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Radioactive Decay and Half-Life, Explained

After reading this you can compute how much of a radioactive sample remains after a given time, find the activity, and read the age of a sample from the fraction of a parent isotope still present.

What decay counts and why it is exponential

A radioactive nucleus is unstable. At any instant it has a fixed probability per unit time of decaying, and that probability does not depend on how old the nucleus is or on how many neighbors have already decayed. A carbon-14 nucleus that has survived 4,000 years is no more "ready" to decay than a fresh one. This memorylessness is the whole story.

Take a large number of nuclei N. If each has the same small chance of decaying in a short time dt, then the number that decay is proportional to how many are present:

\frac{dN}{dt} = -\lambda N

Here \lambda is the decay constant, the probability per unit time that any one nucleus decays, with units of inverse time. The minus sign says the population shrinks. Solving this differential equation gives an exponential, which is the reason every quantity in this article carries a factor of e or a power of one half.

When to use this model and when not to

The exponential law is exact for a single decaying isotope with a large number of nuclei. It works for dating, for medical isotope dosing, and for shielding calculations. Three assumptions matter.

First, you need many nuclei. The law describes an average. With only 10 atoms left, the actual count jumps around and the smooth curve fails. For a microgram of carbon-14 you have on the order of 10^{16} atoms, so the average is excellent.

Second, the decay constant must be truly constant. For ordinary radioactive decay it is, to the precision anyone has measured, unaffected by temperature, pressure, or chemistry.

Third, for dating you must know the starting amount. Carbon-14 dating assumes the ¹⁴C fraction in a living organism matched the atmosphere at the time of death. That assumption breaks for samples older than about 50,000 years (too little ¹⁴C left to measure) and needs calibration for the past few thousand years because atmospheric ¹⁴C has varied.

Do not use a single half-life to date a sample whose parent isotope can enter or leave after formation. Uranium-lead dating works because the rock traps both parent and daughter, so the clock is sealed. An open system leaks and gives a wrong age.

The formula and its intuition

Integrating dN/dt = -\lambda N gives the amount remaining after time t:

N(t) = N_0 \, e^{-\lambda t}

N_0 is the initial amount, N(t) the amount left after time t, and \lambda the decay constant. Any consistent unit works for the amount: number of atoms, grams, or moles, because all of them shrink by the same factor.

The half-life t_{1/2} is the time for the amount to fall to half. Set N(t_{1/2}) = N_0/2 and solve:

t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.6931}{\lambda}

This lets you rewrite the decay law using the more intuitive half-life. After n half-lives the amount is N_0 (1/2)^n:

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

Both forms give the same numbers. The power-of-one-half form is easier to reason about: three half-lives leave one eighth, no matter what you started with. The exponential form is easier to invert for dating.

Activity: what a counter actually reads

A Geiger counter does not see grams. It sees decays per second, called the activity A. Since each decay removes one nucleus, the activity is the rate of change of N:

A = \lambda N = A_0 \, e^{-\lambda t}

Activity follows the same exponential as the amount because it is just \lambda times N. The SI unit is the becquerel (Bq), one decay per second. The older curie (Ci) equals 3.7 \times 10^{10} Bq.

Worked number: 1 gram of cobalt-60 (t_{1/2} = 5.27 yr) contains about 1.0 \times 10^{22} atoms. Convert the half-life to seconds: 5.27 \times 3.156 \times 10^{7} = 1.663 \times 10^{8} s. Then \lambda = 0.6931 / 1.663 \times 10^{8} = 4.17 \times 10^{-9} per second, and A = \lambda N = 4.17 \times 10^{-9} \times 1.0 \times 10^{22} \approx 4.2 \times 10^{13} Bq, roughly 1,100 Ci. A short half-life means a high activity for the same mass.

A worked example: dating with carbon-14

Age from 25 percent of the carbon-14 remaining

This reproduces the demo: mode dating, isotope carbon-14, fraction remaining 25 percent.

  1. Write the fraction remaining as f = N/N_0 = 0.25.
  2. Start from f = e^{-\lambda t} and take the natural log: \ln f = -\lambda t, so t = -\ln f / \lambda.
  3. Use the half-life form instead, which avoids computing \lambda separately: t = t_{1/2} \, \dfrac{\ln f}{\ln(1/2)} = t_{1/2} \, \dfrac{\ln f}{-\ln 2}.
  4. Plug in f = 0.25: \ln 0.25 = -1.386 and \ln 2 = 0.6931, so the ratio is -1.386 / -0.6931 = 2.000.
  5. Multiply by the half-life: t = 5730 \times 2.000 = 11{,}460 years.

