Answer :
When identifying a nuclear reaction that involves alpha emission, it's important to look for a reaction where an alpha particle (denoted as [tex]\({ }_2^4 \text{He}\)[/tex]) is produced as a product. An alpha particle consists of 2 protons and 2 neutrons.
Let's go through each of the given reactions:
1. [tex]\({ }_{92}^{235} U \rightarrow{ }_2^4 \text{He} + { }_{90}^{231} Th\)[/tex]
In this reaction, Uranium-235 ([tex]\(U\)[/tex]) decays into Thorium-231 ([tex]\(Th\)[/tex]) and an alpha particle ([tex]\( { }_2^4 \text{He}\)[/tex]). This is a classic example of alpha emission since the alpha particle is one of the products.
2. [tex]\({ }_{92}^{235} U + { }_0^1 n \rightarrow{ }_{56}^{139} \text{Ba} + { }_{36}^{94} \text{Kr} + 3{ }_0^1 n\)[/tex]
This reaction describes nuclear fission, where Uranium-235 absorbs a neutron and splits into Barium-139 ([tex]\(Ba\)[/tex]), Krypton-94 ([tex]\(Kr\)[/tex]), and releases three neutrons. No alpha particle is produced in this reaction.
3. [tex]\({ }_{34}^{75} \text{Se} \rightarrow{ }_{-1}^0 \beta + { }_{35}^{75} \text{Br}\)[/tex]
This reaction describes beta decay, where Selenium-75 ([tex]\(Se\)[/tex]) decays into Bromine-75 ([tex]\(Br\)[/tex]), releasing a beta particle ([tex]\( \beta \)[/tex]). No alpha particle is involved here.
4. [tex]\({ }_7^{14} \text{N} + { }_2^4 \text{He} \rightarrow{ }_8^{17} \text{O} + { }_1^1 \text{H}\)[/tex]
This reaction involves the fusion of Nitrogen-14 ([tex]\(N\)[/tex]) and an alpha particle ([tex]\( { }_2^4 \text{He}\)[/tex]), producing Oxygen-17 ([tex]\(O\)[/tex]) and a Hydrogen nucleus ([tex]\( { }_1^1 \text{H}\)[/tex]). This is not an example of alpha emission but rather a fusion reaction.
5. [tex]\({ }_{53}^{123} \text{I} \rightarrow{ }_{53}^{123} \text{I} +\)[/tex] energy
This is an isomeric transition where Iodine-123 ([tex]\(I\)[/tex]) releases energy but remains the same isotope (no change in the number of protons or neutrons). There is no alpha particle emitted in this process.
Therefore, the correct example of alpha emission is:
[tex]\({ }_{92}^{235} U \rightarrow{ }_2^4 \text{He} + { }_{90}^{231} \text{Th}\)[/tex]
Let's go through each of the given reactions:
1. [tex]\({ }_{92}^{235} U \rightarrow{ }_2^4 \text{He} + { }_{90}^{231} Th\)[/tex]
In this reaction, Uranium-235 ([tex]\(U\)[/tex]) decays into Thorium-231 ([tex]\(Th\)[/tex]) and an alpha particle ([tex]\( { }_2^4 \text{He}\)[/tex]). This is a classic example of alpha emission since the alpha particle is one of the products.
2. [tex]\({ }_{92}^{235} U + { }_0^1 n \rightarrow{ }_{56}^{139} \text{Ba} + { }_{36}^{94} \text{Kr} + 3{ }_0^1 n\)[/tex]
This reaction describes nuclear fission, where Uranium-235 absorbs a neutron and splits into Barium-139 ([tex]\(Ba\)[/tex]), Krypton-94 ([tex]\(Kr\)[/tex]), and releases three neutrons. No alpha particle is produced in this reaction.
3. [tex]\({ }_{34}^{75} \text{Se} \rightarrow{ }_{-1}^0 \beta + { }_{35}^{75} \text{Br}\)[/tex]
This reaction describes beta decay, where Selenium-75 ([tex]\(Se\)[/tex]) decays into Bromine-75 ([tex]\(Br\)[/tex]), releasing a beta particle ([tex]\( \beta \)[/tex]). No alpha particle is involved here.
4. [tex]\({ }_7^{14} \text{N} + { }_2^4 \text{He} \rightarrow{ }_8^{17} \text{O} + { }_1^1 \text{H}\)[/tex]
This reaction involves the fusion of Nitrogen-14 ([tex]\(N\)[/tex]) and an alpha particle ([tex]\( { }_2^4 \text{He}\)[/tex]), producing Oxygen-17 ([tex]\(O\)[/tex]) and a Hydrogen nucleus ([tex]\( { }_1^1 \text{H}\)[/tex]). This is not an example of alpha emission but rather a fusion reaction.
5. [tex]\({ }_{53}^{123} \text{I} \rightarrow{ }_{53}^{123} \text{I} +\)[/tex] energy
This is an isomeric transition where Iodine-123 ([tex]\(I\)[/tex]) releases energy but remains the same isotope (no change in the number of protons or neutrons). There is no alpha particle emitted in this process.
Therefore, the correct example of alpha emission is:
[tex]\({ }_{92}^{235} U \rightarrow{ }_2^4 \text{He} + { }_{90}^{231} \text{Th}\)[/tex]