Question

The function Q(t)=Q 0 e^ -kt be used to model radioactive decay . represents the quantity remaining after tyears; k is the decay constant, 0.00011 . How long, in years , will it take for a quantity of plutonium -240 to decay to 25% of its original amount?

Answer 1

As an intro, the general solution to \[\frac{dy}{dt}=-ky\] is given as \[y=C e^{-kt}\]
\[\Rightarrow Q(t)=Q_0 e^{-kt}\]
The question is asking for the time it will take for a quantity of Pu-240 to decay to 25% of its original amount. This can be translated to the time it takes for Pu-240 change from \(Q_0\) to \(0.25 Q_0\). Let \(Q(t)=0.25Q_0\)
\[0.25Q_0=Q_0e^{-kt}\]
It is given that \(k=0.00011=1.1 \times10^{-4}\)
\[0.25Q_0=Q_0e^{-(1.1 \times 10^{-4})t}\]
\[\frac{0.25Q_0}{Q_0}=e^{-(1.1 \times 10^{-4})t}\]
\[0.25=e^{-(1.1 \times 10^{-4})t}\]
\[\ln(0.25)=\ln(e^{-(1.1 \times 10^{-4})t})\]
By the properties of logarithms, \(\ln(0.25)=\ln(\frac{1}{4})=\ln(1)-\ln(4)=-\ln(4)\). Also, \(\ln(e^{-(1.1 \times 10^{-4})t})=(-(1.1 \times 10^{-4})(t))\ln(e)=-(1.1 \times 10^{-4})t\)
\[∴-\ln(4)=-(1.1 \times 10^{-4})t\]
I believe now you can isolate t using simple algebra and solve for the time. Round to 2 s.f. Let me know if you have any questions.