Half-life (\(t_{1/2}\)) is the time required for a substance’s quantity to reduce to half of its initial value. This concept describes the rate of decay for various processes and is a fundamental measure in science. Calculating the duration of this decay is an important tool in multiple scientific disciplines. This measurement is widely used in nuclear physics, in medicine to determine drug dosing schedules, and in environmental science for assessing substance stability.
Defining the Half-Life Concept
The concept of half-life is based on exponential decay, where the rate of change is proportional to the amount of substance present. In most applications, such as radioactive decay, the process follows first-order kinetics, meaning the decay rate depends only on the amount of the single decaying substance.
For a first-order process, the half-life is constant. For example, the time required to drop from 100 grams to 50 grams is the same as the time required to drop from 50 grams to 25 grams. This constant nature allows scientists to predict the amount remaining after any given time period.
The half-life is inversely related to the rate constant (\(k\)), which quantifies the probability of decay per unit of time. The relationship is \(t_{1/2} = \ln(2)/k\). A shorter half-life means a larger rate constant and a faster decay process. To calculate the time for any decay, the initial amount (\(N_0\)) and the amount remaining at time \(t\) (\(N_t\)) are needed.
Finding Time Through Simple Half-Life Intervals
The simplest way to find the time elapsed is when the amount remaining is a perfect fraction (e.g., one-half, one-quarter) of the original quantity. This involves counting the number of half-life intervals (\(n\)) that have passed. After one half-life, 50% remains; after two, 25% remains; and after three, 12.5% remains.
This pattern follows a power of two relationship, where the fraction remaining is \((1/2)^n\). If an isotope has a half-life of 20 years and only 1/8th of the original amount is left, three half-lives have occurred. The total time is calculated by multiplying the number of intervals by the half-life duration (3 x 20 years = 60 years).
This interval-counting method is convenient because it does not require logarithms or the rate constant. However, its utility is limited to points where the remaining quantity is an exact fraction corresponding to a whole number of half-lives.
The First-Order Integrated Rate Law
To calculate the time for any percentage of decay, scientists use the first-order integrated rate law, derived from calculus. This law links the amount of substance remaining to the time elapsed. The general form is \(\ln(N_t) = -kt + \ln(N_0)\), where \(\ln\) represents the natural logarithm.
In this equation, \(N_t\) is the amount remaining after time \(t\), and \(N_0\) is the initial amount. The term \(k\) is the rate constant. Natural logarithms are used because they convert the exponential decay relationship into a linear one, allowing precise calculation of either the remaining amount or the time elapsed.
Isolating and Calculating Time (\(t\))
The first-order integrated rate law must be algebraically rearranged to isolate time, \(t\). Starting with \(\ln(N_t) = -kt + \ln(N_0)\), move the initial quantity term: \(\ln(N_t) – \ln(N_0) = -kt\). Using the logarithm rule for quotients, this simplifies to \(\ln(N_t/N_0) = -kt\).
To eliminate the negative sign, the terms inside the logarithm are inverted, resulting in the useful form: \(\ln(N_0/N_t) = kt\). Dividing by the rate constant (\(k\)) isolates the time variable, providing the final equation: \(t = \frac{\ln(N_0/N_t)}{k}\).
Before using this equation to find \(t\), the rate constant (\(k\)) must be known. If only the half-life (\(t_{1/2}\)) is provided, \(k\) is calculated using \(k = \ln(2)/t_{1/2}\). For example, consider a drug with a half-life of 10 hours, meaning \(k = \ln(2)/10 \approx 0.0693 \text{ hr}^{-1}\). If 30% of the initial dose remains, the ratio \(N_0/N_t\) is \(3.33\). Plugging these values into the derived equation yields \(t = \ln(3.33) / 0.0693 \approx 1.203 / 0.0693\), which calculates to approximately 17.36 hours.