Question

Question: The half-lives of ${}^{\mathbf{235}}\mathbf{U}$and ${}^{\mathbf{238}}\mathbf{U}$are $\mathbf{7}\mathbf{.}\mathbf{04}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$years and role="math" localid="1660824319605" $\mathbf{4}.\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$years, respectively, and the present abundance ratio is role="math" localid="1660824295665" ${}^{\mathbf{238}}\mathbf{U}\mathbf{/}{}^{\mathbf{235}}\mathbf{U}\mathbf{=}\mathbf{137}\mathbf{.}\mathbf{7}$. It is thought that their abundance ratio was 1 at some time before our earth and solar system was formed about role="math" localid="1660824359231" $\mathbf{4}.\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$years ago. Estimate how long ago the supernova occurred that supposedly produced all the uranium isotopes in equal abundance, including the two longest lived isotopes, ${}^{\mathbf{235}}\mathbf{U}$and ${}^{\mathbf{238}}\mathbf{U}$.

Short answer

The time of the supernova is $5.9\mathrm{x}{10}^9\mathrm{years}$.

Step-by-step solution

Decay constant of  U238

The decay constant can be calculated from the half-life through the following expression:

$\begin{array}{l}{\mathrm{t}}_{1/2}=\frac{0.693}{{\mathrm{k}}_1}\\ {\mathrm{k}}_1=\frac{0.693}{4.47\times {10}^9}\\ {\mathrm{k}}_1=0.155\times {10}^{-9}\end{array}$

Decay constant of U235

The decay constant can be calculated from the half-life through the following expression:

$\begin{array}{l}{\mathrm{t}}_{1/2}=\frac{0.693}{{\mathrm{k}}_2}\\ {\mathrm{k}}_2=\frac{0.693}{7.04\times {10}^8}{\mathrm{years}}^{-1}\\ {\mathrm{k}}_2=0.98\times {10}^{-9}{\mathrm{years}}^{-1}\end{array}$

Time of occurrence of supernova

${\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{238}\mathrm{U}}={\left({\mathrm{N}}_0\right)}_{{}_{{}^{238}\mathrm{U}}}{\mathrm{e}}^{-\mathrm{k}}$₁t1(Nt)U235=(N0)U235e-k2t2

Dividing equation 1 by equation 2, we get

$\frac{{\left(N_t\right)}_{{}^{238}U}}{{\left(N_t\right)}_{{}^{235}U}}=\frac{{\left(N_0\right)}_{{}^{238}U}}{{\left(N_0\right)}_{{}^{235}U}}{e}^{-\left(k_1-k_2\right)t}$

$\begin{array}{l}137.7={\mathrm{e}}^{-\left(0.155-0.98\right)\times {10}^{-9}\mathrm{t}}\\ \ln 137.7=0.825\times {10}^{-9}\mathrm{t}\\ \mathrm{t}=5.9\times {10}^9\mathrm{years}\end{array}$

The time of the supernova is $5.9\times {10}^9$years.