Nuclear Physics - Half-life

A radioactive sample has an initial activity of $1200 \text{ Bq}$. If the half-life of the isotope is $5 \text{ days}$, calculate the activity remaining after $15 \text{ days}$.

1. Calculate the number of half-lives ($n$): $n = \frac{\text{Total Time}}{\text{Half-life}} = \frac{15 \text{ days}}{5 \text{ days}} = 3$
2. Apply the decay formula: $A = 1200 \times \left( \frac{1}{2} \right)^3$
3. Solve: $A = 1200 \times \frac{1}{8} = 150 \text{ Bq}$

After the first 5 days, activity drops to $600 \text{ Bq}$. After 10 days, it drops to $300 \text{ Bq}$. After 15 days (the third half-life), it reaches $150 \text{ Bq}$.

The count rate from a source is measured as $240 \text{ counts/minute}$. If the background radiation is $40 \text{ counts/minute}$, and the count rate drops to a corrected value of $25 \text{ counts/minute}$ in $12 \text{ hours}$, find the half-life.

1. Calculate the initial corrected activity ($A_0$): $A_0 = 240 - 40 = 200 \text{ counts/minute}$
2. Determine how many halvings occurred to reach $25 \text{ counts/minute}$: $200 \xrightarrow{1} 100 \xrightarrow{2} 50 \xrightarrow{3} 25$ This is $n = 3$ half-lives.
3. Calculate the half-life ($t_{1/2}$): $t_{1/2} = \frac{\text{Total Time}}{n} = \frac{12 \text{ hours}}{3} = 4 \text{ hours}$

First, the background radiation is removed to isolate the source's activity. By halving the initial corrected activity repeatedly, we find that 3 half-lives have passed in the 12-hour period.