Question 2.4.2: A radioactive isotope initially has 40 g of mass. After 10 d...

Because the isotope decays radioactively, we know that its mass M(t ) must have the form M(t ) = M_{0}e^{−kt} . To answer the questions posed, we must first determine the constant k. In the given problem, we know that M_{0} =40 and that M(10) = 39.7. It follows that

39.7 = 40 e^{−10k}

Dividing both sides of the equation by 40, taking natural logs, and solving for k, we find that

=− \frac {1}{10} ln(\frac {39.7}{40})

To compute the half life, we now solve the equation

\frac {M_{0}}{2}=M_{0}e^{−kt}

for t . In particular, we have

20 = 40 e^{\frac {1}{10} ln (\frac {39.7}{40})t}

Dividing by 40 and taking natural logs,

ln (\frac {1}{2})=\frac {1}{10} ln (\frac {39.7}{40})t

so

t=\frac {ln (\frac {1}{2})}{\frac {1}{10} ln (\frac {39.7}{40})}

Thus the half-life of the isotope is approximately 921 days.

Finally, to determine when 1 g of the substance will remain, we simply solve the equation

1 = 40 e^{\frac {1}{10} ln (\frac {39.7}{40})t}

Doing so shows that t ≈ 4900 days.