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?

The answer is a) 7 half-lives and b) 0.78% (or as fraction 1/128) of the starting amount.

a) If half-life of Polonium-210 is 138 days, to calculate how many half-lives occur in 966 days, we will simply divide them: 966/138 = 7
So, 7 half-lives occur in 966 days.

b) To calculate the remaining amount, we will use the formula:

where n is the  number of half-lives, and x is the remaining amount in decimals, and (1/2) is half-life.

We've already found that n = 7, so replace it in the formula:
0.78% = 1/128

The correct answer is d

So3 that’s what i think