Radioactive Decay Law

In one sentence: An unstable sample loses the same fraction of its atoms in each equal stretch of time, so the number left shrinks along a smooth curve.

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

Read it out loud: “N of t equals N zero times e to the power of minus lambda times t. N of t is the number of atoms that have not decayed yet at time t. N zero is the number at the start. e is Euler's number, about 2.718. Lambda is a Greek letter, said LAM-duh, and it is the decay constant, a measure of how quickly each atom tends to decay. t is the time that has passed. The minus sign in the power means the number shrinks as time goes on.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, like $1.0 \times 10^{6}$ and $10^{-4}$: A positive exponent means multiply by ten that many times. $10^{6}$ is 1 followed by six zeros, so $1.0 \times 10^{6}$ is 1,000,000 atoms. A negative exponent means divide by ten that many times. $1.21 \times 10^{-4}$ is 0.000121, which is a number about one ten-thousandth. When multiplying powers of ten, add the exponents; when dividing, subtract them.
Euler’s number $e$ and $e^{x}$: The letter $e$ stands for Euler’s number, about 2.71828, a fixed number like $\pi$ . It appears whenever something changes at a rate proportional to its current size. For a whole number, $e^{2} = e \times e \approx 7.389$ . The power $e^{-x}$ means 1 divided by $e^{x}$ , so a negative power is a small number, and it gets smaller as $x$ grows. Calculators have a key labeled $e^{x}$ or exp.
The Greek letter $\lambda$ (lambda) and units per year: The Greek letter $\lambda$ is said LAM-duh. Here it is the decay constant, the chance per unit of time that one atom decays. If time is in years, $\lambda$ is in units of one over a year, written 1/yr or per year. Multiplying $\lambda$ by a time in years gives a number with no unit, which is why the exponent is a plain number.
The natural logarithm $\ln 2$: The natural logarithm, written $\ln$ , is the power you must raise $e$ to in order to get the number inside. Because $e^{0.693147} \approx 2$ , we have $\ln 2 \approx 0.693147$ . The number $\ln 2$ appears in the half-life formula, and it is the reason a half-life is a natural way to measure decay.
Half-life, $t_{1/2}$: The half-life is the time it takes for half of the atoms in a sample to decay. After one half-life, half of the original atoms remain. After two half-lives, a quarter remains. Carbon-14 has a half-life of about 5,730 years. Every isotope has its own half-life, from a tiny fraction of a second to billions of years.
Unstable atoms and random decay: Some atomic nuclei are unstable. They can break apart or change into a different nucleus, and they release radiation while doing so. This change is called radioactive decay. Each decay is random, so you cannot say when one particular atom will decay, but a large group of atoms follows a steady pattern.

What each symbol means

Symbol	Meaning
$N(t)$	The number of atoms that have not decayed yet at time $t$ . It is a count of atoms, so it has no unit. For the example below, $N(10000) \approx \text{298,300}$ carbon-14 atoms.
$N_0$	The number of undecayed atoms at time zero, when we start counting. It is a count with no unit. The example uses $N_0 = 1.0 \times 10^{6}$ carbon-14 atoms, which is 1,000,000 atoms.
$\lambda$	The decay constant, in units of one over time, such as per year. It is the chance per unit time that one atom decays. For carbon-14 it is about $1.21 \times 10^{-4}$ per year, so roughly 0.012 percent of the atoms decay in a single year.
$t$	The time that has passed since the start, in the same unit as $\lambda$ ’s time. In the example $t$ is measured in years, so $t = 10000$ means 10,000 years.
$e$	Euler’s number, about 2.71828. It is a fixed number with no unit, the natural base for exponential growth and decay.
$t_{1/2}$	The half-life, the time for half of the atoms to decay. It has the same unit as time, such as years. It is linked to the decay constant by $t_{1/2} = \frac{\ln 2}{\lambda}$ , and it is often easier to read from measurements than $\lambda$ .

The formula piece by piece

$N_0$: The starting count of atoms. It sets the size of the sample, and the formula scales the whole sample by the same factor, so a sample twice as large has twice as many atoms left at every time.
$\lambda t$: The decay constant times the time. It is a plain number, because the unit of $\lambda$ , one over time, cancels the unit of time. A larger value means more decay has had time to happen.
$-\lambda t$: The same product with a minus sign. The minus sign makes the exponent negative, so the next step gives a number smaller than 1.
$e^{-\lambda t}$: Euler’s number raised to the negative exponent. This factor is always between 0 and 1. It is the fraction of atoms that survive, and it shrinks steadily as time grows.
$N_0\, e^{-\lambda t}$: Start with the starting count and multiply by the surviving fraction. Each later time gives a smaller count, following the curve of exponential decay.
$N(t) = N_0\, e^{-\lambda t}$: The whole formula in one sentence: the number of atoms left at time t equals the starting number times Euler’s number raised to the power of minus the decay constant times the time. The count falls along a smooth curve and never quite reaches zero.

What the formula is doing, and why

What goes in and what comes out. The inputs are the starting count $N_0$ , the decay constant $\lambda$ , which measures how quickly each atom tends to decay, and the time $t$ . The output $N(t)$ is the number of atoms left. Each atom decays on its own, so the count at a later time is the starting count multiplied by one number, $e^{-\lambda t}$ , which lies between 0 and 1.

Why multiply by $e^{-\lambda t}$ ? Each atom has the same chance per unit time of decaying. An atom does not age or remember how long it has existed. So the fraction that survives shrinks by the same factor in every equal stretch of time. A rule that multiplies the count by the same factor in each equal time step gives an exponential curve, and $e$ is the natural base for that kind of change, the same base used when interest is added at every instant.

