eae excitation with an energy, £, leads to ionization), is given by®: ® n= 6+ (1 — dl + (kw)/ml’, (1) where 6 is the fraction of excitations leading to direct ionization, ky is the rate constant for the preionization of the superexcited states, kouY? is the rate constant for the atomic rearrangement processes leading to neutral products, and y« is the reduced mass for the motion lead- ing to a point in coordinate space subsequent to which preionization is impossible. Platzman assumes that fy is approximately independent of the isotopic composition of the molecule; and we shall make the further assumption that 6 is also independent of isotopic composition. One can then write nm = 6 + (1 — 6)[L + ka/(amp)7? (2) and where the subscripts D and H refer to the deuterated and the protonated molecules, respectively. The maximum value of the yp/ny ratio will then occur when § = 0. Eliminationof kz /ki from equations (2) and (3) with 6 = 0 gives (%),- ; nfo ma + (1 — on) (n/un)!? _ (up/ma)*? (4) 7 1+ oul(up/ta)” — 1)’ where the subscript zero indicates that the ratio is for 6 = 0. The (qn/np )o values will be largest for the largest values of ( p/n)"* and the largest up/y ratios are for motions involving one hydrogen atom moving against the rest of the molecule. The maximum (p/un)”” value depends on the molecule, but a typical value is about 1.4 (for example, it is.1.388 and 1.39 for light molecules like H2O and C.He, and it would be 1.414 for a molecule of infinite mass). Figure 170 shows a plot of (np/nu)o VS. ty for (up/My)*” = 1.40. The experimental observation of an p/ny ratio smaller than the (yp/na)» value at the experimental value of m_ could mean that direct ionization is making a contribution (6 # 0), or it could mean that the effective value of (up/j)"” is less than 1.4. However, it is possible to estimate an upperlimit to the rela- tive contribution of direct ionization by comparing the sms, (8) ; 130 7p 0 VS 1.10 105 1995 m = 6 + (1 — 4)[l + ke/Chwy?)T, (3) observed yp/qq ratio with np/ny ratios caleulated assuming (up/uu)”” = 1.4 and 6 ¥ 0. The calculation is done by eliminating ky/k, from equations (2) and (3), dividing by 7, , and rearranging to give St amenma 1.35 np _ 6(1 — qu) + (uv/un)'(ne — 8) mo Mul (Ho/te)? (ne — 6) + 1 — nal (5) a! 02 O35 04 05 ™H 06 OT o8 a9 1.0 Fic. 170.—Theratio of the ionization yield for the deuterated molecule (yp) to the ionization yield for the protonated molecule (na) when there is no direct ionization (6 = 0) as a funetion of na. The square root of the ratio of the reduced masses (un/un)”? is taken as 1.40. Theresults are presented more compactly bydefining a term, R, by _ (nv/nm) — 1 — Cp/ame 21? (6) where the 7p/ny ratio in the numeratoris from equation (5). The R values are functions of (up/yy)'”, 6, and ma. Figure 171 is a plot of 5/ny vs. R, giving nine curves for the values of ny from 0.1 to 0.9, all calculated for (up/myx)'” = 1.4. Actually, the curves for (6/ny) vs. & are not very sensitive to the choice of (up/3)"”", provided that the same value of (up/my)"” is used in equation (4) as in equation (5). Thus, for (up/py)"” = 1.36, the 6/74 value at a given value of R would be slightly smaller (<0.01 smaller) than the value in Figure 171, and for (up/py)"” = 1.08, the 6/ny value would be smaller (<0.05 smaller) than the value in Figure 171. To illustrate the use of the method consider an example where 9p/ny = 1.07 and ny = 0.25. From Figure 170 we find that (ap/ny)o = 1.273, so that R = 0.07/0.273 = 0.256. From Figure 171 we find that 6/ny = 0.81. This would indicate that at least 19% of the ionization in this example was from the Platzman competitive jonization process. One should use this method with care, however, as there are somedifficulties. One difficulty is the problem of how to handle energy level shifts," which may be real shifts caused by differences in the zero-point energies or may be apparent shifts caused by unfavorable Franck-Condon factors. One method of handling these