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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