The terms in the expression for the frequency ratio v/v, due to the electrons in the inner orbit only are

2 (N-.25)2 (1 + 1/4 B2 + 1/8 B1 +..) — N2(1 + 1/4 B'2 + B'1 +..), (6) where

and N is the atomic number of the chemical element. In this equation the relatively small forces due to electrons in orbits outside of the inner orbit have been neglected. The two series in increasing powers of B represent the correction for the change of mass of the electron with its velocity.

In calculating the velocities of the electrons in the orbits outside of the inner one, I have made several approximations. Firstly, I have assumed that the force acting on an electron, due to the electrons in orbits that are smaller than its own, is the same as it would be, if these electrons were concentrated at the nucleus of the atom. Secondly, I have neglected the part of the force acting on an electron, due to the electrons in orbits that are larger than its own. Thirdly, the calculation of the force acting on an electron in an orbit, say at A in the figure, due to the electrons in the other orbit BC of the pair I have made by means of two distinct approximations. In the first of these I have assumed that this force equals what it would be, if half the electrons in the orbit BC were concentrated at B, the nearest point in it to A, and half at C, the furthest point from A. This gives only a rough estimate of the terms in the equation representing the ratio v/v, due to the electrons in the various orbits outside of the inner one. As, however, all these terms add up to only 15 or 20 per cent of the value of the term due to the innermost orbit (expression 6), the error thereby introduced does not appear to be enormous. In the second approximate computation of the force acting on the electron at A, due to the electrons in the orbit BC, I have assumed the electricity of the electrons in BC to be uniformly distributed along the orbit. The results of these computations will be given in a subsequent note. They do not differ much from those obtained from the first approximate computation.

If we calculate the radii of the orbits, we find that, in general, the distances between the electrons in a pair of orbits and the electrons in another pair of orbits are greater in comparison with those radii than as represented in the figure. Hence the influence of electrons in outer orbits is small, and that of electrons in inner orbits approximates to what it would be, if they were at the nucleus.

To calculate the velocity of an electron, say at A, we first calculate the angle a in the figure as follows: Let N' be the total number of electron

charges inside the orbits A B C D. N' then equals the atomic number of the chemical element N less the number of electrons inside the orbits A B C D. Assuming that half the charge in the orbit BC is concentrated at B and the other half at C, equating to zero the horizontal components of the forces due to the nuclear charge and to the electrons in and inside of the orbits ABCD and reducing we get the equation

Equating the radial components of the forces acting on an electron at A to the centripetal acceleration, multiplying by a2, and substituting the value of N' from equation 7, we get the equation

The velocities of the electrons other than the two in the inner orbit are small as compared with the velocity of light, and the correction for the change of the mass of an electron with its velocity is, therefore, negligible. Neglecting this change and combining equation 8 with the angular momentum law (equation 2) we get for the kinetic energy of the electron 22e1m, n (-tan3 a h272 8

1/2 mv2

=

Sn) 2

where the angle a is given by equation 7, and s,, by equation 9.

(10)

In systems of the kind considered, the potential energy of the electrons equals twice their kinetic energy with the negative sign before it.3 Hence, the total energy, kinetic plus potential, is minus the kinetic energy. Substituting the expression for the total energy of all the electrons with one electron removed from the inner orbit, and with this electron in place in equation 3, and dividing by hvo, we get

2(−N.25)2 (1+11 ẞ2 + 1 / 8 ß1 + . . . ) − N2 (1 + 11ẞ'2 + 1/8 ẞ'1 + ..)

V

=

να

as the expression for the critical absorption frequency of the chemical element divided by the Rydberg constant.

Table 1 contains the calculated and the observed values of the ratio v/vo. The observed values I have taken from table 2 of the report on Data Relating to X-Ray Spectra which I compiled for the National Research Council. Column 4 in the table contains the observed values, and column 3 the values calculated according to the following scheme:

It appears that, except in the case of the elements magnesium and sulphur, the calculated values in column three do not differ from the observed values by as much as two per cent. Considering the fact that the equations contain no undetermined constants after the distribution law has been fixed and considering the fact that various influences have been neglected, such as magnetic forces, forces due to electrons in outer orbits, the possible influence of electrons which may be forming bonds with other atoms, etc., the close agreement between the calculated and observed values appears extraordinary.

