Question

Question: Using the result of problem 29 and the accepted age of the earth,$\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{x}{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{years}$, calculate the ratio ${}^{\mathbf{238}}\mathbf{U}\mathbf{/}{}^{\mathbf{235}}\mathbf{U}$at the time the earth was formed.

Short answer

The ratio${}^{238}\mathrm{U}/{}^{235}\mathrm{U}$ when the earth was formed was 3.17.

Step-by-step solution

Time period of decay

The supernova event took place $5.9\times {10}^9$years ago.

The accepted age of the earth is $4.5\times {10}^9$years.

The time difference between the supernova and formation of earth$=\left(5.9-4.5\right)\times {10}^9\mathrm{years}=1.4\times {10}^9\mathrm{years}$

Ratio of U238/U235

To calculate the ${}^{238}\mathrm{U}/{}^{235}\mathrm{U}$ratio at time of earth formation, we have to estimate the amount of ${}^{238}\mathrm{U}$and ${}^{235}\mathrm{U}$decayed through a time period of$1.4\times {10}^9$years.

Therefore,

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

Dividing equation 1 by equation 2, we get:

$\frac{{\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{238}\mathrm{U}}}{{\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{235}\mathrm{U}}}=\frac{{\left({\mathrm{N}}_0\right)}_{{}^{238}\mathrm{U}}}{{\left({\mathrm{N}}_0\right)}_{{}^{235}\mathrm{U}}}{\mathrm{e}}^{-\left({\mathrm{k}}_1-{\mathrm{k}}_2\right)\mathrm{t}}\left(3\right)$

Since the relative abundance of ${}^{238}\mathrm{U}$and ${}^{235}\mathrm{U}$isotope was the same during the supernova,$\frac{{\left({\mathrm{N}}_0\right)}_{{}^{238}\mathrm{U}}}{{\left({\mathrm{N}}_0\right)}_{{}^{235}\mathrm{U}}}=1$

Therefore,

localid="1660882426957" $$\begin{gathered} \frac{{\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{238}\mathrm{U}}}{{\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{235}\mathrm{U}}}={\mathrm{e}}^{-\left({\mathrm{k}}_1-{\mathrm{k}}_2\right)\mathrm{t}} \\ \frac{{\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{238}\mathrm{U}}}{{\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{235}\mathrm{U}}}=e^{-\left(0.155-0.98\right)\times {10}^{-9}{\mathrm{years}}^{-1}\times 1.4\times {10}^9\mathrm{years}} \\ \frac{{\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{238}\mathrm{U}}}{{\left({\mathrm{N}}_{\mathrm{t}}\right)}_{{}^{235}\mathrm{U}}}=3.17 \end{gathered}$$