The A B C of Atoms

THE mathematics involved in this theory is so simple that only a very slight acquaintance with elementary dynamics is required in order to understand it. Let us consider an electron revolving in a circle about the nucleus. Let m be the mass of the electron, a the radius of its orbit, ω its angular velocity. Also let e be the (negative) charge on the electron and the (positive) charge on the nucleus. Then according to elementary dynamics, the centrifugal force of the electron in its orbit is ma ω² while the force attracting it to the nucleus is e²/a² by Coulomb’s Law. These two must be equal, so that ma ω² = e²/a² qquad (1) So far, we have been proceeding on traditional lines. But we come now to the application of the quantum theory. The kinetic energy of the electron is 1/2 ma ω²; the potential energy is -e²/a. In virtue of the above equation, e²/a is double 1/2 ma ω², so that the total energy is equal to the kinetic energy with its sign changed. The impulse corresponding to ω is ma² ω, and this has to be taken round one complete circuit of the orbit. This yields the value 2 π ma² ω, which must be put equal to a multiple of h, say nh, where n is an integer. Thus we have the equation 2 π ma² ω = n h qquad (2) Now m and e and h are known; thus (1) and (2) determine a and ω as soon as n is fixed. We have a = n²h²/4 π²me², quad ω = (8 π³ me⁴)/n³h³ The smallest possible orbit is got by putting n = 1; thus its radius is a, where a₁ = h²/4 π²me² The next possible radius is a₂ = 4h²/4 π²me² = 4a₁ The kinetic energy in the nth orbit is tfrac12 ma²ω² = tfrac12m((2 π e²)/nh)² Since the total energy is the kinetic energy with its sign changed, the loss of energy in passing from the kth to the nth orbit is tfrac12m((2 π e²)/h)² (1/n² - 1/k²) If this transition is to give rise to a wave of frequency υ, we must have tfrac12m((2 π e²)/h)² (1/n² - 1/k²) = h υ by the principle of quanta. That is to say υ is given by the equation υ = m (2 π² e⁴)/h³ (1/n² - 1/k²) If c is the velocity of light, this gives a wave-number υ/c. Now the empirical formula for the wave-numbers of the lines of the hydrogen spectrum is R (1/n² - 1/k²) where R is Rydberg’s constant. This shows that, if our theory is right, we ought to have R = (2 π² me⁴)/h³c By substituting the observed values for m, e, h, and c, it is found that this equation is satisfied. This was perhaps the most sensational evidence in favour of Bohr’s theory when it was first published.