Question 28.QE.3: Determine the missing product in the following reactions. (a......
Determine the missing product in the following reactions.
\text { (a) }{ }_2^4 \mathrm{He}+{ }_6^{12} \mathrm{C} \rightarrow{ }_7^{15} \mathrm{~N}+\text { ? }
\text { (b) }{ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\text { ? }
\text { (c) }{ }_0^1 \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+?+2{ }_0^1 \mathrm{n}
\text { (d) }{ }_{55}^{137} \mathrm{Cs} \rightarrow \text { ? }+{ }_{-1}^0 e
Represent mathematically To determine the products we need to use the rules that the A and Z numbers remain constant throughout the reaction.
(a) 4 + 12 = 15 + A
2 + 6 = 7 + Z
(b) 2 + 3 = 4 + A
1 + 1 = 2 + Z
(c) 1 + 235 = 140 + A + 2 ⋅ 1
0 + 92 = 54 + Z + 2 ⋅ 0
(d) 137 = A + 0
55 = Z + (-1)
Solve and evaluate Solving for A and Z in each case, we find the following:
(a) A = 1, Z = 1. The unknown product must be ahydrogen nucleus { }_1^1 \mathrm{H} .
(b) A = 1, Z = 0. The unknown product must be a neutron { }_0^1 \mathrm{n} .
(c) A = 94, Z = 38. The unknown product must be a strontium nucleus { }_{94}^{38} \mathrm{Sr}.
(d) A = 137, Z = 56. The unknown product must be a barium nucleus { }_{56}^{137} \mathrm{Ba} .
Try it yourself: Determine the missing product in this reaction: { }_6^{14} \mathrm{C} \rightarrow ?+{ }_{-1}^0 e
\text { Answer: }{ }_7^{14} \mathrm{~N} \text {. }