From Measurements to a Half-Life
A half-life describes how long a decaying quantity takes to fall to half its value. Sometimes a half-life is already known; in other situations, measurements are available and the half-life must be estimated from them.
Suppose a quantity is measured at two different times. If its decay follows a constant continuous relative rate, the ratio of the later measurement to the earlier one tells us how quickly it changed over the time interval. From that ratio we can estimate the rate constant, then find how long the fitted model takes to halve the quantity.
The measurements need not include the initial amount at time zero. The method works with two positive measurements at any two times, provided we interpret the result as an estimate under an exponential-decay model.
Estimating the Rate from Two Measurements
Let the earlier measurement be \(Q_1=Q(t_1)\), and the later measurement be \(Q_2=Q(t_2)\), where \(t_2>t_1\). Assume both measurements are positive and the quantity follows one constant-rate exponential model between those times. The model gives the ratio
Taking natural logarithms and solving for \(k\) gives the estimated continuous rate constant. When \(Q_2<Q_1\), the logarithm of the ratio is negative, so the estimated \(k\) is negative, as required for decay.
To obtain a half-life, ask how much time \(H\) the fitted model needs to reduce the quantity to half its value. Over that interval, the ratio is \(1/2\), so \(e^{kH}=1/2\). Since \(k<0\), the resulting \(H\) is positive.
Why the Method Works
The derivation uses the same model for both measurements. In particular, it does not require knowing the amount at time zero. To see why the unknown starting amount disappears, write the model relative to the first measurement.
Under a constant rate \(k\), the quantity after \(t_2-t_1\) time units is \(Q_1e^{k(t_2-t_1)}\).
Since this model must give \(Q_2\) at \(t_2\), we have \(Q_2=Q_1e^{k(t_2-t_1)}\).
Divide by \(Q_1>0\), take natural logarithms, and divide by \(t_2-t_1>0\). This gives \(k=\ln(Q_2/Q_1)/(t_2-t_1)\).
Set the model’s ratio over a time \(H\) equal to one half: \(e^{kH}=1/2\). Taking logarithms gives \(kH=-\ln 2\), hence \(H=\ln 2/(-k)\).
This argument also explains why the time difference matters: the measurements describe change over \(t_2-t_1\), not necessarily over a time interval starting at zero. It explains why their ratio matters too: dividing removes the amount at the earlier time, leaving the relative change that determines the continuous rate.
Worked Examples
Worked Example: A Fourfold Decrease in Ten Days
A sample contains \(800\) mg at day \(3\) and \(200\) mg at day \(13\). Estimate its continuous decay rate and half-life.
First record the time interval and measurement ratio:
The rate estimate is
Therefore the estimated half-life is
Check: Five days at this rate gives a factor \(e^{(-0.138629)(5)}=e^{-0.693147}=0.5\), so the fitted model halves the amount in \(5\) days. Ten days gives \(e^{(-0.138629)(10)}=e^{-1.38629}=0.25\), matching \(200/800=0.25\). Thus the estimate reproduces both the observed fraction and the inferred half-life.
Worked Example: A Non-Exact Measurement Ratio
A liquid contains \(750\) mg of a substance at hour \(2\) and \(300\) mg at hour \(10\). Estimate the half-life, rounding the final result to three decimal places.
The elapsed time is \(10-2=8\) hours. The observed fraction remaining is
Thus the estimated rate constant is
Using the estimated rate, the half-life is
Check: Using the unrounded rate, the model’s ratio after eight hours is \(e^{(-0.1145363415)(8)}=e^{-0.9162907}\approx0.400\), the observed fraction. Over the estimated half-life, the exponent is approximately \((-0.1145363415)(6.05177)=-0.693147\), so the model’s ratio is \(e^{-0.693147}=0.500\). The estimate is therefore consistent with both measurements and the definition of half-life.
Worked Example: Measurements Not Taken at Time Zero
A decaying quantity is \(120\) g at day \(5\) and \(45\) g at day \(13\). Find the estimated half-life. Round the final result to three decimal places.
The interval is \(13-5=8\) days. The fraction remaining over that interval is
Calculate the rate constant:
Then calculate the half-life:
Check: The fitted model predicts a ratio of \(e^{(-0.122603657)(8)}=e^{-0.980829}\approx0.375\) over the measured eight days, which is \(45/120\). Over \(5.6536\) days, its ratio is \(e^{(-0.122603657)(5.6536)}\approx e^{-0.693147}=0.500\). The measurements begin at day \(5\), but the ratio-based calculation still estimates the time required to halve from any current amount.
Conditions and Interpretation
The formula estimates a decay half-life only when the later measurement is smaller than the earlier one, and both measurements are positive. If the two measurements are equal, the formula gives \(k=0\), which describes no change over that interval; it does not yield a finite decay half-life. If the later measurement is larger, the fitted rate is positive and describes growth rather than decay.
Two measurements determine a rate constant for an exponential model through those observations. They do not, by themselves, show that the quantity truly follows a constant relative rate at all other times. With measurement error or a process that changes rate, the estimate is an approximation. Additional measurements can help assess whether a single exponential model is a reasonable description.
Units must also be kept consistent. If the elapsed time is measured in days, \(k\) is in inverse days and \(H\) is in days. A rate constant expressed per hour must not be combined directly with a time interval in days.
Common Pitfall: Using the Difference Instead of the Ratio
For the \(800\) mg and \(200\) mg measurements ten days apart, a tempting but incorrect calculation is
This is an average absolute change in amount per day, not the continuous relative rate constant \(k\). Its units are mg/day, whereas a rate constant in an exponential model has units of inverse days. Substituting \(-60\text{ mg/day}\) into \(H=\ln 2/(-k)\) would not even produce a time unit.
The correct calculation uses a ratio and its logarithm:
The ratio measures the fraction remaining; taking its logarithm converts that multiplicative change into the constant rate used by the exponential model.
Check Your Understanding
For each question, pay attention to the ratio, the elapsed time, and the units of the result.
- A quantity falls from \(640\) mg to \(160\) mg over \(12\) hours. What are the estimated rate constant and half-life?
- Why can measurements at two times other than time zero still determine an exponential rate constant?
- A sample is \(90\) g at day \(4\) and \(60\) g at day \(10\). Write the expressions for the estimated rate constant and half-life, and state the units.
- What does it indicate about decay if the later positive measurement is greater than the earlier one?
- Explain why the difference of two measured amounts divided by elapsed time is not the rate constant in the exponential model.
