Exponential Decay: Half-Life Formula

Half-life shows up whenever a quantity shrinks by the same fraction over and over, rather than by a fixed amount. Radioactive isotopes, medication levels in the bloodstream, and even the charge left in a capacitor all behave this way. The half-life, usually written \(T\) or \(T_{1/2}\), is the amount of time it takes for the quantity to fall to exactly half of whatever value it currently has. It does not matter when you start the clock, after one half-life, half of the current amount is always gone.

This idea is a special case of the exponential decay you meet in exponential growth and decay by a factor and exponential growth and decay by percentage. Half-life problems are just decay problems where the decay factor is fixed at one half over a known time span.

If \(N_0\) is the starting amount, \(T\) is the half-life, and \(N(t)\) is the amount remaining after time \(t\), then:

\( N(t) = N_0 \left(\frac{1}{2}\right)^{t/T} \)

Notice what happens when \(t = T\): the exponent becomes 1, so \(N(T) = N_0 \times \frac{1}{2}\), exactly half the starting amount, just as expected. When \(t = 2T\), the exponent is 2, giving \(N_0 \times \frac{1}{4}\), and so on. Each additional half-life cuts the remaining amount in half again.

Half-life problems are often written using the continuous decay model \(N(t) = N_0 e^{-kt}\), where \(k\) is the decay rate constant. Setting \(N(T) = \frac{N_0}{2}\) and solving gives the useful relationship:

\( T = \frac{\ln(2)}{k} \quad \)or equivalently\( \quad k = \frac{\ln(2)}{T} \)

This means the half-life and the decay constant are just two different ways of describing the same rate of decay. If a problem gives you \(k\), you can find \(T\) directly, and if it gives you \(T\), you can find \(k\). Both forms of the equation, the base one half version and the base \(e\) version, describe identical decay curves once you convert between \(T\) and \(k\).

A sample of iodine-131 has a half-life of 8 days. If a hospital starts with 40 mg, how much is left after 24 days?

Substitute \(N_0 = 40\), \(T = 8\), and \(t = 24\) into the formula:

\( N(24) = 40 \left(\frac{1}{2}\right)^{24/8} = 40 \left(\frac{1}{2}\right)^{3} = 40 \times \frac{1}{8} = 5 \)

After 24 days, only 5 mg of the original sample remains. Since 24 days is exactly 3 half-lives, the amount was cut in half three times: 40 to 20 to 10 to 5.

A 100 gram sample decays to 25 grams after 12 hours. What is the half-life?

Start with the formula and substitute the known values:

\( 25 = 100 \left(\frac{1}{2}\right)^{12/T} \)

Divide both sides by 100:

\( \frac{1}{4} = \left(\frac{1}{2}\right)^{12/T} \)

Since \(\frac{1}{4} = \left(\frac{1}{2}\right)^2\), the exponents must match:

\( 2 = \frac{12}{T} \quad \Rightarrow \quad T = 6 \)

The half-life is 6 hours. When the exponent does not simplify so neatly, you take the logarithm of both sides instead. That technique is covered in solving logarithmic equations, and it works on any half-life equation, no matter how the numbers turn out.

The most frequent error is placing \(t\) and \(T\) in the wrong spots in the exponent. Remember the exponent is always time elapsed divided by the half-life, \(t/T\), never the reverse. Another common slip is forgetting that after exactly one half-life the amount is halved, not reduced by some other fraction, so it is always worth checking a simple case like \(t = T\) before trusting a longer calculation. It also helps to be comfortable reading and sketching decay curves in general, which you can review in graphing exponential functions.

Half-life describes a constant time interval over which any remaining amount is cut in half. The formula \(N(t) = N_0 (1/2)^{t/T}\) lets you find the amount remaining at any time, and rearranging it (often with logarithms) lets you solve for the elapsed time or for the half-life itself. The same decay can also be written with the constant \(k\), since \(T = \ln(2)/k\).

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