Half-Life — Cobalt-60 Source (Worked Example)

Medium

Problem

A radiotherapy unit holds 8.0 g of cobalt-60 (half-life 5.27 years). How much Co-60 remains after 15.81 years?

Given

N₀ = 8.0 g
t½ = 5.27 years
t = 15.81 years

Find

The remaining mass N.

Principle

Three half-lives elapse; each multiplies the remainder by ½.

Formula

N = N_0\left(\frac{1}{2}\right)^{t/t_{1/2}}

N = N₀(1/2)^(t/t½)

Solution

Half-lives: $n = 15.81/5.27 = 3$
Apply: $N = 8.0 \times (1/2)^3$
Calculate: $N = 8.0/8 = 1.0\ \mathrm{g}$

1.0 g

Unit check

Grams × (dimensionless fraction) = grams. The result is a mass — dimensionally correct.

What this means physically

1.0 g — the source is now 1/8 as active, so treatment times must be recalculated or the source replaced. Clinics track this decay precisely for dosing.

Common mistake

Assuming linear decay to zero. After 3 half-lives 1/8 remains — never zero; decay is asymptotic.