chemical physics of discharges - Argonne National Laboratory
chemical physics of discharges - Argonne National Laboratory chemical physics of discharges - Argonne National Laboratory
92 closely follows 1 / in ~ dependence ~ on [N]. *, L a l/Ta. Discussion of the ~ last possibility will be deferred. In the former case L must be very large indeed and [N2(A3xt)] must be essentially in a quasi-steady state with its production rate, & T~ refers to the decay of P. Since [N2(A3Et)] is not strongly deactivated by pure N2, L would certainly not increase when the energy of the system dissipates as would be necessary if P did not decay. Because the decay of [N2(A3c:)] appears to remain exponential for all values of [N] and because Ta is a smooth function of [N] then L > 1/T (max) if the data are; interpreted as the decay of P with [N2(A3xi)] in a quasf-steady state. For this to be true some species capable of deactivating N2(A31t) must be generated in the discharge and remain constant for times long compared to those where [N2(A3C$)] is detectable. The most abundant chemical species generated by the discharge is N; the most abundant excited molecule is Nz(A3L:) -- except possibly for the 3A state of N2 I which has not yet been directly detected. Only atomic nitrogen appears to be a realistic choice for the species which decays slowly and would keep L >> l/Ta. i. I . I However, a difficulty arises in the interpretation with L >> l/Ta if we consider the observed behavior of [N2(A3 :>I when the discharge is turned on. The rise of 1 [N2(A3xz)] to a steady state is describable by If this is substituted into Eq. (3). we obtain 3 + If -1/~~ = I,, then P = L [N2(A E,)] = Po, a constant in time. When t = 0, P = Po = ([N2(A3~~)]o/Ta) if L # (19~~); and if P = Po + P', we obtain If L >> 1/~~, P' is positive; and if L > 1 / ~ 1 ~ is the only one tenable during the excitation discharge. 4 Since Po follows the discharge transients, it must become zero when the dis- charge is turned off, so that P changes rapidly compared with Ta from P + P' 1 to P'. But this is inconsistent with the interpretation of the decay of [82(A3$)1 , previously given on the assumption that L >> l / because ~ ~ this rapid fall in P would be reflected in a rapid fall of [N2(A3C$)] which is not observed. Hence this contradiction forces L = l/ta and P = Po, a constant, during the discharge and I L = l/ra and P = 0 after the discharge. During the excitation discharge a steady state Is reached, and j > I ! I 4
93 where the bar indicates values constant in time. Then indexing all quantities with a subscript 1 when atoms are not added and with N when they are added, we obtain where the last equality uses the results of Fig. 8. Hence 4 involves [N] to a higher power than does sN. Thus T~ cannot refer to the decay of [Nl. This is consistent with the usual slow decay of [N]. Since 4 is essentially proportional to [NI2, these results imply that FN is proportional to [N]. If this additional production of N2(A3Z$) utilizes nitrogen atoms on a one-for-one basis, [N] must decrease in passing through the discharge, &, where At is an average residence time, where L[NlOut represents the rate of atom removal, and where mixing is rapid compared with reaction. This leads to or which is the general form observed. From Fig. 9 we can obtain L (using At %50 sec), and from Fig. 8 with a slope ' of (KbPo[N2]/L), we can also obtain L if KT,P,[N~] can be found. A comparison of L derived from Fig. 9 and Fi . 8 is then possible. Since P = N2(A3L;fi>]/~,, absolute measurements of [N2(A3$)] derived from I(VK) = ( [N2(A3~)]/Tr) L where is the radiative lifetime of the excited state give T~P. T~ The absolute intensity of the 0,5 Vegard-Kaplan band in the discharge bulb can , be roughly obtained by measuring its signal and calibrating the spectrometer response using the NO y (0,3) band excited by the chemiluminescent association of N and 0 in conjunction with measurements of [N] and [O] and the known rate coefficient for I these pro~esses.~ In general, the comparisons of L derived from Figs. 8 and 9 show ' that if nitrogen atoms are consumed in the excitation of N2(A3c), then several I hundred times more atoms need to disappear than in fact are destroyed. This implies I that nitrogen atoms are not consumed but catalyze the production of N2(A3~t) in conjunction with other species generated in the discharge. > A similar argument can be used to show that atomic nitrogen is not consumed in 1 destroying N2(A3c). For example, there are often as many N~(A3zt)~ $olecules as \ N atoms; these would be seriously depleted during the decay of N2(A zU) and thus would lead to a nonexponential behavior. , i '1 3 + The short lifetime of N2(A &,> in the discharge (Fig. 4) cannot be due to atoms because their concentration, as inferred from downstream chemiluminescence measurement, ' is too small. Furthermore, the pressure dependence of T~ is incompatible with atom
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92<br />
closely follows 1 / in ~ dependence ~ on [N]. *, L a l/Ta. Discussion <strong>of</strong> the<br />
~<br />
last possibility will be deferred. In the former case L must be very large indeed<br />
and [N2(A3xt)] must be essentially in a quasi-steady state with its production<br />
rate, & T~ refers to the decay <strong>of</strong> P. Since [N2(A3Et)] is not strongly<br />
deactivated by pure N2, L would certainly not increase when the energy <strong>of</strong> the<br />
system dissipates as would be necessary if P did not decay. Because the decay <strong>of</strong><br />
[N2(A3c:)] appears to remain exponential for all values <strong>of</strong> [N] and because Ta<br />
is a smooth function <strong>of</strong> [N] then L > 1/T (max) if the data are; interpreted as<br />
the decay <strong>of</strong> P with [N2(A3xi)] in a quasf-steady state. For this to be true<br />
some species capable <strong>of</strong> deactivating N2(A31t) must be generated in the discharge<br />
and remain constant for times long compared to those where [N2(A3C$)] is detectable.<br />
The most abundant <strong>chemical</strong> species generated by the discharge is N; the most<br />
abundant excited molecule is Nz(A3L:) -- except possibly for the 3A state <strong>of</strong> N2 I<br />
which has not yet been directly detected. Only atomic nitrogen appears to be a<br />
realistic choice for the species which decays slowly and would keep L >> l/Ta.<br />
i.<br />
I<br />
. I<br />
However, a difficulty arises in the interpretation with L >> l/Ta if we consider<br />
the observed behavior <strong>of</strong> [N2(A3 :>I when the discharge is turned on. The rise <strong>of</strong> 1<br />
[N2(A3xz)] to a steady state is describable by<br />
If this is substituted into Eq. (3). we obtain<br />
3 +<br />
If -1/~~ = I,, then P = L [N2(A E,)] = Po, a constant in time. When t = 0,<br />
P = Po = ([N2(A3~~)]o/Ta) if L # (19~~); and if P = Po + P', we obtain<br />
If L >> 1/~~, P' is positive; and if L > 1 / ~ 1 ~<br />
is the only one tenable during the excitation discharge. 4<br />
Since Po follows the discharge transients, it must become zero when the dis-<br />
charge is turned <strong>of</strong>f, so that P changes rapidly compared with Ta from P + P' 1<br />
to P'. But this is inconsistent with the interpretation <strong>of</strong> the decay <strong>of</strong> [82(A3$)1 ,<br />
previously given on the assumption that L >> l / because ~ ~ this rapid fall in P<br />
would be reflected in a rapid fall <strong>of</strong> [N2(A3C$)] which is not observed. Hence this<br />
contradiction forces L = l/ta and P = Po, a constant, during the discharge and I<br />
L = l/ra and P = 0 after the discharge.<br />
During the excitation discharge a steady state Is reached, and<br />
j<br />
><br />
I<br />
!<br />
I<br />
4
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