QB_PH_SEC9_014: Nuclear Models & Decays

A radioactive sample has an activity of $A_0 = 800\text{ Bq}$ at $t = 0$. If its decay constant is $\lambda = 0.0693\text{ min}^{-1}$, what is its activity (in Bq) after $t = 30\text{ minutes}$? (Round off to the nearest integer)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Radioactive Decay Law**: $A(t) = A_0 e^{-\lambda t}$ 2. **Half-Life Calculation**: $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.69315}{0.069315\text{ min}^{-1}} = 10.0\text{ min}$ 3. **Number of Half-Lives**: $n = \frac{t}{T_{1/2}} = \frac{30\text{ min}}{10\text{ min}} = 3$ 4. **Activity Evaluation**: $A(30) = A_0 \left(\frac{1}{2}\right)^n = 800\text{ Bq} \times \left(\frac{1}{2}\right)^3 = \frac{800}{8} = 100\text{ Bq}$ 5. **Official Exact Integer**: 100.