Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half-lives of polonium-210 occur in 966 days? How much polonium is in the sample 966 days later?
To find the number of half-lives that occur in 966 days, we can divide the total time elapsed by the half-life of polonium-210.
1. Convert the half-life of polonium-210 to the same unit as the total time elapsed:
- 138 days
2. Divide the total time elapsed by the half-life:
- 966 days ÷ 138 days ≈ 7
Therefore, there are approximately 7 half-lives of polonium-210 in 966 days.
Now, to determine the amount of polonium in the sample 966 days later, we can use the concept of half-life.
1. Calculate the fraction of polonium-210 remaining after each half-life:
- After the first half-life, half of the original sample remains (50%).
- After the second half-life, half of the remaining sample from the first half-life remains (25%).
- After the third half-life, half of the remaining sample from the second half-life remains (12.5%).
- And so on...
2. Calculate the amount of polonium remaining after each half-life:
- Multiply the original amount of polonium by the fraction remaining after each half-life.
First half-life: 31 g × 0.50 = 15.5 g remaining
Second half-life: 15.5 g × 0.50 = 7.75 g remaining
Third half-life: 7.75 g × 0.50 = 3.875 g remaining
Continue this process for the number of half-lives calculated earlier (7 in this case).
Fourth half-life: 3.875 g × 0.50 = 1.9375 g remaining
Fifth half-life: 1.9375 g × 0.50 = 0.96875 g remaining
Repeat until you reach the desired number of half-lives (7 in this case).
Sixth half-life: 0.96875 g × 0.50 = 0.484375 g remaining
Seventh half-life: 0.484375 g × 0.50 = 0.2421875 g remaining
After 7 half-lives, there will be approximately 0.2421875 g of polonium-210 remaining in the sample, 966 days later.