Question

1. cobalt - 60, which has a half - life of 5.3 years, is used in medical radiology. a sample of 60 mg of the material is present today.

a) write an equation to relate the amount of cobalt - 60 remaining and the number of half - life periods. 1t&i
b) what amount will be present in 10.6 years? 2t&i
c) approximately how many years will it take for cobalt - 60 to decay to 12.5% of its initial amount? 2t&i

Explanation:

Step1: Write the general formula for half - life

The general formula for the amount of a substance remaining after \(n\) half - life periods is \(A = A_0(\frac{1}{2})^n\), where \(A_0\) is the initial amount and \(A\) is the amount remaining. Given \(A_0=60\) mg, the equation is \(A = 60(\frac{1}{2})^n\).

Step2: Find the number of half - life periods for part (b)

The half - life \(T = 5.3\) years. For \(t = 10.6\) years, the number of half - life periods \(n=\frac{t}{T}=\frac{10.6}{5.3}=2\).
Substitute \(n = 2\) into the formula \(A = 60(\frac{1}{2})^n\). So \(A=60\times(\frac{1}{2})^2=60\times\frac{1}{4}=15\) mg.

Step3: Find the number of half - life periods for part (c)

We want \(A = 0.125A_0\). Substitute into \(A = A_0(\frac{1}{2})^n\), we get \(0.125A_0=A_0(\frac{1}{2})^n\). Since \(A_0
eq0\), we can divide both sides by \(A_0\) to get \(0.125 = (\frac{1}{2})^n\). Since \(0.125=\frac{1}{8}=(\frac{1}{2})^3\), so \(n = 3\).
The time \(t=nT\), with \(n = 3\) and \(T = 5.3\) years, \(t=3\times5.3 = 15.9\) years.

Answer:

a) \(A = 60(\frac{1}{2})^n\)
b) \(15\) mg
c) \(15.9\) years