Question

A certain radioactive isotope has leaked into a small stream. Three hundred days after the leak, 11% ofthe original amount of the substance remained. Determine the half-life of this radioactive isotope.

Answer 1

By the law of radioactive decay, the amount of a certain radioactive isotope as a function of time is given by

$N(t)=N_0e^{-\lambda t}$

Where N_0 represents the initial amount of the substance, N represents the amount of the substance given a time t, and lambda is the decay constant.

This isotope that has leaked, started with an amount N_0, and after 300 days remained only 11% of it, which means that

$N(300)=0.11N_0$

If we evaluate 300 in our function, we're going to have

$N(300)=N_0e^{-300\lambda}$

If we compare those two results, we have

$0.11N_0=N_0e^{-300\lambda}$

Dividing both sides by the initial amount.

$0.11=e^{-300\lambda}$

Solving for lambda, we have

$\begin{gathered} 0.11=e^{-300\lambda} \\ \ln (0.11)=-300\lambda \\ \lambda=-\frac{300}{\ln(0.11)} \\ \lambda=135.914198185\ldots \end{gathered}$

The decay rate and the half life of a isotope are related by the following formula

$\lambda=\frac{\ln2}{t_{1/2}}$

Where t_(1/2) is the half life of the isotope.

Using our value for lambda, we have

$\begin{gathered} 135.914198185\ldots=\frac{\ln2}{t_{1/2}} \\ t_{1/2}=\ln 2\cdot135.914198185\ldots \\ t_{1/2}=94.20854327\approx94 \end{gathered}$

The half life of this substance is approximately 94 days.