Radioactive Activity — Technetium-99m Tracer (Worked Example)

Hard

Problem

A medical dose contains 1.0 × 10¹² atoms of technetium-99m (half-life 6.0 hours) for a scan. What is its activity in becquerels?

Given

N = 1.0 × 10¹² atoms
t½ = 6.0 h = 21,600 s

Find

The activity A.

Principle

A = λN: short half-life means a large λ and high activity per atom — why Tc-99m is ideal for imaging.

Formula

A = \lambda N,\quad \lambda = \frac{\ln 2}{t_{1/2}}

A = λN, λ = ln2/t½

Solution

Convert: $t_{1/2} = 6 \times 3600 = 21{,}600\ \mathrm{s}$
$\lambda = 0.693/21600 = 3.21 \times 10^{-5}\ \mathrm{s^{-1}}$
$A = (3.21 \times 10^{-5})(1.0 \times 10^{12}) = 3.21 \times 10^7\ \mathrm{Bq}$

3.21 × 10⁷ Bq (32.1 MBq)

Unit check

s⁻¹ × atom count = s⁻¹ = Bq. Decays per second — the becquerel, correct.

What this means physically

32.1 MBq — a typical diagnostic dose. It decays fast enough to limit patient exposure yet emits enough gamma rays for a clear image: the 6-hour half-life is a medical sweet spot.

Common mistake

Reporting in Bq but forgetting how big mega is. 3.21 × 10⁷ Bq = 32.1 MBq — use prefixed units for readability.