Get insightful responses to your questions quickly and easily on IDNLearn.com. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.
Sagot :
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]
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]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
