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?

So we want to know how many half-lives of polonium does occur in 1104 days if it has half-life of 138 days and how much of polonium will remain in the sample after 1104 days if the initial mass of the sample was 30 grams. First lets answer how many half lives do occur in 1104 days: 1104/138=8. So the formula for half-life is N=N0 * (1/2) ^-t/T where N0 is the initial quantity of the sample, t is the overall time and T is one half-life. Now we input the numbers in the equation and get: N=30g*2^-8=0.117 grams

One Half-Life of polonium = 138 days

We have to find the number of half-lives that occur in 1104 days. For this we simply divide 1104 by 138.

Number of half-lives in 1104 days = 1104/138 = 8 half-lives

Thus, in 1104 days polonium-210 undergoes 8 half-lives i. e. it is halved 8 times.

Initial Amount of sample = 30 gram

The amount remaining after 8 half-lives will be = 30 x (0.5) ⁸ = 0.1171875 grams.

This means, after 1104 days, 30 gram sample of polonium-210 will be reduced to 0.1171875 grams only.