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# Which nuclear equation represents beta decay? 1. $\quad{ }_{13}^{27} \mathrm{Al}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{15}^{30} \mathrm{P}+{ }_{0}^{1} \mathrm{n}$ 2. $\quad{ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \mathrm{He}$ 3. $\quad{ }_{89}^{227} \mathrm{Ac} \rightarrow{ }_{90}^{227} \mathrm{Th}+{ }_{- 1}^{0} \mathrm{e}$ 4. $\quad{ }_{18}^{40} \mathrm{Ar} \rightarrow{ }_{19}^{40} \mathrm{~K}+{ }_{- 1}^{0} \mathrm{e}$

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Step 1
: Identify the possible beta decay equation

where $${}_{-1}^{0} e$$ is the electron and $\bar{v}$ is the antineutrino.
Betadecayisatypeofradioactivedecaywhereaneutronisconvertedintoaproton,anelectron,andanantineutrino.Thisprocesscanberepresentedbythenuclearequation:01n→11p+−10e+vˉBeta decay is a type of radioactive decay where a neutron is converted into a proton, an electron, and an antineutrino. This process can be represented by the nuclear equation: _{0}^{1} n \rightarrow{} _{1}^{1} p + {} _{- 1}^{0} e + \bar{v}

Step 2
: Analyze the given nuclear equations

Let′sanalyzethegivennuclearequationsandseewhichonerepresentsbetadecay.Option1:1327Al+24He→1530P+01nThisisnotabetadecayequation,asitinvolvesthefusionofAluminum−27andHelium−4toformPhosphorus−30andaneutron.Option2:92238U→90234Th+24HeThisisanalphadecayequation,whereUranium−238decaysintoThorium−234andanalphaparticle(Helium−4nucleus).Option3:89227Ac→90227Th+−10eThisisabetadecayequation,whereActinium−227decaysintoThorium−227andanelectron.Option4:1840Ar→1940 K+−10eThisisabetadecayequation,whereArgon−40decaysintoPotassium−40andanelectron.Let's analyze the given nuclear equations and see which one represents beta decay. Option 1: _{13}^{27} \mathrm{Al}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{15}^{30} \mathrm{P}+{ }_{0}^{1} \mathrm{n} This is not a beta decay equation, as it involves the fusion of Aluminum- 27 and Helium- 4 to form Phosphorus- 30 and a neutron. Option 2: _{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \mathrm{He} This is an alpha decay equation, where Uranium- 238 decays into Thorium- 234 and an alpha particle (Helium- 4 nucleus). Option 3: _{89}^{227} \mathrm{Ac} \rightarrow{ }_{90}^{227} \mathrm{Th}+{ }_{- 1}^{0} \mathrm{e} This is a beta decay equation, where Actinium- 227 decays into Thorium- 227 and an electron. Option 4: _{18}^{40} \mathrm{Ar} \rightarrow{ }_{19}^{40} \mathrm{~K}+{ }_{- 1}^{0} \mathrm{e} This is a beta decay equation, where Argon- 40 decays into Potassium- 40 and an electron.

Final Answer

The nuclear equations that represent beta decay are options 3 and 4: 1. $\quad{ }_{89}^{227} \mathrm{Ac} \rightarrow{ }_{90}^{227} \mathrm{Th}+{ }_{- 1}^{0} \mathrm{e}$ 2. $\quad{ }_{18}^{40} \mathrm{Ar} \rightarrow{ }_{19}^{40} \mathrm{~K}+{ }_{- 1}^{0} \mathrm{e}$

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