Half-life, activity and background radiation (Physics HL)

The big idea: The half-life is the time it takes for half the radioactive nuclei to decay. In that same time the activity and the count rate also halve.

Listen to a Geiger counter: Hold a Geiger counter near a fresh radioactive sample and it clicks like mad. Come back tomorrow and the clicks have thinned out. Decay is random — you can't say when one nucleus will go — but on average a sample gets weaker with time.

Random and spontaneous: Random: you cannot predict which nucleus will decay next, or when. Spontaneous: nothing outside the nucleus (temperature, pressure, chemical bonds) changes how likely it is to decay. With huge numbers of nuclei, though, the half-life is perfectly predictable.

New words, plainly: Activity = how many nuclei decay each second, measured in becquerel (Bq), where 1 Bq = 1 decay per second.

Count rate = how many of those decays a detector actually records each second (clicks per second).

Half-life = the time for the activity (or count rate) to fall to half.

Halving keeps going: Every half-life that passes, you halve again — not subtract a fixed amount.

So after 1, 2, 3 half-lives the count rate is 1/2, 1/4, 1/8 of the start. It flattens but never reaches zero.

At SL you only deal with whole numbers of half-lives, so you never need the exponential equation. You just halve, halve, halve, once for each half-life that has passed.

The rule (learn this — it is NOT in the data booklet): After n whole half-lives, the count rate (or activity) is the start value multiplied by (1/2)ⁿ.

Work out n = total time ÷ half-life, then halve that many times.

Subtract the background FIRST: A detector always records some background radiation (from rocks, cosmic rays, etc.) even with no source.

The true count rate from the source = measured count rate − background count rate. Correct for background before you start halving.

Finding a half-life from a graph

Background unknown? Use two readings: If the reading is R₁ now and R₂ after n half-lives, with an unknown background b: (R₁ − b) × (1/2)ⁿ + b = R₂. Solve for b. Example: 72 s⁻¹ now and 40 s⁻¹ one half-life later: (72 − b)/2 + b = 40, so b = 8 s⁻¹.

From activity to a current: In an ionisation chamber each alpha makes (alpha energy ÷ energy per ion pair) ion pairs. Ion pairs per second = alphas per second into the chamber × ion pairs per alpha; multiply by e for the current. Example: 1.0 × 10⁴ alphas s⁻¹, each 5.0 MeV, 25 eV per pair: 2.0 × 10⁵ pairs each, current = 2.0 × 10⁹ × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹⁰ A.

How this is tested — a classic Paper 1A one-step calculation that also turns up in Paper 2:

The classic trap: Forgetting the background, or subtracting a fixed amount each half-life instead of halving.

Ratios stay put: If two sources have the same half-life, after the same time both have halved the same number of times.

The ratio of their activities is unchanged — e.g. 5 : 1 stays 5 : 1.