Certainly! Let's examine each reaction to identify which one is an example of a chain reaction.
1. First Reaction:
[tex]\[ {}_{34}^{75} Se \rightarrow {}_{-1}^0 \beta + {}_{35}^{75} Br \][/tex]
This reaction is an example of beta decay. In beta decay, a neutron in an unstable nucleus is converted into a proton while emitting a beta particle ([tex]\(\beta\)[/tex]). This is a type of radioactive decay and does not initiate further reactions.
2. Second Reaction:
[tex]\[ {}_{92}^{235} U + {}_0^1 n \rightarrow {}_{56}^{142} Ba + {}_{36}^{91} Kr + 3 {}_0^1 n \][/tex]
In this reaction, a Uranium-235 ([tex]\( {}^{235}_{92}U \)[/tex]) nucleus captures a neutron and undergoes fission. The fission process splits the nucleus into smaller nuclei (Barium-142 ([tex]\( {}^{142}_{56}Ba \)[/tex]) and Krypton-91 ([tex]\( {}^{91}_{36}Kr \)[/tex])) and produces additional neutrons (three neutrons in this case). These neutrons can further induce fission in other [tex]\( {}^{235}_{92}U \)[/tex] nuclei. This process can continue, leading to a self-sustaining series of reactions, commonly known as a chain reaction.
3. Third Reaction:
[tex]\[ {}_{92}^{235} U \rightarrow {}_2^4 He + {}_{90}^{231} Th \][/tex]
This reaction shows alpha decay, where Uranium-235 decays to Thorium-231 by emitting an alpha particle ([tex]\( {}_2^4 He \)[/tex]). This decay process does not produce additional particles that can continue the reaction.
4. Fourth Reaction:
[tex]\[ {}_7^{14} N + {}_2^4 He \rightarrow {}_8^{17} O + {}_1^1 H \][/tex]
This reaction represents a nuclear fusion process in which a nitrogen-14 nucleus reacts with a helium-4 nucleus to form an oxygen-17 nucleus and a proton. This specific reaction does not lead to a chain reaction since the products do not further induce more reactions.
5. Fifth Reaction:
[tex]\[ {}_{53}^{123} I \rightarrow {}_{53}^{123} I + \text{ energy} \][/tex]
This reaction shows a process where iodine-123 ([tex]\( {}_{53}^{123} I \)[/tex]) releases energy. There are no products that will continue to react further, hence, it is not a chain reaction.
Thus, the second reaction:
[tex]\[
{}_{92}^{235} U + {}_0^1 n \rightarrow {}_{56}^{142} Ba + {}_{36}^{91} Kr + 3 {}_0^1 n
\][/tex]
is an example of a chain reaction. The neutrons produced in this reaction can propagate further reactions in other uranium-235 nuclei, creating a self-sustaining process.
Therefore, the answer is:
[tex]\[ 2 \][/tex]
