Question

Consider the following process of decays
${ }_{92}^{234} U \rightarrow{ }_{90}^{230} Th +{ }_{2}^{4} He ; T_{1 / 2}=250000$ years
${ }_{90}^{230} Th \rightarrow{ }_{88}^{226} Ra +{ }_{2}^{4} He ; T_{1 / 2}=80000$ years
${ }_{88}^{226} Ra \rightarrow{ }_{86}^{222} Rn +{ }_{2}^{4} He ; T_{1 / 2}=1600$ 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

${ }^{234} U \rightarrow{ }^{230} Th \rightarrow{ }^{226} Ra \rightarrow{ }^{222} Rn$
$\left.\frac{d N}{d t}\right|_{\text {Formation of } Ra }=\left.2 \frac{d N}{d t}\right|_{\text {Formation of } Rn }$
$\left.\frac{d N}{d t}\right|_{\text {Decay of } Th }=\left.2 \frac{d N}{d t}\right|_{\text {Decay of } Ra }$
$\lambda_{1} N_{ Th }=2\left(\lambda_{2} N_{ Ra }\right)$
$\frac{N_{ Th }}{N_{ Ra }}=\frac{2 \lambda_{2}}{\lambda_{1}}=2 \times \frac{T_{1}}{T_{2}}$
$=\frac{2 \times 80000}{1600}=\frac{100}{1}$