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