Let us consider an electron revolving in a circle about the nucleus. Let be the mass of the electron, a the radius of its orbit, its angular velocity. Also let 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 while the force attracting it to the nucleus is [Pg 160] by Coulomb’s Law. These two must be equal, so that 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 ; the potential energy is . In virtue of the above equation, is double , so that the total energy is equal to the kinetic energy with its sign changed. The impulse corresponding to is , and this has to be taken round one complete circuit of the orbit. This yields the value , which must be put equal to a multiple of , say , where is an integer. Thus we have the equation Now and and are known; thus (1) and (2) determine and as soon as is fixed. We have [Pg 161] The smallest possible orbit is got by putting ; thus its radius is , where The next possible radius is The kinetic energy in the orbit is Since the total energy is the kinetic energy with its sign changed, the loss of energy in passing from the to the orbit is If this transition is to give rise to a wave of frequency , we must have by the principle of quanta. That is to say is given by the equation [Pg 162] If is the velocity of light, this gives a wave-number . Now the empirical formula for the wave-numbers of the lines of the hydrogen spectrum is where is Rydberg’s constant. This shows that, if our theory is right, we ought to have By substituting the observed values for , , , and , 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.