Radioactive Decay IVPs from Half-Life Data

From Half-Life Data to a Decay IVP

A radioactive sample is often described by two measurements: its mass at a chosen starting time and its half-life. These data determine a model for the sample’s mass at later times. Once the model is set up, it can answer two practical questions: how much mass remains after a specified time, and when the mass reaches a specified threshold.

Let \(M(t)\) be the sample’s mass at time \(t\), and let \(M_0\) be its mass when the clock starts. The standard constant-rate radioactive-decay model assumes that the rate of change of mass is proportional to the current mass, with a negative rate because the mass decreases.

The half-life \(T\) is the positive time it takes the mass to become half of its initial value. It must be expressed in the same time units as \(t\). For this model, the half-life determines \(k\), so the starting mass and half-life together specify the IVP’s solution.

Why the Half-Life Determines the Model

For positive mass, separate variables in the differential equation and integrate from the initial time \(0\) to a time \(t\). This gives a direct derivation of the model’s solution.

Exponentiating gives \(M(t)=M_0e^{-kt}\). At the half-life, \(M(T)=M_0/2\). Substituting this condition into the solution yields the value of \(k\).

Thus, the half-life form of the solution is \(M(t)=M_0e^{-(\ln 2)t/T}\), equivalently \(M(t)=M_0\,2^{-t/T}\). The exponent \(t/T\) counts the number of half-lives elapsed, including when that number is not an integer.

To check the formula, differentiate it and evaluate it at the initial time:

It therefore satisfies both the differential equation, with \(k=\ln 2/T\), and the initial condition. The half-life condition also follows directly: \(M(T)=M_0e^{-\ln 2}=M_0/2\).

Calculating Remaining Mass

When the elapsed time is known, substitute it and the half-life into the model. Check units before calculating: if the half-life is in years, the elapsed time must also be in years. The initial mass and the resulting mass have the same units.

As a check on the scale of an answer, mass must decrease as elapsed time increases under this model. After exactly one half-life, the formula gives half the initial mass; after two half-lives, it gives one quarter.

Finding the Time to Reach a Threshold

Suppose the target mass is \(M_*\), and we want the elapsed time when the sample reaches it. Solve the model equation for \(t\). For a target between zero and the initial mass, the resulting time is nonnegative:

This formula applies to a positive target \(M_*\) with \(M_*\leq M_0\). If \(M_*=M_0\), the time is \(0\). If \(0<M_*<M_0\), the time is positive. A target of zero is never reached in finite time by this exponential model, and a target greater than \(M_0\) cannot be reached at a nonnegative time.

When the ratio \(M_0/M_*\) is not a power of \(2\), use the natural logarithm form of the formula. The logarithm gives an exact expression for the time; a decimal approximation is useful when the time is needed numerically.

Reaching a Threshold Versus Falling Below It

Exponential decay approaches zero without reaching it at any finite time. This matters when a question asks when a sample falls below a threshold rather than when it reaches the threshold. For \(0<M_*<M_0\), the model reaches \(M_*\) at the time \(t_*=T\log_2(M_0/M_*)\). It is equal to the threshold at \(t=t_*\), and it is strictly below the threshold only for \(t>t_*\).

A related check is whether the target is within the model’s range for \(t\geq0\). Since \(2^{-t/T}\leq1\) for \(t\geq0\), the mass never exceeds \(M_0\). A negative time obtained by using the threshold equation for a larger target describes a backward extrapolation, not a future elapsed time.

A Reliable Calculation Process

Write down the initial mass, half-life, elapsed time or target, and all units before substituting. Then select the model equation that matches the question.

Summary

A radioactive-decay IVP is determined by its starting mass and half-life. The resulting model gives the remaining mass at any nonnegative time, while rearranging it gives the time to reach a positive threshold no greater than the initial mass. Always check units and whether a threshold can be reached in forward time. For a strict “below” threshold, the sample must be observed after, not merely at, the time of equality.