Repeated Half-Life Calculations

Introduction

Half-life calculations often look like subtraction problems, but they are repeated multiplication by one-half. The amount lost in each interval changes because each interval halves what remains. A clear table of time, remaining parent fraction, parent mass and activity prevents a common error: taking the same absolute amount away at every step. It also makes it easier to work backward from a measured remainder.

Core explanation

For n complete half-lives, the expected remaining fraction of one parent nuclide is (1/2)ⁿ. If the initial parent count is N₀, then N = N₀(1/2)ⁿ. Under the same simple conditions, parent mass m = m₀(1/2)ⁿ and parent activity A = A₀(1/2)ⁿ. These formulas use the same fraction because mass is proportional to the number of remaining parent nuclei and, for one nuclide, activity is proportional to that number.

First turn elapsed time into a count of half-life intervals: n = t/t₁⁄₂. For example, if t = 18 years and t₁⁄₂ = 6 years, n = 3. Then the remaining fraction is (1/2)³ = 1/8. Starting with 64 mg of parent isotope gives 8 mg after 18 years. The amount that decayed is 64 − 8 = 56 mg of original parent, not “56 mg of total sample mass lost.” Daughter atoms may still be present and can have different masses or further decay.

A halving table can display the same arithmetic: 64 → 32 → 16 → 8 mg. Write the time above each step as 0, 6, 12 and 18 years. Each transition uses the currently remaining value. If you instead subtract 32 mg at all three steps, the second step reaches zero, contradicting the half-life rule.

Working backward reverses the operation. If 5 mg of parent remains after two half-lives, the initial amount was 5 × 2 × 2 = 20 mg. This is valid only if the given 5 mg refers to undecayed parent and the sample was not replenished or chemically separated in a way that changes the accounting. A problem might give a present activity instead; then double it once for each earlier half-life to recover the stated initial activity.

Fraction decayed equals 1 minus fraction remaining for a simple original-parent population. After three half-lives, 1/8 remains and 7/8 has transformed. It is easy to report 1/8 as “decayed” by accident, so label each result. For a simple closed system with a stable daughter and no daughter initially, the number of daughter nuclei produced equals the number of original parents that decayed. In an actual chain, a daughter can itself transform, so the final daughter population requires further information.

The repeated-halving method works exactly at whole-number multiples of a stated half-life in the ideal model. For 2.5 half-lives, a rule of “two halvings and then another half” is wrong: half of a half-life does not halve the remaining amount again. The exponential expression N = N₀(1/2)^(t/t₁⁄₂) handles noninteger n. A later page develops that connection; for now, use whole-number intervals when the exercise is designed for repeated halving.

Detector data needs care. A background-subtracted count rate may track parent activity if detection conditions remain the same. Raw gross counts include background and therefore do not simply halve toward zero. If a source-related detector rate begins at 160 counts/s and the background is 10 counts/s, the total gross rate after two half-lives would be about 40 + 10 = 50 counts/s, not 160/4 = 40 counts/s, assuming efficiency and geometry stay fixed.

Keep time units consistent. Convert months to years or hours to minutes before dividing t by the half-life. The numerical calculation may be easy, but an unnoticed unit mismatch can make the result physically absurd. Also, do not mix half-lives of two different isotopes in one repeated-halving table.

Step-by-step reasoning

1. Identify the parent isotope, its half-life and the elapsed time with matching units. 2. Calculate the number of half-lives n = elapsed time ÷ half-life. 3. Halve the remaining parent count, mass or activity n times, or use (1/2)ⁿ. 4. Label the remaining result; subtract from the initial amount only if the decayed amount is asked. 5. For detector counts, subtract or restore background appropriately rather than halving gross counts blindly.

Visual explanation

Make a four-column table with headings time, half-lives elapsed, parent fraction and activity. Fill rows 0, T, 2T and 3T with 1, 1/2, 1/4 and 1/8, alongside 800, 400, 200 and 100 Bq. Draw arrows down the last two columns labelled “divide the previous row by 2,” not “subtract a fixed amount.”

Real-world analogy

A shop that sells half of its remaining stock each day has 100, then 50, then 25 items in an idealised example. It does not sell 50 each day. That matches repeated halving, although radioactive decay is random and cannot be controlled like sales.

Real-world example

Suppose an archival sample contains a radioactive isotope with a known half-life and a measured parent amount. Repeated halving provides a first-pass estimate of how much parent existed at an earlier time, provided the starting condition and closed-system assumptions are justified. Radiometric dating needs those assumptions as well as the arithmetic.

Why?

Why does the absolute number decaying per half-life decrease? Each surviving nucleus continues to have the same decay probability over a given interval, but fewer parent nuclei remain to undergo decay. A constant fraction of a shrinking population is a shrinking absolute count.

Common misconception

“Three half-lives leave half of the initial amount minus three equal losses.” Each interval halves what remains: after three intervals the share is 1/8, not a linear subtraction pattern. Write a table before calculating if the language is confusing.

Worked example

A source contains 96 mg of a parent isotope and has initial activity 480 Bq. Its half-life is 4 days. After 12 days, n = 12/4 = 3. The parent mass sequence is 96 → 48 → 24 → 12 mg, and activity is 480 → 240 → 120 → 60 Bq. The remaining fraction is 1/8. The original parent mass that has transformed is 96 − 12 = 84 mg, equivalent to 7/8 of the original. These calculations do not say the source's total material mass has fallen by 84 mg.

Quick check

1. A parent sample has 80 mg initially and a 5-hour half-life. What remains after 15 hours? Answer: Three half-lives pass, so 80 → 40 → 20 → 10 mg of parent remains.

Exam focus

Show n = t/t₁⁄₂ with consistent units and label “remaining” versus “decayed.” Apply repeated halving to parent count, parent mass or pure-parent activity. If a count rate includes background, separate that constant contribution before applying the fraction.

Advanced insight

At whole-number n, the repeated-halving table is a special case of the exponential law. The same law can be inverted: t = t₁⁄₂ log₂(N₀/N). That expression is useful when the remaining fraction is not a simple power of one-half, although a logarithm is unnecessary for common exam examples such as one-quarter or one-eighth remaining.

Summary

After n complete half-lives, a simple parent sample retains the expected fraction (1/2)ⁿ of its original nuclei, parent mass and activity. Each interval halves the remainder. Work backward by doubling, and treat detector background, daughter products and time units explicitly.

Practice questions

1. A parent isotope has a 3-year half-life. What fraction remains after 9 years? Answer: Three half-lives have passed, so (1/2)³ = 1/8 remains. 2. A present sample has 6 mg of parent after two half-lives. What was its original parent mass? Answer: 6 × 2² = 24 mg, assuming no parent was added or removed. 3. A source-related detector rate is 120 counts/s and background is 8 counts/s. What gross rate is expected after one half-life under unchanged conditions? Answer: The source-related rate becomes 60 counts/s, so gross rate is about 60 + 8 = 68 counts/s.