Alpha Decay Find the daughter nuclei when the [latex]{ }^{208} Po \text { and }{ }^{238} U[/latex] and undergo alpha decay. In each case, write the full reaction.

The daughter following alpha decay has two fewer protons and two fewer neutrons than the parent, dropping its atomic number by 2 and mass number by 4. Polonium has Z = 84, so [latex]{ }^{208} Po \text { has } Z=82( lead , Pb )[/latex], and is the lead isotope with mass number 204. Thus the reaction is [latex]{ }^{208} Po \rightarrow{ }^{204} Pb +{ }^{4} He[/latex]. [latex]\text { Similarly, the parent }{ }^{238} U \text { has } Z=92 \text { and } A=238 \text {. So the daughter has } Z=90[/latex] (thorium, Th), and is the isotope with A = 234: [latex]{ }^{238} U \rightarrow{ }^{234} Th +{ }^{4} He[/latex]. REFLECT An essential feature here is that the total number Z of protons and total number A - Z of neutrons don't change. This implies conservation of electric charge, one of the fundamental conservation laws (Chapter 20). Mass, however, is not precisely conserved, because of the mass change [latex]\Delta m=E / c^{2}[/latex] associated with energy E released in the reaction.

Question 25.4: Alpha Decay Find the daughter nuclei when the ^208Po ^238U a...

Essential College Physics

Alpha Decay

Find the daughter nuclei when the ${ }^{208} Po \text { and }{ }^{238} U$ and undergo alpha decay. In each case, write the full reaction.

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The daughter following alpha decay has two fewer protons and two fewer neutrons than the parent, dropping its atomic number by 2 and mass number by 4. Polonium has Z = 84, so ${ }^{208} Po \text { has } Z=82( lead , Pb )$ , and is the lead isotope with mass number 204. Thus the reaction is

${ }^{208} Po \rightarrow{ }^{204} Pb +{ }^{4} He$ .

$\text { Similarly, the parent }{ }^{238} U \text { has } Z=92 \text { and } A=238 \text {. So the daughter has } Z=90$ (thorium, Th), and is the isotope with A = 234:

${ }^{238} U \rightarrow{ }^{234} Th +{ }^{4} He$ .

REFLECT An essential feature here is that the total number Z of protons and total number A – Z of neutrons don’t change. This implies conservation of electric charge, one of the fundamental conservation laws (Chapter 20). Mass, however, is not precisely conserved, because of the mass change $\Delta m=E / c^{2}$ associated with energy E released in the reaction.