Define the half-life of a particular isotope as the time taken for half the nuclei of that isotope in any sample to decay; recall and use this definition in simple calculations, which might involve information in tables or decay curves (calculations

5.2.4 Half‑life

Learning objective

Define the half‑life of a radioactive isotope and use the definition in simple calculations involving tables of data or decay‑curve information. (In all Cambridge IGCSE/AS‑Level questions the activity given is already corrected for background radiation.)

Definition

The half‑life (t_½) of a particular isotope is the constant time required for the number of undecayed nuclei (or the activity or the mass) to be reduced to one‑half of its original amount.

Mathematically, if N₀ is the initial number of nuclei (or the initial activity/mass) and N is the amount after time t, then for one half‑life

Key points to remember

• The half‑life is a constant property of the isotope; it does not depend on the amount of material you start with.
• After n half‑lives the fraction remaining is \(\left(\frac12\right)^{n}\).
• Half‑lives can range from fractions of a second to billions of years.
• All exam calculations ignore background radiation – the activity given is already the net activity of the sample.
• Keep the units of time consistent (e.g. days with days, seconds with seconds). Convert first if necessary.
• Round the final answer to the same number of significant figures as the least‑precise datum in the question.

Using a known half‑life in calculations

1. Identify the half‑life t_½ from the question (table, statement or decay curve).
2. Determine how many half‑lives have elapsed:

   • Count the number of times the sample could be halved, or
   • Use the formula \(\displaystyle n = \frac{t}{t_{½}}\) where t is the elapsed time (same units as t_½).

3. Calculate the remaining fraction:

   \[

   \text{remaining fraction}= \left(\frac12\right)^{n}

   \]

4. Find the required quantity (number of nuclei, mass or activity):

   \[

   N = N_{0}\left(\frac12\right)^{n},\qquad

   m = m_{0}\left(\frac12\right)^{n},\qquad

   A = A_{0}\left(\frac12\right)^{n}

   \]

5. Round the answer to the correct number of significant figures.

Exam‑style formulae

Deriving the half‑life from data

Worked example – perfect halving

Problem: The activity of a sample is recorded as follows.

1. From 0 min to 10 min the activity falls from 800 Bq to 400 Bq – a halving.
2. Therefore the elapsed time (10 min) is the half‑life: t_½ = 10 min.
3. Check with another pair: 800 Bq → 200 Bq is a quarter, i.e. two half‑lives.

   \(\;2t{½}=20\;{\rm min}\;\Rightarrow\;t{½}=10\;{\rm min}\) – consistent.

Worked example – non‑ideal data (requires logarithms)

Problem: An unknown isotope gives the following activities.

Because the activity does not halve exactly, use the logarithmic formula.

1. Choose two points, e.g. 0 s (1 200 Bq) and 30 s (530 Bq).
2. Apply \(\displaystyle t{½}= t\,\frac{\log{10}2}{\log{10}(A{0}/A)}\):

   \[

   t{½}=30\;\text{s}\times\frac{\log{10}2}{\log_{10}(1200/530)}

   \approx 30\times\frac{0.3010}{0.3541}=25.5\;\text{s}

   \]

3. Check with the second interval (30 s → 60 s):

   \[

   t{½}=30\;\text{s}\times\frac{0.3010}{\log{10}(530/235)}\approx 25.4\;\text{s}

   \]

   The two results agree to within experimental error; the half‑life is therefore \(\boxed{25.5\;\text{s}}\) (2 s sf).

What if background radiation isn’t ignored?

Real‑world applications (why the half‑life matters)

• ≈ 10 days – ¹³¹I: medical tracer for thyroid scans (gives a dose but disappears quickly).
• ≈ 5 730 years – ¹⁴C: carbon‑dating of archaeological samples.
• ≈ 1 hour – ²⁴¹Am in smoke alarms: long enough for a steady ionising source, short enough for safe disposal.
• ≈ 4.5 × 10⁹ years – ²³⁸U: dating of rocks and the age of the Earth.

Interpreting decay curves

A decay curve plots the remaining quantity (nuclei, mass or activity) against time. The curve is exponential and passes through the point where the quantity has fallen to 50 % of its initial value at t = t_½.

• Locate the 50 % point on the vertical axis → read off the corresponding time → that is the half‑life.
• Successive points at 25 %, 12.5 %, … correspond to 2, 3, … half‑lives.

Common pitfalls

• Thinking the half‑life is the time for a *fixed* number of nuclei to disappear – it is always half of whatever remains at that moment.
• Mixing units (days for the half‑life but seconds for the elapsed time). Convert first.
• For non‑integer numbers of half‑lives, remember the fractional exponent:

  \[

  N = N{0}\left(\frac12\right)^{t/t{½}}

  \]

• Neglecting the “ignore background radiation” instruction – the activity given in the question is already net activity.

Practice worksheet (step‑by‑step)

Complete the following questions using the exam‑style formulae. Show every step and keep track of units.

1. Mass after a known time – A 5.0 g sample of an isotope has a half‑life of 10 days. What mass remains after 30 days?
   

   Solution outline:

   • Number of half‑lives: \(n = 30/10 = 3\).
   • Remaining fraction: \((1/2)^{3}=1/8\).
   • Mass remaining: \(5.0\;\text{g}\times\frac18 = 0.63\;\text{g}\) (2 sf).

2. Elapsed time from activity – The activity of a ¹⁴C sample is \(2.0\times10^{5}\) Bq. After how many years will the activity be \(2.5\times10^{4}\) Bq? (Half‑life of ¹⁴C = 5 730 y.)
   

   Solution outline:

   • Use \(t = t{½}\dfrac{\log{10}(A{0}/A)}{\log{10}2}\).
   • \(t = 5\,730\;\text{y}\times\dfrac{\log_{10}(2.0\times10^{5}/2.5\times10^{4})}{0.3010}

     =5\,730\;\text{y}\times\dfrac{0.9031}{0.3010}\approx 17\,200\;\text{y}\) (3 sf).

3. Half‑life from a decay curve – From a decay curve the activity drops to 25 % of its initial value after 40 min. What is the half‑life?
   

   Solution outline:

   • 25 % = \((1/2)^{2}\) → two half‑lives have elapsed.
   • \(t_{½}=40\;\text{min}/2 = 20\;\text{min}\).

4. Half‑life from non‑ideal data – Use the data below to calculate the half‑life.

   Time (s)	Activity (Bq)
   0	1 200
   30	600
   60	300

   
   Solution: each 30 s interval halves the activity, so \(t_{½}=30\;\text{s}\). (If the numbers were not exact, the logarithmic formula would be used.)

Summary

The half‑life is a fundamental, constant property of a radioactive isotope. By recognising that each half‑life reduces the quantity by a factor of ½, students can:

• read a half‑life directly from a table or decay curve,
• calculate the remaining amount after any time using \(\displaystyle Q = Q{0}\left(\frac12\right)^{t/t{½}}\),
• determine the elapsed time or the half‑life itself with the logarithmic (exam) formulae,
• avoid common mistakes – keep units consistent, ignore background radiation (exam data are already corrected), and apply correct significant figures.

Mastering these steps enables quick, accurate answers to all Cambridge IGCSE/AS‑Level half‑life questions.