Two radioactive elements A and B have half-lives of 100 and 50 years respectively. Samples of A and B initially contain equal amounts/number of atoms. What is the ratio of the number of the remaining atoms of A to that of B after 200 years?
-
4 : 1
- 2 : 1
- 1 : 1
- 1 : 2
- 1 : 4
The correct answer is: A
Explanation
Given that a mass N of elements A and B exists, then:
For element A, after 100 years, we have A of mass N/2 remaining
After 200 years, we have A of mass \(\frac{1}{2}\times\frac{N}{2}\) = \(\frac{N}{4}\)
For element B, after 50 years, we have B of mass N/2 remaining
After 100 years, we have B of mass \(\frac{1}{2}\times\frac{N}{2}\) = \(\frac{N}{4}\)
After 150 years, we have B of mass \(\frac{1}{2}\times\frac{N}{4}\) = \(\frac{N}{8}\)
After 200 years,we have B of mass \(\frac{1}{2}\times\frac{N}{8}\) = \(\frac{N}{16}\)
Ratio: \((\frac{N}{4} : \frac{N}{16})\)
=\(4 : 16\) = \(1 : 4\)