# Review of the isotope effect in the hydrogen spectrum

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1 Review of the isotope effect in the hydrogen spectrum 1 Balmer and Rydberg Formulas By the middle of the 19th century it was well established that atoms emitted light at discrete wavelengths. This is in contrast to a heated solid which emits light over a continuous range of wavelengths. The difference between continuous and discrete spectra can be readily observed by viewing both incandescent and fluorescent lamps through a diffraction grating. Around 1860 Kirchoff and Bunsen discovered that each element has its own characteristic spectrum. The next several decades saw the accumulation of a wealth of spectroscopic data from many elements. What was lacking though, was a mathematical relation between the various spectral lines from a given element. The visible spectrum of hydrogen, being relatively simple compared to the spectra of other elements, was a particular focus of attempts to find an empirical relation between the wavelengths of its spectral lines. In 1885 Balmer discovered that the wavelengths λ n of the then nine known lines in the hydrogen spectrum were described to better than one part in a thousand by the formula λ n = 3646 n 2 /(n 2 4) Å (angstrom), (1) where n = 3, 4, 5,... for the various lines in this series, now known as the Balmer series. (1 angstrom = m.) The more commonly used unit of length on this scale is the nm (10 9 m), but since the monochromator readout is in angstroms, we will use the latter unit throughout this write-up. In 1890 Rydberg recast the Balmer formula in more general form as 1/λ n = R(1/2 2 1/n 2 ) (2) where again n = 3, 4, 5,... and R is known as the Rydberg constant. For reasons to be explained later, its value depends on the mass of the nucleus of the particular isotope of the atom under consideration. The value for hydrogen from current spectroscopic data is R H = cm 1 [1], and the current value for an infinitely heavy nucleus is R = (73) cm 1 [2]. A generalization of the Rydberg formula to 1/λ = R H [( 1 n 2 1 ) ( )] 1 n 2 2 (3) suggested that other series may exist with n 1 taking on the values 1, 3, 4, 5,..., subject to the restriction that n 2 > n 1. This proved to indeed be the case. In 1906 Lyman measured the ultraviolet spectrum of hydrogen and found the series, now bearing his name, corresponding to n 1 = 1. In 1908 Paschen measured the infrared spectrum of hydrogen and discovered the series, now bearing his name, corresponding to n 1 = 3. Figure 1 shows some of the lines in these series. The vertical axis in this figure is linear in energy (given in units of electron-volts, or ev, at the right), so you can get some sense of the relative energies of the different series. 1

2 Figure 1: Energy level diagram of atomic hydrogen showing the transitions giving rise to the Lyman (IR), Balmer (visible) and Paschen (UV) spectra. 2

4 That Bohr s theory resulted in an expression for the Rydberg constant in terms of fundamental constants was a great triumph as it allowed for evaluation of the above expression and comparison to the value derived from spectroscopic data. Then current values for m, e, h and c indicated good agreement between R H calculated from Bohr s theory and experimentally derived values. But doubters of Bohr s theory remained, with skepticism on the continent being particularly strong. His brother Harald wrote Bohr in the fall of 1913 from Göttingen, Germany, saying that the physicists there considered Bohr s model to be too bold and too fantastic. Rutherford himself was skeptical, asking Bohr, How does the electron decide what frequency it is going to vibrate (radiate) at when it passes from one stationary state to another? It seems to me that you would have to assume the electron knows beforehand where it is going to stop. And skeptics were buttressed by new experimental results. Measurements of the spectral lines of singly ionized helium (one electron orbiting around a nucleus with two protons) showed that the ratio of the wavelengths of corresponding helium and hydrogen lines (same n i and n f ) was not 1:4 as predicted by Bohr s formula, but rather 1: (The ratio 1:4 results from increasing Z from one to two in the above expressions for the electron energy.) Bohr quickly revised his postulates, now requiring that the total angular momentum, the combined angular momentum of both the electron and the nucleus, be quantized in units of h. The result of this requirement is that in the expression for the electron energy, the electron mass m must be replaced by the reduced mass µ, µ = mm/(m + m) (6) where M is the mass of the nucleus. This result is derived below. The expression for the radiation energy of hydrogen-like ions (i.e., one electron orbiting Z protons) is now E i E f = hc λ = µ X Z2 e 4 [ 1 (4πε 0 ) 2 2 h 2 1 ] n 2 (7) i where µ X is the reduced mass of atom X. According to Bohr s revised theory, the wavelength ratios between corresponding hydrogen and helium lines should be 4µ He /µ H. It turns out that µ He µ H = , (you may wish to verify this) resulting in complete agreement between theory and observation. When Einstein learned of this latest success of the Bohr theory, he responded: This is an enormous achievement. The theory of Bohr must then be right. The inclusion of the nuclear mass M into the equation for the spectral energies indicates that different isotopes of the same element should show different spectra, since these atoms would have different reduced mass. The existence of the isotope of hydrogen, deuterium, whose nucleus contains one neutron and one proton, was discovered by Harold Urey and his co-workers around 1930 using spectroscopic methods after they successfully produced samples of deuterium-enriched hydrogen by a distillation process [4]. Harold Urey was awarded the Nobel Prize in Chemistry in 1934, for his discovery of heavy hydrogen. n 2 f 3 Theory of the reduced-mass correction Consider a classical system of two particles of mass M and m interacting via central forces (i.e., Coulomb or gravity). Let the position vector between the two particles be r. When no external 4

