Half-life is a fundamental concept used across various scientific disciplines, from physics and chemistry to medicine and environmental science. It is defined as the time required for a quantity of a substance to fall to half its initial value. Although often associated with radioactive decay, the principle applies to any process exhibiting exponential decay, such as the metabolism of drugs or the cooling of an object. This period remains constant, making it a reliable unit for calculating decay. This article provides step-by-step methods and formulas to solve common half-life problems.
The Conceptual Foundation of Half-Life Calculations
The simplest method for solving half-life problems relies on a step-by-step decay cycle, which is effective when the elapsed time is a whole number of half-lives. This conceptual approach avoids complex formulas by illustrating the halving process directly. After one half-life period, exactly half (50%) of the original quantity remains.
After a second half-life, the remaining 50% is again halved, leaving 25% of the initial amount. This process continues, with the amount halving with each subsequent period. For instance, after three half-lives, 12.5% remains, and after four half-lives, 6.25% remains.
Consider a substance with a half-life of five years. If you start with 100 grams, after five years (one half-life), 50 grams will be left. After ten years (two half-lives), the amount is reduced to 25 grams.
Solving for Remaining Mass or Elapsed Time
For situations where the elapsed time is not a whole number of half-lives, the exponential decay formula is required for precise calculation. This formula links the amount remaining to the initial amount, the time elapsed, and the half-life period. The standard half-life equation is \(N(t) = N_0 \cdot (1/2)^{t/T}\), where \(N(t)\) is the remaining quantity, \(N_0\) is the initial quantity, \(t\) is the elapsed time, and \(T\) is the half-life period.
The most straightforward application is determining the remaining mass, \(N(t)\), when all other variables are known. For instance, if a 400-gram sample has a half-life of 10 years, the amount remaining after 15 years is calculated by substituting the values. This results in \(400 \cdot (1/2)^{15/10}\), which simplifies to \(400 \cdot (1/2)^{1.5}\) and yields the exact remaining mass.
A more complex problem involves finding the elapsed time, \(t\), when the initial amount, final amount, and half-life are known. Since \(t\) is in the exponent, logarithms must be used. First, rearrange the main equation to isolate the exponential term: \(N(t)/N_0 = (1/2)^{t/T}\).
Apply a logarithm (natural log or log base 10) to both sides. A property of logarithms allows the exponent to be moved down as a multiplier, resulting in \(\log(N(t)/N_0) = (t/T) \cdot \log(1/2)\). The final step isolates \(t\) by multiplying by \(T\) and dividing by \(\log(1/2)\).
The rearranged formula to solve for time is \(t = T \cdot (\log(N(t)/N_0) / \log(1/2))\). This logarithmic rearrangement is necessary, for example, when determining the age of a bone fragment containing 20% of its original Carbon-14.
Determining the Half-Life Period
A distinct type of half-life problem requires solving for the half-life period, \(T\). This occurs when the initial amount (\(N_0\)), the final amount (\(N(t)\)), and the elapsed time (\(t\)) are all known. Since \(T\) is part of the exponent, this calculation also necessitates the use of logarithms.
Start by rearranging the decay formula to \(N(t)/N_0 = (1/2)^{t/T}\). Applying the logarithm to both sides yields \(\log(N(t)/N_0) = (t/T) \cdot \log(1/2)\). The goal is to isolate \(T\) from the exponent.
First, isolate the time ratio \(t/T\) by dividing both sides by \(\log(1/2)\), resulting in \(t/T = \log(N(t)/N_0) / \log(1/2)\). The entire right side of the equation is now a single calculated number. The final algebraic step involves moving \(T\) out of the denominator and isolating it. The resulting formula to solve for the half-life period is \(T = t \cdot (\log(1/2) / \log(N(t)/N_0))\).