the half - life of cobalt - 60 is 5 years. how old is a sample of cobalt - 60 if only one - eighth of the…

Explanation:

Step1: Recall the half - life formula

The amount of a radioactive substance $N$ at time $t$ is given by $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N_0$ is the initial amount, $T_{1/2}$ is the half - life, and $t$ is the time elapsed. We know that $N=\frac{1}{8}N_0$ and $T_{1/2}=5$ years.

Step2: Substitute values into the formula

Substitute $N = \frac{1}{8}N_0$ and $T_{1/2}=5$ into $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$. We get $\frac{1}{8}N_0=N_0(\frac{1}{2})^{\frac{t}{5}}$. Divide both sides by $N_0$ (since $N_0\neq0$), so $\frac{1}{8}=(\frac{1}{2})^{\frac{t}{5}}$.

Step3: Rewrite $\frac{1}{8}$ as a power of $\frac{1}{2}$

Since $\frac{1}{8}=\frac{1}{2^3}=(\frac{1}{2})^3$, the equation becomes $(\frac{1}{2})^3=(\frac{1}{2})^{\frac{t}{5}}$.

Step4: Solve for $t$

If $a^m=a^n$, then $m = n$. So, $3=\frac{t}{5}$. Multiply both sides by 5 to get $t = 15$ years.

Answer:

15 years