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Suppose a laboratory has a 30 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?

A. 9; 0.06 g

B. 8; 2,070 g

C. 8; 0.12 g

D. 8; 3.75 g

User Jairaj
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1 Answer

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Answer: the amount of polonium remaining in the sample 1104 days later is approximately 0.12 g, and the answer is C

Explanation:

To determine how many half-lives of polonium-210 occur in 1104 days, we can divide 1104 by the half-life of 138 days:

1104 / 138 ≈ 8

Therefore, there are approximately 8 half-lives of polonium-210 in 1104 days.

To determine how much polonium is in the sample 1104 days later, we can use the fact that the amount of polonium remaining after n half-lives is given by:

Amount remaining = initial amount × (1/2)^n

where n is the number of half-lives that have occurred.

In this case, we know the initial amount is 30 g and the number of half-lives is 8. So we can calculate the amount of polonium remaining as:

Amount remaining = 30 g × (1/2)^8 ≈ 0.117 g ≈ 0.12 g (rounded to two decimal places)

Therefore, the amount of polonium remaining in the sample 1104 days later is approximately 0.12 g, and the answer is C.

User AndreiC
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