I have also calculated the absorption frequencies, assuming a distribution according to the scheme

The calculated values based on this scheme do not differ very much from those contained in column 3 of the table. It appears, therefore, that the calculation of the K critical absorption frequencies does not furnish a very sensitive method of determining exactly what the distribution of electrons is. It is important to note, however, that two electrons have a quantum number of one, sixteen electrons have a quantum

number of two, thirty-six electrons have a quantum number of three, etc. On examining the data closely we find a systematic deviation of the calculated from the observed values. This deviation becomes most marked in the elements of low atomic number. It lies in the same direction as the well-known difference between the observed and calculated ionizing potentials of helium.

In order to see whether a better representation of facts can be obtained by supposing that the two inner electrons revolve in opposite directions in separate orbits, I have made the calculations on that basis. The only difference in formula 11 occurs in the terms representing the innermost orbit. The first two terms of equation 11 must be replaced by the single term.

(N − 1)2 (1 + 14 ß'2 + 1/8 ß11 +.....)

(12)

Column 5 of the table contains the results of the computations. It appears from the data of column 5 that the observed and calculated values differ from each other by amounts up to 41/2%. A systematic variation exists which increases as the atomic weight decreases, and which lies in the opposite direction to that represented by the data of column 3.

1 Science, May 21, 1920.

2 Fall Meeting of the National Academy of Sciences, 1920; Physic. Rev., March, 1921, p. 431.

3 This can be proved easily for the particular case in question. For a proof of the theorem in a more general form see A. Sommerfeld, Atombau und Spektrallinien, Appendix 5.

ON THE CALCULATION OF THE X-RAY ABSORPTION FREQUENCIES OF THE CHEMICAL ELEMENTS (SECOND NOTE)

BY WILLIAM DUANE

JEFFERSON PHYSICAL LABORATORY, HARVARD UNIVERSITY

Communicated July 25, 1921

In a note presented to the National Academy of Sciences1 I have given some computations of the K critical absorption frequencies of the chemical elements based on the Rutherford-Bohr theory of the structure of atoms and the mechanism of radiation. In these computations I have assumed that the electrons were distributed in circular orbits, which did not lie in planes passing through the nucleus of the atom.

In order to estimate roughly the forces exerted on an electron in one orbit (A in the figure of the previous note) due to the electrons in the parallel orbit, I assumed that they were the same as if the charges of the electrons in the orbit BC were concentrated, half at the nearest point

in the orbit to A, and half at the furthest point in the orbit from A. In the computations presented in this note I assume that the forces are the same as if the electricity of the electrons in the orbit BC were uniformly distributed along the orbit. As in the previous note, I neglect the forces acting on an electron at A, due to the electrons in the orbits that are larger than the orbits A B C D, and assume that the forces due to the electrons in the orbits smaller than A B C D are the same as if these electrons were concentrated at the nucleus of the atom.

The terms in the expression for the ratio of the critical absorption frequency to the Rydberg constant, v/v。, due to the two innermost electrons, are the same as in the previous case (expressions 6 and 12 of the previous note). To get the correction term due to the electrons in the orbits outside of the inner orbit we proceed as follows: Equating to zero the horizontal components of the forces acting on an electron at A due to the nucleus and to the electrons inside and in the orbits A B C D, we get the equation

where is half the angle made by the radius of the orbit BC drawn to any point in the orbit with the vertical.

Similarly, equating the centripetal acceleration of the electron at A to the centripetal force acting on it, due to the nucleus, to the electrons inside the orbits A B C D and to the electricity of the orbit BC, and reducing, we get the equation

and where s, is given by equation 9 of the previous note.

Combining equation 14 with equation 2 of the previous note, representing the angular momentum law, we get an expression similar to equation 10 for the kinetic energy of the electron at A. Taking the sum of these expressions for all the electrons with one of the electrons of the inner orbit removed and with it in place, and substituting in formula 3, representing the frequency law, we get the equation for v/v。

1

2(N-.25) 2(1 + 1/2 ẞ2 + 1/8 ẞ1 +..)−N2(1 + 1 ẞ'2 + 1/8 B14 + ..)