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

Published by Patrick Mutisya · 14 days ago

IGCSE Physics 0625 – Half‑life

5.2.4 Half‑life

Objective

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 will not include background radiation).

Definition

The half‑life, denoted \$t{1/2}\$, of a radioactive isotope is the time required for half of the original number of undecayed nuclei, \$N0\$, to disappear. After one half‑life the remaining number of nuclei \$N\$ is

\$N = \frac{N_0}{2}.\$

After \$n\$ half‑lives the remaining fraction is \$\left(\tfrac12\right)^n\$, so

\$N = N_0\left(\frac12\right)^n.\$

Key Relationships

  1. Number of half‑lives elapsed: \$n = \dfrac{t}{t_{1/2}}\$, where \$t\$ is the elapsed time.

  2. Remaining nuclei after time \$t\$: \$N = N0\left(\frac12\right)^{t/t{1/2}}\$.

  3. Activity (decays per second) is proportional to the number of nuclei: \$A = \lambda N\$, where \$\lambda = \dfrac{\ln 2}{t_{1/2}}\$.

Typical Half‑life \cdot alues

IsotopeHalf‑lifeCommon Use in IGCSE
\$^{14}\$C5,730 yearsCarbon dating (historical contexts)
\$^{35}\$S87.5 daysTracer experiments
\$^{60}\$Co5.27 yearsRadiation source for experiments
\$^{226}\$Ra1,600 yearsBackground radiation discussion

Using Decay Curves

A decay curve plots the activity (or number of nuclei) against time. The curve is exponential and passes through the point where the activity has fallen to half its initial value after one half‑life.

Suggested diagram: Sketch of a typical decay curve showing \$t_{1/2}\$ where the activity has fallen to 50 % of its initial value.

Worked Examples

Example 1 – Direct use of the definition

A sample contains \$8.0\times10^{6}\$ nuclei of \$^{35}\$S. After 87.5 days (one half‑life) how many nuclei remain?

  1. Identify \$t{1/2}=87.5\text{ d}\$ and \$N0=8.0\times10^{6}\$.

  2. After one half‑life \$N = N_0/2 = 4.0\times10^{6}\$ nuclei.

Example 2 – Multiple half‑lives

A \$^{14}\$C sample has an initial activity of \$A_0 = 200\,\$Bq. What is the activity after 17,190 years?

  1. Determine the number of half‑lives: \$n = \dfrac{t}{t_{1/2}} = \dfrac{17\,190\text{ yr}}{5\,730\text{ yr}} = 3\$.

  2. Activity after \$n\$ half‑lives: \$A = A_0\left(\frac12\right)^n = 200\left(\frac12\right)^3 = 200\times\frac18 = 25\,\$Bq.

Example 3 – Solving for time

A \$^{60}\$Co source initially has an activity of \$A_0 = 1000\,\$Bq. After how long will the activity be \$125\,\$Bq?

  1. Set up the decay relation: \$A = A0\left(\frac12\right)^{t/t{1/2}}\$.

  2. Insert values: \$125 = 1000\left(\frac12\right)^{t/5.27\text{ yr}}\$.

  3. Divide both sides by 1000: \$0.125 = \left(\frac12\right)^{t/5.27}\$.

  4. Recognise \$0.125 = \left(\frac12\right)^3\$, so \$t/5.27 = 3\$.

  5. Therefore \$t = 3\times5.27\text{ yr} = 15.81\text{ yr}\$.

Common Pitfalls

  • Confusing half‑life with the time for a fixed number of decays; the definition always refers to “half the nuclei”.

  • For calculations, always use the same units for time (e.g., days, years) as the given half‑life.

  • Remember that background radiation is ignored in IGCSE calculations; use the given initial activity directly.

Quick Revision Checklist

  • Can you state the definition of half‑life in words?

  • Do you know how to write \$N = N0(1/2)^{t/t{1/2}}\$?

  • Can you read a half‑life from a table and apply it to a problem?

  • Are you comfortable converting “number of half‑lives” into elapsed time and vice‑versa?