81.3k views
0 votes
Chlorine-38 has a half-life of 37.2 minutes. If a sample that has been in storage for 111.6 minutes contains 2.03 x 10¹⁷ atoms of chlorine-38, how many atoms of the radioisotope was present when it was put into storage?

User Solr
by
7.5k points

1 Answer

4 votes

Final answer:

To determine the initial amount of chlorine-38 in a sample after three half-lives, multiply the remaining amount by two, three times. The sample initially had 8.12 x 10¹⁷ atoms of chlorine-38.

Step-by-step explanation:

The question asks about calculating the original number of chlorine-38 atoms in a sample, based on its half-life and the remaining quantity after a certain period of time. To find out how many atoms were present when the sample was put into storage, we use the concept of half-lives. The half-life of chlorine-38 is given as 37.2 minutes, and the sample has been stored for 111.6 minutes, which is exactly three half-lives (111.6 / 37.2 = 3).

After one half-life, the amount of a radioactive substance is halved. Therefore, after three half-lives, the remaining amount is halved three times. To find the initial amount, we can reverse this process:

  • 1st half-life: 2.03 x 10¹⁷ atoms → initial amount (x)
  • 2nd half-life: 2 x (2.03 x 10¹⁷) = 4.06 x 10¹⁷ atoms
  • 3rd half-life: 2 x (4.06 x 10¹⁷) = 8.12 x 10¹⁷ atoms

Therefore, the sample contained 8.12 x 10¹⁷ atoms of chlorine-38 when it was initially put into storage.

User Alfredox
by
8.2k points