The half-life of a radioactive substance is 3200 years. Find the quantity $Q(t)$ of the substance left at time $t > 0$ if $Q(0)=20$ $grams$?



The general exponential decay function is defined as:

$Q(t) = Q_0e^{-k(t-t_0)}$

$Q_0$  is the initial quantity, $-k$ is the “proportionality constant”, $t_0$ is the initial time, and $t$ would be any time duration.

For radioactive decay problems, $-k$ is treated as the “decay constant

Since $\tau$, or the “half-life,” is the amount of time at which a radioactive material’s quantity is reduced to half, we can turn $Q(t)$ into,

$Q(\tau) = \frac{Q_0}{2} = Q_0e^{-k(\tau-t_0)}$

$\frac{Q_0}{2} = Q_0e^{-k\tau}$

where $t_0 = 0$ and $t = \tau$.

We then solve for $k$ by canceling like terms and taking the natural logarithm of the equation:

$\ln(\frac{1}{2}) = \ln(e^{-k\tau})$

((Recall that $\ln(\frac{a}{b}) = \ln(a)-\ln(b)$ and $\ln(1) = 0$))

$-\ln(2) = -k\tau$

$k = \frac{ln(2)}{\tau}$

Actual solution

With these in mind, for Exercise 4.1.1, only algebra would be needed.

Given that $t_0 = 0$ and $Q(0) = 20$ $grams$,

$Q(t) = 20e^{-k(t-0)}$

And since $\tau = 3200$ $years$, solving for $k$ would yield

$k = \frac{ln(2)}{3200}$

Therefore, the quantity over time of a 20 gram substance with a 3200 years half-life can be found using,

$Q(t) = 20e^{-\frac{ln(2)}{3200}t}$

We don’t simply use $Q(t) = Q_0e^{-k\tau}$ or $Q(t) = 20e^{-k(3200)}$ as the solution because the resulting equation will NOT give us different values of $Q(t)$ at $t>0$, only at $t = \tau$. Remember that $\tau$ is just a value of $t$.

Solution by Brian and Jian Hui

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