Question

refer to the radioisotope cobalt - 60, which has a half - life of 5.3 years. an initial sample of cobalt - 60 has an activity of 12,400 cpm. how much time has passed if the activity of the sample has dropped to 775 cpm? a 789 years b 21.2 years c 15.9 years d 84.8 years

Explanation:

Step1: Calculate the number of half - lives

The formula for radioactive decay is \(N = N_0(\frac{1}{2})^n\), where \(N\) is the final activity, \(N_0\) is the initial activity, and \(n\) is the number of half - lives.
We know that \(N_0=12400\) cpm and \(N = 775\) cpm.
So, \(\frac{N}{N_0}=(\frac{1}{2})^n\), \(\frac{775}{12400}=(\frac{1}{2})^n\), \(\frac{1}{16}=(\frac{1}{2})^n\), and \(n = 4\).

Step2: Calculate the time passed

The half - life \(T_{1/2}=5.3\) years.
The time passed \(t=n\times T_{1/2}\).
Since \(n = 4\) and \(T_{1/2}=5.3\) years, \(t=4\times5.3\) years \(=21.2\) years.

Answer:

B. 21.2 years