Suppose a laboratory has a 38 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half-lives of polonium-210 occur in 1104 days? How much polonium is in the sample 1104 days later?
1104/138 = 8 half-lives
after 8 half-lives, 1/256 remains, or 0.148g
How many half-lives of polonium-210 occur in 1104 days? 1104/138 = 8.0 half-lives
How much polonium is in the sample 1104 days later? 38 g * (1/2)^8 = 0.148 g
Thank you both very much.
8; 2,622 g
8; 0.11
8';0.148
To find out how many half-lives of polonium-210 occur in 1104 days, we need to divide the total elapsed time by the half-life.
So, we divide 1104 days by the half-life of polonium-210, which is 138 days:
1104 days / 138 days = 8
Therefore, there are 8 half-lives of polonium-210 that occur in 1104 days.
Now let's calculate the remaining amount of polonium-210 in the sample 1104 days later.
Each half-life reduces the amount of polonium-210 by half. After one half-life, the remaining amount would be half of the original sample (38 g / 2 = 19 g).
After two half-lives, the remaining amount would be half of the remaining amount after the first half-life (19 g / 2 = 9.5 g).
We can continue this pattern to find the remaining amount after 8 half-lives:
Remaining amount after 3 half-lives = 9.5 g / 2 = 4.75 g
Remaining amount after 4 half-lives = 4.75 g / 2 = 2.375 g
Remaining amount after 5 half-lives = 2.375 g / 2 = 1.1875 g
Remaining amount after 6 half-lives = 1.1875 g / 2 = 0.59375 g
Remaining amount after 7 half-lives = 0.59375 g / 2 = 0.296875 g
Remaining amount after 8 half-lives = 0.296875 g / 2 ≈ 0.14844 g
Therefore, the amount of polonium-210 in the sample 1104 days later is approximately 0.14844 g.