5 forces act on the system, the center of mass R will be at rest with respect to some inertial reference frame S. The center of mass is defined as R = M X + m x M + m, (8) where X and x locate the particles of mass M and m with respect to the origin of S, and x = X + r. Since the center of mass is at rest with respect to S, a physically equivalent reference frame S has its origin at the center of mass itself; in S, the center of mass of the system R = M X + m x M + m = 0. (9) Hence M X = m x. It is still true that r = x X, since x and X are related to x and X by a simple translation. These two equations can be solved to yield In S the energy of the two particle system is given by X = m r, (10) m + M x M = r. (11) m + M E = 1 2 mẋ M X 2 + V (r), (12) where the dot notation indicates a derivative with respect to time and V (r) is the potential energy between the two particles. Upon taking derivatives of Eqs. (10) and (11) and substituting the result into the energy equation one obtains E = 1 ( ) mm ṙ 2 + V (r). (13) 2 m + M Clearly, the expression in parenthesis in the above equation is the reduced mass µ, Eq. (6). An advantage of expressing the energy in terms of µ is that one transforms the problem of two particles of masses m and M into a problem of one particle of mass µ. A similar calculation shows that one can write the angular momentum about the center of mass in a similar way: L ( = x m x ) ( + X M X ) = r µ r. (14) In the Bohr model and in the subsequent Schrödinger theory, these classical results translate directly into their quantum equivalents. 4 The effect of the reduced mass on the spectrum From a measurement of the wavelength difference for a given transition that produces a spectral line, you can derive the ratio of the masses of the hydrogen and deuterium atoms. The results for the three visible lines (the α, β and γ lines) can then be combined into a final result for the mass ratio. 5

6 The spectral lines are produced by a gas discharge lamp. In our setup, the lamp contains hydrogen enriched with deuterium so that both spectra are produced simultaneously. Enrichment of the hydrogen is necessary as the natural abundance of deuterium is about.015%. From the Rydberg formula corrected for reduced mass, Eq. (7), we see that the wavelength of light emitted from a particular transition is inversely proportional to the reduced mass: 1/λ X µ X. (15) Because of this proportionality, we see that the nuclear masses of hydrogen and deuterium, M H and M D, are related to the different wavelengths of light produced by each isotope: λ H λ D λ H = 1/µ H 1/µ D 1/µ H = 1 M H/M D 1 + M H /m. (16) The mass M H is, of course, just the mass of the proton M p, and the ratio of the proton and electron masses is well known, M p /m = Combining this with the results of your measurements, the ratio of the masses of the two nuclei, M H /M D can easily be derived. By adding in electron masses, the mass ratio of the two isotopes can be derived and compared to established values, References M hydrogen M deuterium = = H. Haken and H. Wolf, The Physics of Atoms and Quanta, Springer, New York, 1996, p National Institute for Standards and Technology website. Go to and search for physical constants or the Rydberg constant. 3. R. Eisberg and R. Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles, 2d ed., Wiley, New York, 1985, p H. C. Urey, F. G. Brickwedde, and G. M. Murphy, Phys. Rev. 40, 1 (1932). Prepared by J. Stoltenberg and D. Pengra HD_mass_theory.tex -- Updated 27 March

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