Question

Consider the following process of decays
${}_{92}^{234}U\to {}_{90}^{230}Th+{}_2^{4}He;T_{1/2}=250000$ years
${}_{90}^{230}Th\to {}_{88}^{226}Ra+{}_2^{4}He;T_{1/2}=80000$ years
${}_{88}^{226}Ra\to {}_{86}^{222}Rn+{}_2^{4}He;T_{1/2}=1600$ years .
After above process has occurred for a long time, a state is reached where every two thorium atoms formed from ${}_{92}^{234}U$, one decomposes to form ${}_{88}^{226}Ra$ and for every two ${}_{88}^{226}Ra$ formed, one decomposes to form ${}_{86}^{222}Rn$. Calculate the ratio of number of nuclei of ${}_{90}^{230}Th$ to ${}_{88}^{226}Ra$ at this state.

Answer 1

Answer: 100

${}^{234}U\to {}^{230}Th\to {}^{226}Ra\to {}^{222}Rn$
${\frac{dN}{dt}|}_{\text{Formation of }Ra}={2\frac{dN}{dt}|}_{\text{Formation of }Rn}$
${\frac{dN}{dt}|}_{\text{Decay of }Th}={2\frac{dN}{dt}|}_{\text{Decay of }Ra}$
${\lambda }_1{N}_{Th}=2\left({\lambda }_2{N}_{Ra}\right)$
$\frac{N_{Th}}{N_{Ra}}=\frac{2{\lambda }_2}{{\lambda }_1}=2\times \frac{T_1}{T_2}$
$=\frac{2\times 80000}{1600}=\frac{100}{1}$