The answer is clean because 25 percent is exactly two halvings: 1 \to 1/2 \to 1/4. Two half-lives of 5,730 years give 11,460 years. Any fraction that is not a power of one half needs the logarithm, but the method is the same.

To check any dating result fast, count halvings. 50 percent is one, 25 percent is two, 12.5 percent is three. If your answer is not close to n \times t_{1/2} for the nearest whole n, you have a unit error somewhere.

Reading and interpreting the curve

The decay curve never reaches zero. It approaches it. After 10 half-lives you have (1/2)^{10} = 1/1024, about 0.098 percent left. That is why carbon-14 dating tops out near 50,000 years, roughly 8.7 half-lives, where under 0.25 percent remains and the measurement drowns in noise.

The curve halves every 5,730 years. The marker at 11,460 years sits at exactly 25 percent, the demo answer.

Notice the shape. Early on the sample loses a lot fast because there are many nuclei. Late on it barely changes because few remain. Equal drops in fraction do not mean equal time: going from 100 to 50 percent takes one half-life, but 50 to 25 percent takes another full half-life even though the absolute drop is smaller.

With a half-life of 5,730 years, the fraction remaining is (1/2)^{t/5730}: 50 percent at 5,730 years, 25 percent at 11,460 years, 12.5 percent at 17,190 years. A shorter half-life pulls the whole curve to the left, so the same fraction is reached sooner.

Common mistakes

The errors below account for most wrong answers.

Mixing time units
If the half-life is in years and the elapsed time is in days, the ratio t/t_{1/2} is meaningless. Iodine-131 has a half-life of 8.02 days. A "2 year" exposure is 730/8.02 \approx 91 half-lives, so essentially nothing is left. Convert both to the same unit first.
Confusing half-life with mean lifetime
The mean lifetime is \tau = 1/\lambda = t_{1/2}/\ln 2 \approx 1.443 \, t_{1/2}. For carbon-14 that is 8,267 years, not 5,730. Use half-life in the halving formula and \tau only where the mean is genuinely wanted.
Forgetting the sign in the log
Since f \lt 1, \ln f is negative. The age t = -\ln f / \lambda is positive only because of the leading minus. Drop it and you get a nonsensical negative age.
Reading percent as a decimal or vice versa
25 percent is f = 0.25. Entering 0.25 as "0.25 percent" would put the sample at about 8.6 half-lives, near 49,300 years instead of 11,460.

Related tools

Radioactive decay sits inside modern physics, so several neighbors connect to it. The energy released per decay shows up as gamma photons you can size with the Photon Energy Calculator. Fast decay products approach light speed, where the Special Relativity Calculator governs their observed lifetimes. When you report an age from measured activity, feed the counting uncertainty through the Error Propagation Calculator, and compare a computed age against a reference date with the Percent Error Calculator.

Frequently asked questions

Why does temperature not change a half-life?

Decay happens inside the nucleus, where energies are millions of times larger than the thermal energy available at any ordinary temperature. Heating a sample to 1,000 K adds about 0.1 eV of thermal energy per particle, far too little to touch a process governed by MeV nuclear energies. Half-lives measured across wide temperature and pressure ranges stay constant to measurement precision.

Can I date something younger than a few hundred years with carbon-14?

Only roughly. In 200 years just 1 - 0.5^{200/5730} = 2.4 percent of the ¹⁴C decays, so the fraction remaining is 97.6 percent. Measuring that small a drop precisely is hard, and modern fossil-fuel and bomb-test effects on atmospheric ¹⁴C complicate recent dates. Other methods suit the last few centuries better.

What is the difference between the amount remaining and the activity?

The amount is how many nuclei are left. The activity is how fast they decay, equal to \lambda N. Both fall off with the same exponential, so the fraction of activity remaining equals the fraction of atoms remaining. A detector reads activity, so dating from a Geiger count uses the activity ratio directly.

How many half-lives until a sample is "gone"?

It is never mathematically zero, but 10 half-lives leave about 0.1 percent and 20 leave about 0.0001 percent. For safety and disposal, roughly 10 half-lives is the common rule of thumb for a source dropping to a negligible level.

Does the starting amount affect the half-life?

No. The half-life depends only on the isotope. Whether you start with 1 gram or 1 kilogram, half is gone after one half-life. This follows directly from the halving form, where N_0 cancels when you compute the fraction remaining.

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