Half-Life Arithmetic & Decay

You’ll be able to

Compute the remaining quantity of a radioisotope after a given time
Determine an elapsed time or a half-life from decay data
Explain why radioactive decay is a random process with a statistical law

Constant fraction, not constant amount

Radioactive decay removes a constant fraction per unit time, not a constant amount. After one half-life half the original nuclei remain; after two, a quarter; after three, an eighth. This is why the quantity never reaches exactly zero and why the graph is an exponential curve rather than a straight line. The half-life is a fixed property of the isotope, unaffected by temperature, pressure or chemical state.

The decay law

N = N₀ (½)^(t/T) · fraction remaining = (½)^(number of half-lives) · number of half-lives = t / T

For whole numbers of half-lives, halving repeatedly is faster and less error-prone than using the exponent.

Working backward

Given a fraction remaining, count how many halvings reach it: 1/2 is one, 1/4 is two, 1/8 is three, 1/16 is four. Then elapsed time = that count × half-life. If the fraction is not a power of a half, the exponential form is needed. This backward direction is how radiometric dating works: measure the remaining fraction of a known isotope and read off the elapsed time.

Decay is random

No individual nucleus can be predicted — it has a fixed probability of decaying per unit time and no memory of how long it has already existed. The exponential law is statistical, emerging from enormous numbers of independent events. Two consequences worth stating: a sample of only a few atoms shows large fluctuations from the smooth curve, and there is no sense in which an old nucleus is "due" to decay. This is genuinely different from a mechanical timer, and questions sometimes probe whether students understand that.

Activity falls with the same half-life

Activity is decays per second, measured in becquerels. It is proportional to the number of undecayed nuclei present, so it follows the same exponential law with the same half-life. So a question can give you activities instead of masses and the arithmetic is unchanged — a sample whose activity has fallen to one eighth has been through three half-lives, exactly as if you had counted nuclei.

Worked example

A sample of an isotope with a half-life of 8.0 days has an initial activity of 640 Bq. Find the activity after 24 days, and the time for the activity to fall to 40 Bq.

1.24 days ÷ 8.0 days = 3 half-lives.
2.Halving three times: 640 → 320 → 160 → 80 Bq.
3.For the second part: 640 → 320 → 160 → 80 → 40 requires four halvings.
4.Four half-lives × 8.0 days = 32 days.

Answer: Activity is 80 Bq after 24 days, and it takes 32 days to fall to 40 Bq. Counting halvings is faster and safer than exponentials whenever the numbers work out to whole half-lives, which on the exam they usually do.

Watch out

Three half-lives leaves one eighth, not one third and not zero. And two half-lives do not remove all the material — the fraction remaining is (½)^n, which never reaches zero for any finite n.

Checkpoint

After 4 half-lives, the fraction of the original nuclei remaining is:

Checkpoint

A sample's activity falls from 800 Bq to 100 Bq in 18 hours. Its half-life is:

Checkpoint

The half-life of a radioactive isotope is unaffected by:

On the exam

Count halvings rather than reaching for the exponential whenever the elapsed time is a whole multiple of the half-life. It is faster, self-checking, and exam numbers are almost always chosen to make it work.

Answer the 3 checkpoints as you read.