Chapter 30, Problem 52P

To determine

Part a:

The resulting intermediate daughter nuclei if the naturally occurring radioactive decay sequences start with ${}_{90}{}^{232}Th$.The first five decays are $\alpha ,\beta ,\beta ,\alpha ,\alpha$?

Answer to Problem 52P

Solution:

${}_{88}{}^{228}Ra$, ${}_{89}{}^{228}Ac$, ${}_{90}{}^{228}Th$, ${}_{88}{}^{224}Ra$ and ${}_{86}{}^{220}Rn$

Explanation of Solution

${}_{90}{}^{232}Th\to {}_{88}{}^{228}Ra+{}_2{}^4\alpha$; ${}_{88}{}^{228}Ra\to {}_{89}{}^{228}Ac+{\beta }^{-}$; ${}_{89}{}^{228}Ac\to {}_{90}{}^{228}Th+{\beta }^{-}$;

${}_{90}{}^{228}Th\to {}_{88}{}^{224}Ra+{}_2{}^4\alpha$; ${}_{88}{}^{224}Ra\to {}_{86}{}^{220}Rn+{}_2{}^4\alpha$

To determine

Part b:

The resulting intermediate daughter nuclei if the naturally occurring radioactive decay sequences start with ${}_{92}{}^{235}U$.The first five decays are $\alpha ,\beta ,\alpha ,\beta ,\alpha$?

Answer to Problem 52P

Solution:

${}_{90}{}^{231}Th$, ${}_{91}{}^{231}Pa$, ${}_{89}{}^{227}Ac$, ${}_{90}{}^{227}Th$ and ${}_{88}{}^{223}Ra$

Explanation of Solution

${}_{92}{}^{235}U\to {}_{90}{}^{231}Th+{}_2{}^4\alpha$; ${}_{90}{}^{231}Th\to {}_{91}{}^{231}Pa+{\beta }^{-}$; ${}_{91}{}^{231}Pa\to {}_{89}{}^{227}Ac+{}_2{}^4\alpha$;

${}_{89}{}^{227}Ac\to {}_{90}{}^{227}Th+{\beta }^{-}$; ${}_{90}{}^{227}Th\to {}_{88}{}^{223}Ra+{}_2{}^4\alpha$