Question 28.QE.8: Alpha and beta decay The following nuclei undergo different ......
Alpha and beta decay
The following nuclei undergo different types of radioactive decay. Determine the daughter nucleus for each and write an equation representing each decay reaction. \text { (a) }{ }_{94}^{239} \mathrm{Pu} \text { alpha decay, (b) }{ }_{58}^{144} \mathrm{Ce} \text { beta-minus decay, (c) and } { }_{30}^{65} \mathrm{Zn} \text { beta-plus decay. The latter produces a positron. }
Represent mathematically
(a) We have to subtract A = 4 and Z = 2 \text { (the }{ }_2^4 \mathrm{He} \text { alpha } particle) from the { }_{94}^{239} \mathrm{Pu} . Thus, the daughter nucleus has A = 239 – 4 = 235 and Z = 94 – 2 = 92, which is uranium-235.
(b) We have to subtract A = 0 and Z = -1 \text { (the }{ }_{-1}^0 e \text { beta } minus particle, an electron) from the { }_{58}^{144} \mathrm{Ce} . Thus, the daughter nucleus has A = 144 – 0 = 144 and Z = 58 – 1-12 = 59, which is praseodymium-144.
(c) We have to subtract A = 0 and Z = +1 \text { (the }{ }_1^0 e beta plus particle, a positron) from the { }_{30}^{65} \mathrm{Zn} . Thus, the daughter nucleus has A = 65 – 0 = 65 and Z = 30 – 1 = 29, which is copper-65.
Solve and evaluate The reactions are then:
\text { (a) }{ }_{94}^{239} \mathrm{Pu} \rightarrow{ }_{92}^{235} \mathrm{U}+{ }_2^4 \mathrm{He}+\text { energy }
\text { (b) }{ }_{58}^{144} \mathrm{Ce} \rightarrow{ }_{59}^{144} \mathrm{Pr}+{ }_{-1}^0 e+\bar{\nu}+\text { energy }
\text { (c) }{ }_{30}^{65} \mathrm{Zn} \rightarrow{ }_{29}^{65} \mathrm{Cu}+{ }_1^0 e+\nu+\text { energy }
“Energy” here means the kinetic energy of the products and possibly one or more gamma rays.
Try it yourself: Identify the daughter nucleus and write a reaction equation for the beta-minus decay of { }_{53}^{131} \mathrm{I}
Answer: { }_{53}^{131} \mathrm{I} \rightarrow{ }_{54}^{131} \mathrm{Xe}+{ }_{-1}^0 e+\bar{\nu}+\text { energy. }