Question

A compound containing 3 mCi of Sulfur-35 (half-life = 87 days)is stored in a laboratory. How much radioactivity will there be after 348 days?

1.

0.1875 mCi

2.

0.75 mCi

3.

0.375 mCi

4.

0.1875 mCi

5.

0.094 mCi

Answer 1

answer : 0.1875 mCi

half-life t_1/2 = 87 days

time = 348 days

t = n t_1/2

348 = n x 87

n = 4

remaining radioactive compound = initial compound / 2ⁿ

                                                                             = 3 / 2^4

                                                     = 0.1875 mCi