What fraction of the ${}^{235}U\left(t_{1/2}=7.0\times {10}^{8}yr\right)$created when Earth was formed would remain after$2.8\times {10}^{9}yr$ ?

The half-life of a radioactive substance is defined as the amount of time required for the decay of half the initial amount of a substance. Half-life is denoted by $t_{1/2}$.

Radioactive decay is a first-order reaction. The expression for half-life is given below.

$t_{1/2}=\frac{0.693}{k}$

Where, k is the decay constant.

Half-life time, $t_{l/2}$ of ${}^{235}U=7.0\times {10}^{8}yr$

Time, $t=2.8\times {10}^{9}yr$

The rate constant, k, is calculated as follows:

$$\begin{gathered} t_{1/2}=\frac{\ln 2}{k} \\ k=\frac{\ln 2}{t_{1/2}} \\ =\frac{0.693}{\left(7.0\times {10}^{8}yr\right)} \\ =9.9\times {10}^{-10}y{r}^{-1} \end{gathered}$$

Radioactive decay is a first order reaction.The fraction of ${}^{\text{235}}\text{U}$ that would still be on the planet after $2.8\times {10}^{9}yr$ years can be calculated using the following equation:

$$\begin{gathered} \left(\frac{N_t}{N_0}\right)=-kt \\ \ln \left(\frac{N_t}{N_0}\right)=-\left(9.9\times {10}^{-10}y{r}^{-1}\right)\left(2.8\times {10}^{9}yr\right) \\ \left(\frac{N_t}{N_0}\right)=-2.772 \\ \left(\frac{N_t}{N_0}\right)= e^{-2.772} \\ \left(\frac{N_t}{N_0}\right)=0.0625. \end{gathered}$$

Therefore, the fraction of ${}^{235}U$ that would remain on earth after $2.8\times {10}^{9}yr$ is 0.0625 .