Half-life

A single nucleus decays at a random moment — no one can predict which second. But a sample holds billions of nuclei, and statistics takes over: in a fixed time, always half of whatever is currently there decays. That fixed time is the half-life, $t_{1/2}$. Carbon-14’s is 5,730 years; fluorine-18’s is 110 minutes; uranium-238’s is 4.5 billion years. Same rule, wildly different clocks.

In short

• After one half-life $\tfrac{1}{2}$ is left, after two $\tfrac{1}{4}$, after three $\tfrac{1}{8}$. Each half-life removes half of what is left, not half of the original.
• After ~10 half-lives ($\tfrac{1}{1000}$ left) a source has mostly faded.
• Heat, pressure, and chemistry basically cannot change it. Nuclear energies are millions of eV; chemistry deals in a few eV.

Everything below is one formula, used forwards and backwards:

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

$N_0$ is what you start with, $N$ is what remains after time $t$. The exponent $n = t / t_{1/2}$ is simply the number of half-lives that elapsed — whole or fractional.

Worked problem 1 — whole half-lives, going forwards

Start with $80\ \mathrm{g}$ of carbon-14 ($t_{1/2} = 5{,}730$ years). How much remains after $17{,}190$ years?

First say it in words: each 5,730-year block halves whatever survived the previous block. Three blocks fit into 17,190 years, so the sample halves three times: $80 \to 40 \to 20 \to 10\ \mathrm{g}$.

As a formula: $n = 17{,}190 / 5{,}730 = 3$, and $N = 80 \times (\tfrac{1}{2})^3 = 80 \times \tfrac{1}{8} = 10\ \mathrm{g}$.

The classic trap: if a half-life is 10 years, is everything gone after 20? No — after two half-lives $\tfrac{1}{4}$ is still there ($100 \to 50 \to 25$). Halving never reaches zero; it just keeps halving the remainder.

Try it. Iodine-131 ($t_{1/2} = 8$ days), $96\ \mathrm{MBq}$ to start. After 24 days ($n = 3$): $96 \times \tfrac{1}{8} = 12\ \mathrm{MBq}$.

Worked problem 2 — counting half-lives backwards

A phosphorus-32 sample ($t_{1/2} = 14$ days) drops from $200$ counts/min to $25$. How much time passed?

Reason it out first: $25$ is $\tfrac{1}{8}$ of $200$, and $\tfrac{1}{8} = (\tfrac{1}{2})^3$, so three half-lives elapsed. Three blocks of 14 days is $3 \times 14 = 42$ days.

In symbols: $N/N_0 = 25/200 = 0.125 = (\tfrac{1}{2})^n$, so $n = 3$ and $t = 3 \times 14 = 42$ days. Whenever the remaining fraction is a clean power of $\tfrac{1}{2}$ ($\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \tfrac{1}{16}$…), you can read $n$ off by counting without any logs.

Try it. A source falls to $\tfrac{1}{16}$ of its activity. That is $(\tfrac{1}{2})^4$, so $n = 4$ half-lives. With $t_{1/2} = 4$ hours, that is 16 hours.

Worked problem 3 — fractional half-lives, going forwards

Most real times are not whole blocks. Fluorine-18 has $t_{1/2} = 110$ min. After $240$ min, how much of a $60\ \mathrm{MBq}$ dose is left for the PET scan?

Compute the exponent: $n = 240 / 110 \approx 2.18$ half-lives. Then $N = 60 \times (\tfrac{1}{2})^{2.18}$. Since $(\tfrac{1}{2})^{2.18} \approx 0.22$, about $60 \times 0.22 \approx 13.2\ \mathrm{MBq}$ remains. In words: a little more than two halvings happened, so a little less than a quarter survives — the calculator just makes “a little less” precise. Any scientific calculator does $(\tfrac{1}{2})^{2.18}$ directly with the $x^y$ key.

Try it. Same isotope, $90$ min elapsed ($n \approx 0.82$). $(\tfrac{1}{2})^{0.82} \approx 0.57$, so ~57% remains — more than half, as expected for less than one half-life.

Worked problem 4 — solving for time with logs

An iodine-131 spill ($t_{1/2} = 8$ days) must decay to $12\%$ before cleanup ends. How long is the wait?

Set up the sentence first: we need the $n$ such that $(\tfrac{1}{2})^n = 0.12$. Because $n$ is trapped in the exponent, take logs of both sides — base 10 is fine:

$0.12 = (\tfrac{1}{2})^n$ $\log_{10}(0.12) = n \cdot \log_{10}(0.5)$ $n = \frac{\log_{10}(0.12)}{\log_{10}(0.5)} \approx \frac{-0.921}{-0.301} \approx 3.06$

Then convert blocks to days: $t = 3.06 \times 8 \approx 24.5$ days. Sanity-check it: $12\%$ sits between $\tfrac{1}{8}$ ($12.5\%$, 3 half-lives) and $\tfrac{1}{16}$, so $n$ just over 3 is exactly right — the logs only refined what the fractions already suggested.

If your class writes $N = N_0 e^{-\lambda t}$ instead, it is the same formula rearranged, with $\lambda = \ln 2 / t_{1/2}$. Halving form and exponential form always agree: $(\tfrac{1}{2})^{t/t_{1/2}} = e^{-\lambda t}$.

Try it. A cobalt-60 source ($t_{1/2} = 5.27$ years) must fall to $5\%$. $n = \log(0.05)/\log(0.5) \approx 4.32$ half-lives, so $t \approx 4.32 \times 5.27 \approx 22.8$ years.

Why the clock is trustworthy

Uranium-238’s 4.5-billion-year half-life means half your atoms are thorium-234 (decaying on down the chain toward lead-206) after one block. Because heat, pressure, and compounds barely touch nuclear energies — except ~0.1% electron-capture nudges — that clock runs steady enough to date rocks and carbon-date artifacts. Randomness per atom, reliability per gram: that is the whole idea of half-life.

That finishes the nuclear-decay lesson. Four decays, the math to balance them, what happens inside, and how fast.