Question

Suppose the half- life of a certain radioactive substance is 12 days and there are 18 grams present initially. Find the time when there will be 20% of the initial substance remaining.

Answer 1

Answer:

27.86 days

Explanation:

The amount, A(t) of a radioactive substance left after time t is modeled by the equation below:

$A(t)=A_o\mleft(\frac{1}{2}\mright)^{\frac{t}{t_{(0.5)}}}$

Given:

• Initial Substance, Ao = 18 grams

,

• Half-Life, t(0.5) = 12 days

,

• Amount, A(t)=20% of 18 grams

Substitute into the formula above:

$\begin{gathered} 20\%\text{ of 18}=18_{}\mleft(\frac{1}{2}\mright)^{\frac{t}{12}} \\ 0.2\times18=18\mleft(\frac{1}{2}\mright)^{\frac{t}{12}} \\ 3.6=18\mleft(\frac{1}{2}\mright)^{\frac{t}{12}} \end{gathered}$

We solve the equation for t below.

$\begin{gathered} \text{Divide both sides by 18} \\ \frac{3.6}{18}=\frac{18(\frac{1}{2})^{\frac{t}{12}}}{18} \\ 0.2=\mleft(\frac{1}{2}\mright)^{\frac{t}{12}} \\ \text{Take the logarithm of both sides} \\ \log (0.2)=\log \mleft(\frac{1}{2}\mright)^{\frac{t}{12}} \\ \implies\frac{t}{12}\log (0.5)=\log (0.2) \\ \text{Divide both sides by log(0.5)} \\ \frac{t}{12}=\frac{\log(0.2)}{\log(0.5)} \\ \text{Multiply both sides by 12} \\ t=12\times\frac{\log(0.2)}{\log(0.5)} \\ t=27.86\text{ days} \end{gathered}$

After 27,86 days, there will be 20% of the initial substance remaining.