The photolysis of N₂O at 1470 Å

The photolysis of pure N 2 O, N 2 O and N 2 , and N 2 O and C 3 H 6 mixtures at 1470 Å and room temperature has been studied to determine the relative importance of the primary processes. The results are equation image where ϕ{O( 1 D)} = 0.515 represents both the O( 1 D) produced in the primary act and that produced by collisional quenching of O( 1 S); ϕ{N 2 ( 3 Σ)} = 0.084 represents only that portion of N 2 ( 3 ϕ) which dissociates N 2 O on deactivation; and ϕ{O( 1 S)} = 0.38 – ±{N( 2 D)} represents only that portion of O( 1 S) which enters into chemical reaction with N 2 O. If the reaction of O( 1 S) with N 2 O yields only N 2 and O 2 as products, which seems likely from potential‐energy curve considerations then ±{O( 1 S)} = 0.135 ± 0.06 and ϕ{N( 2 D)} = 0.245 ± 0.06. Young and coworkers [4] have found from spectroscopic observations that the total quantum yield of O( 1 S) is about 0.5. Thus it can be concluded that collisional removal of O( 1 S) by N 2 O yields mainly O( 1 D) with chemical reaction being less important. Furthermore, most of the O( 1 D) is produced this way, and the true primary yield of O( 1 D) is about 0.15. The metastable N( 2 D) is not deactivated by N 2 O, but is removed by chemical reaction to produce N 2 and NO. The results further indicate that N 2 ( 3 Σ) dissociates N 2 O at least 80% of the time during quenching. The relative efficiency of N 2 O compared to N 2 is about 2 for the removal of O( 1 D). O( 1 S) is removed about 90 times as efficiently by C 3 H 6 as by N 2 O.