6. cobalt - 60, which has a half - life of 5.3 years, is used in medical radiology. a sample of 60 mg of the…

Explanation:

Step1: Determine decay - formula

The general formula for radioactive decay in terms of half - life is $A = A_0\left(\frac{1}{2}\right)^n$, where $A$ is the amount remaining, $A_0$ is the initial amount, and $n$ is the number of half - life periods.

Step2: Solve part (a)

The initial amount $A_0 = 60$ mg. The equation relating the amount of cobalt - 60 remaining $A$ and the number of half - life periods $n$ is $A=60\left(\frac{1}{2}\right)^n$.

Step3: Solve part (b)

The half - life of cobalt - 60 is $t_{1/2}=5.3$ years. In $t = 10.6$ years, the number of half - life periods $n=\frac{10.6}{5.3}=2$. Substitute $A_0 = 60$ mg and $n = 2$ into the decay formula: $A=60\times\left(\frac{1}{2}\right)^2=60\times\frac{1}{4}=15$ mg.

Step4: Solve part (c)

We want $A = 0.125A_0$. Substitute into the decay formula $A = A_0\left(\frac{1}{2}\right)^n$: $0.125A_0=A_0\left(\frac{1}{2}\right)^n$. Since $A_0\neq0$, we can divide both sides by $A_0$ to get $0.125=\left(\frac{1}{2}\right)^n$. Since $0.125=\frac{1}{8}=\left(\frac{1}{2}\right)^3$, then $n = 3$. Since each half - life is $t_{1/2}=5.3$ years, the time $t=3\times5.3 = 15.9$ years.

Answer:

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