Question

If 14g of a radioactive substance are present initially and 9yr later only 7g remain, how much of the substance will be present after 13 yr?

Answer 1

To begin with, let us first write out the formula for the exponential form

`y=b^(rt)`

Given the initial parametrs

`y=14b^t^{}`

Thus

`145^9=7`

`b^(-9)=(1)/(2)`

Using exponential laws

`b=\frac{1}{((1)/(2))^{(1)/(-(1)/(9))}}`

W will soon see that

`b=2^{(1)/(9)}`

Hence after 13 years, the substance left will be;

`\begin{gathered} 14(2^{(1)/(9)})^(-13)=5.144070729 \\ \approx5.144g \end{gathered}`