24.103 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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 $2.8\times {10}^{9}yr$ ?

Step-by-Step Solution

Answer

Fraction of  ${}^{235}U$ that would remain on earth after $2.8\times {10}^{9}yr$  is 0.625 .

Step 1: Half-life

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.

Step 2: Calculation of decay constant from half-life.

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:

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

 

Step 3: Find the fraction of U 235 that would remain on earth after 2 . 8 × 10 9 y r

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 .