Question

Sodium-24 has a half-life of 15 hours. If a sample of sodium-24 has an original activity of 500 Bq, what will its activity be after 60 hours?

Answer 1

We know that after 15 hours the half of initial number of atoms will decay. Therefore, we can write (see https://en.wikipedia.org/wiki/Half-life#Formulas_for_half-life_in_exponential_decay)

$N(T) = N(0)\cdot \left( \dfrac12\right)^{\frac{T}{T_{1/2}}},$ where $T_{1/2}$ is the half-life.

Activity is proportional to the number of atoms not decayed yet. So, we should determine fraction $\dfrac{N(T)}{N(0)} = \dfrac{N(60)}{N(0)} = \left( \dfrac12\right)^{\frac{60}{15}} = \left( \dfrac12\right)^4 = \dfrac{1}{16}.$

Therefore, the activity after 60 hours will be $\dfrac{1}{16}$ of the original activity or $500\cdot\dfrac{1}{16} = 31.25$ Bq.