seen that the magnitude of the correction is

            C ~ 1836 / 1837 = 0.99946.

In other words, the magnitude of the deviation is very small.  Note that because the correction C is less than unity, all the energy levels have in fact increased, since they have become less negative.

Although the magnitude of the correction for rotation about the centre of mass is small, it did nevertheless play a part in the discovery of deuterium.  Deuterium is an isotope of hydrogen in which the nucleus contains one proton and one neutron.  It occurs most commonly in the form of “heavy water”, which finds applications in a number of fields.

Because of the difference in the masses of the nuclei between normal hydrogen (single proton) and deuterium (single proton and single neutron), it follows that the atomic spectra of hydrogen and deuterium are very slightly different, and this difference can be detected if sufficiently sensitive spectroscopic equipment is used.

For deuterium, the correction factor is given by

            C ~ 3672 / 3673 = 0.99973,

where we have noted that the mass of the neutron is virtually identical to that of the proton.

Now, for hydrogen, the transition between the n = 3 and n = 2 energy levels has a wavelength of 656.3 nm.  Thus, in deuterium, it will have a wavelength of

            656.3 × 0.99946 / 0.99973 = 656.1 nm.

Although the difference is only 0.2 nm, it is possible to resolve spectra at this level of resolution.  Indeed, it was minute differences of this type that were responsible for the discovery of deuterium in 1932.