Radioactive Decay from Half-Life Data

From Half-Life Data to a Decay Model

Radioactive decay is modeled by a quantity decreasing at a rate proportional to the amount that remains. A sample with twice as many radioactive nuclei has, in this model, twice the rate of decay. The proportionality makes the amount decrease exponentially, but laboratory data are often reported in terms of a more directly interpretable quantity: the half-life.

A half-life tells you how long it takes for half of the radioactive material to remain. It is not the same as the decay constant. The half-life has units of time; the decay constant has units of inverse time. Converting correctly between them is essential when building or using a model.

Here \(M(t)\) is the mass of radioactive material at time \(t\), measured from the chosen starting time. We assume \(M(0)=M_0>0\), no additional radioactive material enters the sample, and the decay constant stays constant over the period modeled.

Deriving the Half-Life Formula

The equation \(\frac{dM}{dt}=-kM\) says that the rate of change is negative and proportional to the amount present. To find the amount at time \(t\), separate variables and integrate. Since the starting amount is positive, the exponential solution remains positive, so division by \(M\) is valid.

Integrating with respect to time gives \(\ln M=-kt+C\), where \(C\) is a constant. Applying the initial condition \(M(0)=M_0\) gives \(C=\ln M_0\). Exponentiating then gives the exponential decay model.

At one half-life, \(t=T\), the defining condition is \(M(T)=M_0/2\). Substituting that condition into the model determines \(k\).

Taking natural logarithms gives \(\ln(1/2)=-kT\). Since \(\ln(1/2)=-\ln 2\), the decay constant is \(k=\ln 2/T\). Substituting this into the exponential model gives a second, often convenient, version.

The two forms express the same model. The \(e^{-kt}\) form uses the decay constant directly; the \(2^{-t/T}\) form makes the number of half-lives visible. If \(t=nT\), where \(n\) is the number of elapsed half-lives, then \(M(t)=M_0\,2^{-n}\). For example, after three half-lives, \(2^{-3}=1/8\) of the initial mass remains.

Worked Examples

Using the Model to Find Time

When the initial mass \(M_0\), half-life \(T\), and remaining mass \(M\) are known, taking logarithms of the model gives a direct formula for elapsed time:

For decay from a positive initial mass with no added material, the target mass must satisfy \(0<M\leq M_0\) for a nonnegative elapsed time. If \(M=M_0\), the formula gives \(t=0\). If \(0<M<M_0\), it gives a positive time. The ideal model approaches zero without reaching it at any finite time, so a target mass of exactly zero does not give a finite solution.

The formula also offers a useful interpretation: the elapsed number of half-lives is \(\log_2(M_0/M)\). A logarithm is needed when the remaining fraction is not an exact power of \(1/2\), as in the \(100\)-to-\(35\) milligram example.

Keep the Decay Constant and Half-Life Distinct

The half-life \(T\) measures time. The decay constant \(k\) measures the proportional rate of decrease per unit time. They are related, but they are not interchangeable:

A larger \(k\) means faster decay and therefore a shorter half-life. A longer half-life corresponds to a smaller \(k\). This relationship is also a dimensional check: \(\ln 2\) is dimensionless, so dividing it by a time gives \(k\) units of inverse time.

A Common Half-Life Error

Suppose an isotope has a half-life of \(10\) years. A common error is to use \(k=10\), then write \(M(t)=M_0e^{-10t}\). This calculation both reverses the relationship between \(T\) and \(k\) and gives the exponent the wrong units: \(10\) years is a time, not an inverse-time decay constant.

The correction is to use \(k=\ln 2/T\):

The corrected model is \(M(t)=M_0e^{-0.069315t}\) when \(t\) is measured in years. At \(t=10\) years, the exponent is approximately \(-0.69315\), so the remaining fraction is \(e^{-0.69315}\approx0.5000\), as the half-life requires. Always check the model at \(t=T\): it must give \(M(T)=M_0/2\).