and since the angle of deviation is , it follows that Now it is evident from the method used in Appendix E that if there are atoms per cubic centimeter of a metal foil of thickness , and if each atom has a radius , then the probability that a particle of size small in comparison with will pass through one of these atoms in shooting through the foil is given by [Pg 280] Similarly the probability that it will pass within a distance of the center of an atom is If this probability is small in comparison with unity, it represents the fraction of any given number of particles shooting through the foil which will actually come within a distance of the nucleus of an atom of the foil. The fraction of the total number which will strike within radii and is given by differentiation as but from equation (57) Therefore the fraction which is deflected between the angles and is given by integration as It was this fraction of a given number of -particles shot into the foil which Geiger and Marsden found by direct count by the scintillation method to be deflected through the angles included between any assigned limits and . Since and are known, could be at once obtained. It was found to vary with the nature of the atom, being larger for the heavy atoms [Pg 281] than for the lighter ones, and having a value for gold of . This is then an upper limit for the size of the nucleus of the gold atom. As soon as has thus been found for any atom, equation (56) can be solved for , since , , and are all known. It is thus that the number of free positive electrons in the nucleus is found to be roughly half the atomic weight of the atom, and that the size of the nucleus is found to be very minute in comparison with the size of the atom. [Pg 282] APPENDIX G BOHR’S THEORETICAL DERIVATION OF THE VALUE OF THE RYDBERG CONSTANT The Newtonian equation of a circular orbit of an electron rotating about a central attracting charge , at a distance , with a rotational frequency , is The kinetic energy of the electron is . The work required to move the electron from its orbit to a position at rest at infinity is . If we denote this quantity of energy by , it is seen at once that If we combine this with (37), p. 213, there results at once Upon change in orbit the radiated energy must be [Pg 283] and, if we place this equal to , there results the Balmer formula (34), p. 210, in which Since for hydrogen , we have and from (60) [Pg 284] APPENDIX H A. H. COMPTON’S THEORETICAL DERIVATION OF THE CHANGE IN THE WAVE-LENGTH OF ETHER-WAVES BECAUSE OF SCATTERING BY FREE ELECTRONS Imagine, as in Fig. 42A, that an X-ray quantum of frequency is scattered by an electron of mass . The momentum of the incident ray will be , where is the velocity of light and is Planck’s constant, and that of the scattered ray is at an angle with the initial momentum. Fig. 42