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Radioactive Decay Calculator

Enter initial amount and half-life to visualise how a radioactive sample decays over time.

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U-238 takes 14 steps to reach lead

Real isotopes rarely decay directly to something stable — U-238 passes through about 14 sequential decays (through thorium, radium, radon, polonium and more) before finally reaching stable Pb-206, with half-lives ranging from billions of years down to microseconds. When an intermediate step is much shorter-lived than its parent, the chain reaches secular equilibrium: every daughter's activity locks onto the parent's decay rate, which is why a single exponential like the one graphed here still accurately describes the parent's decline even though the full chain has over a dozen steps.

Random atoms, predictable crowds

Each unstable nucleus decays at a moment no one can predict — there is no countdown ticking inside it, only a fixed probability per second. Yet with trillions of atoms, that randomness averages into the smooth curve you see here. The half-life is simply the time for a sample to lose half its atoms, so after 3 half-lives only (1/2)3 = 1/8 remains, no matter how much you started with.

The 'constant' that isn't quite constant

Textbooks say decay rates ignore temperature, pressure and chemistry — and for alpha and beta decay that's almost perfectly true. But isotopes that decay by electron capture, like beryllium-7, borrow an electron from their own atom, so their rate depends on the electron density at the nucleus. Chemists have shifted Be-7's half-life (~53 days) by around 1% just by changing its chemical compound or squeezing it under pressure.

Problem solved in full

  1. A Carbon-14 sample sitting for five 5730 years half-life 5 steps

    Carbon-14 has a half-life of 5730 years; a sample has been sitting for five of them. Find what is left, and find where this smooth exponential stops describing anything real.

    1. The decay constant is not the half-life — it is the probability per year that any one nucleus decays, and the two are related by ln 2 because that is the factor a half-life removes.

      λ=ln⁡2t1/2=0.6931475730=1.21×10−4 yr−1\lambda=\frac{\ln2}{t_{1/2}}=\frac{0.693147}{5730}=1.21\times10^{-4}\ \mathrm{yr^{-1}}

    2. Five half-lives is five halvings. The fraction left is a power of a half, and the exponential form gives the same number.

      NN0=(12)5=0.03125=3.125%\frac{N}{N_0}=\left(\tfrac12\right)^{5}=0.03125=3.125\%

    3. From a thousand nuclei, 31.25 remain. And there is the problem: a quarter of a nucleus does not exist, and the smooth curve has quietly stopped being a description of the sample.

      N=1000×0.03125=31.25N=1000\times0.03125=31.25

    4. What actually remains is a random number with a mean near 31. Decay is a Poisson process, so the spread is the square root of the count — about ±18% on a single measurement, and there is no averaging it away without more sample.

      σ=31=5.6⟹±18% on a single count\sigma=\sqrt{31}=5.6\quad\Longrightarrow\quad \pm18\%\ \text{on a single count}

    5. The elapsed time is five half-lives, or 28,650 years. Double that and only one nucleus in a thousand is left, which is why radiocarbon dating runs out around fifty thousand years — not because the physics changes, but because the signal falls below what can be distinguished from contamination.

      t=5×5730=28 650 yr;(12)10=0.098% at 57 300 yrt=5\times5730=28\,650\ \mathrm{yr};\qquad \left(\tfrac12\right)^{10}=0.098\%\ \text{at}\ 57\,300\ \mathrm{yr}

    Answer

    The tool prints λ = 1.21 × 10⁻⁴ per year, 3.125% remaining, 31.25 nuclei and an elapsed time of 28,650 years. Steps 3 and 4 are where a physics answer becomes a laboratory one. Exponential decay is an average over a population, and its accuracy comes from the population being astronomically large — a gram of carbon holds about 5 × 10²² atoms, so the fluctuations are invisible. Run the same law down to tens of nuclei and the noise is the answer, which is the reason a dating laboratory quotes an uncertainty in years rather than a date.

    N(t)=N0e−λt: a smooth law over a countable populationN(t)=N_0e^{-\lambda t}:\ \text{a smooth law over a countable population}

Learning path

Clocks in the rock

References (2)
Further reading (2)

· Reviewed against published sources. Corrections welcome.