Question

1. the mass of cobalt - 60 in a sample is found to have decreased from 0.800g to 0.200g in a period of 10.5 years. from this information calculate the half - life of cobalt - 60. show your work and box the answer.

Explanation:

Step1: Use radioactive - decay formula

The radioactive - decay formula is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N$ is the final amount, $N_0$ is the initial amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. We know that $N_0 = 0.800g$, $N = 0.200g$, and $t = 10.5$ years. Substitute these values into the formula: $0.200=0.800(\frac{1}{2})^{\frac{10.5}{T_{1/2}}}$.

Step2: Simplify the equation

First, divide both sides of the equation by $0.800$: $\frac{0.200}{0.800}=(\frac{1}{2})^{\frac{10.5}{T_{1/2}}}$. So, $0.25 = (\frac{1}{2})^{\frac{10.5}{T_{1/2}}}$. Since $0.25=\frac{1}{4}=(\frac{1}{2})^2$, we have the equation $(\frac{1}{2})^2 = (\frac{1}{2})^{\frac{10.5}{T_{1/2}}}$.

Step3: Solve for the half - life

Since the bases are the same ($\frac{1}{2}$), the exponents must be equal. So, $2=\frac{10.5}{T_{1/2}}$. Cross - multiply to get $2T_{1/2}=10.5$. Then, solve for $T_{1/2}$: $T_{1/2}=\frac{10.5}{2}=5.25$ years.

Answer:

$5.25$ years