Question

Suppose a laboratory has a 30 g sample of polonium-210. The half-life of polonium-210 is about 138 days. a. How many half-lives of polonium-210 occur in 1104 days? b. How much polonium is in the sample 1104 days later?

9; 0.06 g
8; 0.12 g
8; 2,070 g
8; 3.75 g

Answer 1

Explanation:

the decay or "half-life" of radioactive particles is really a big joke by nature, in my opinion.

it takes that much time for half of the amount to decay, so, naturally, one wound assume that after a second interval then the other half is gone.

but far from it. no, only the half of the originally remaining half is then gone (the half of a half is 1/2 × 1/2 = 1/4). and after another interval the half of the half of the half is then gone. and so on.

so, this creates a geometric sequence with the common factor (or ratio) if 1/2.

that means every term is created by multiplying the previous term by 1/2.

a.

how many "half-lives" in 1104 days ?

well, with every 138 days such a half-life is over.

so, the answer is how often 138 fits into 1104 :

1104 / 138 = 8

there are 8 "half-lives" of polonium-210 in 1104 days.

b.

how much polonium-210 is in the sample after 1104 days ?

since we have 8 half-lives, we need to multiply the original amount 8 times by 1/2. in other words by 1/2⁸.

2⁸ = 256

so, our calculation is

30 g × 1/256 = 0.1171875 g ≈ 0.12 g

so, the second answer option is correct.