Why the minus sign? The exponent $-\lambda t$ is negative, so $e^{-\lambda t}$ equals 1 divided by $e^{\lambda t}$ . As time grows, $e^{\lambda t}$ grows, and the fraction shrinks. The count never falls below zero, and it never reaches zero either. It gets closer and closer, which is why a sample is never completely gone in the formula.

The half-life. Ask when the surviving fraction equals one half. Setting $e^{-\lambda t} = \frac{1}{2}$ gives $t_{1/2} = \frac{\ln 2}{\lambda}$ , the half-life. After one half-life the count is half of $N_0$ . After two half-lives it is a quarter. Each half-life multiplies the count by one half, so the same law can be written as $N(t) = N_0 \left(\frac{1}{2}\right)^{t / t_{1/2}}$ . Half-lives are often easier to read from measurements than $\lambda$ .

What the law does not say. The law is statistical. It predicts how many atoms remain in a large sample, not when one particular atom will decay. Each decay time is random. For nearly all radioactive decays, ordinary changes in temperature and pressure hardly change the rate, because the decay happens inside the nucleus.

An everyday picture

Imagine a concert hall where each person has the same chance of leaving in any given minute. In the first stretch of time, about half of the crowd leaves. In the next stretch, half of those still inside leave, and so on. The crowd shrinks by the same fraction in each equal interval. That steady shrinking is exactly the behavior of the decay law.

Worked example: Carbon-14 after 10,000 years

Carbon-14 has a half-life of about 5,730 years. Start with $N_0 = 1.0 \times 10^{6}$ carbon-14 atoms, which is 1,000,000 atoms. Find how many remain after $t = 10000$ years.

Find the decay constant from the half-life. The formula needs $\lambda$ , but we were given the half-life. Use $\lambda = \frac{\ln 2}{t_{1/2}}$ , which comes from asking when the surviving fraction equals one half. Here $\ln 2 = 0.693147$ , so $\lambda = \frac{0.693147}{5730} \approx 1.2097 \times 10^{-4}$ per year. The factor $10^{-4}$ means the number is about one ten-thousandth, or 0.00012097, per year.
Write the exponent. Multiply the decay constant by the time: $\lambda t = 1.2097 \times 10^{-4} \times 10000$ . Since $10000 = 10^{4}$ , the powers of ten meet, and we add their exponents: $-4 + 4 = 0$ . That leaves $1.2097 \times 10^{0} = 1.2097$ , a plain number with no unit.
Take the exponential. We need $e^{-1.2097}$ , which equals 1 divided by $e^{1.2097}$ . A calculator, using its $e^{x}$ key, gives $e^{-1.2097} \approx 0.2983$ . This means about 29.8 percent of the atoms are still there after 10,000 years.
Multiply by the starting number. $N = 1.0 \times 10^{6} \times 0.2983$ . Multiplying 1.0 by $10^{6}$ leaves $10^{6}$ . Moving the decimal point one place to the right makes 0.2983 into 2.983, so the power of ten drops by one: $0.2983 \times 10^{6} = 2.983 \times 10^{5}$ . That is 298,300 atoms.
Check the answer with half-lives. Ten thousand years is $10000 / 5730 \approx 1.745$ half-lives. Each half-life multiplies the count by one half, so the fraction left is $0.5^{1.745} \approx 0.2983$ . This matches the exponential calculation, so the number is correct.
Read the result in plain words. About 298,300 atoms remain, which is about 30 percent of the start. About 70 percent have decayed. The fraction lies between one half (after one half-life) and one quarter (after two half-lives), which is what we expect after 1.745 half-lives.

Answer: About 298,300 carbon-14 atoms remain after 10,000 years, roughly 30 percent of the starting sample. About 70 percent have decayed. This is a prediction for a large sample, not for any single atom.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Using the half-life where the decay constant belongs. The formula needs $\lambda$ , so convert the half-life first with $\lambda = \frac{\ln 2}{t_{1/2}}$ .
Thinking all the atoms are gone after two half-lives. After one half-life half remains, and after two half-lives a quarter remains. Atoms keep decaying, following the same curve, and no fixed time empties the sample.
Mixing time units. If the half-life is in years, the time $t$ and the decay constant $\lambda$ must also be in years. A mismatch makes the exponent wrong by a large factor.
Treating the formula as a prediction for one atom. It describes large numbers of atoms, because each atom’s decay time is random.

Where you will meet it

Radiocarbon dating, which uses the decay of carbon-14 to estimate the age of material that was once alive.
Nuclear medicine, where the decay of short-lived tracer isotopes sets how much activity remains over time, so that doses can be planned.
Nuclear waste planning, where isotope half-lives set how long a material must be stored before it has decayed enough to handle safely.

Questions people ask

Does a half-life mean half the atoms are gone after that time?

Yes, for a large sample. After one half-life about half of the original atoms have decayed. During the next half-life, half of the atoms still left decay, which leaves a quarter of the original sample.

Can the count reach zero?

In the formula it never does. The curve gets closer and closer to zero. In a real sample the count is a whole number, so the last atom decays at some finite time.

Does temperature or pressure change the half-life?

For nearly all radioactive decays, no. Decay happens inside the nucleus, and ordinary changes in temperature and pressure hardly affect the nucleus.

Quantum Tunneling Probability: A quantum particle can sometimes pass through a barrier that is too high for it to climb over, and this formula gives a rough estimate of how often.
Energy-Time Uncertainty Relation: A state that lasts only a short time cannot have one exact energy, so the light it gives off spreads over a small range of colors.
Planck-Einstein Relation (E = hf): E = hf says each packet of light carries energy in proportion to its frequency: higher frequency means more energy per packet, and brightness does not change it.