Answer:
The half-life of cobalt-60 in this sample is 5.25 years.
Explanation:
Step One: How many half-live(s) in 10.5 years?
The mass of a radioactive element drop by one-half every half-life. This mass will drop by another one-half after another half-life, and so on so forth. In other words,
[tex]\begin{array}{c|c}\text{Number of half-live(s) elapsed}& \dfrac{\text{Current Mass}}{\text{ Initial Mass}}\\0 & 1 \\ 1 & 0.5 \\ 2 & 0.25 \\ \dots & \dots \end{array}[/tex].
For this sample:
[tex]\dfrac{\text{Current Mass}}{\text{ Initial Mass}} = \rm \dfrac{0.200\; g}{0.800\; g} = 0.25[/tex].
Referring to the table, two half-lives have elapsed before the mass drop to 0.200 grams. There are two half-lives in 10.5 years.
Step Two: What's the length of each half-life?
There are two half-lives in 10.5 years. The length of each half-life will be
[tex]\displaystyle \rm \frac{1}{2} \times 10.5 = 5.25\;years[/tex].
Alternative approach to step one: find the number of half-live(s) in 10.5 years with a scientific calculator.
In case the ratio between current mass and initial mass is between two entries, the tabular approach will no longer work. The number of half-lives in the 10.5 years can still be found using logarithm.
[tex]\displaystyle \text{Number of half-lives} = \log_\frac{1}{2}{\left(\frac{\text{Current Mass}}{\text{Final Mass}}\right )} = \frac{\ln{\left(\dfrac{\text{Current Mass}}{\text{Final Mass}}\right )}}{\ln{\left(\dfrac{1}{2}}\right)}}[/tex].
For this question,
[tex]\begin{aligned}\displaystyle \text{Number of half-lives} &= \frac{\ln{\left(\dfrac{\text{Current Mass}}{\text{Final Mass}}\right )}}{\ln{\left(\dfrac{1}{2}}\right)}}\\ &=\frac{\ln{\left(\dfrac{0.200}{0.800}\right)}}{\ln{\left(\dfrac{1}{2}}\right)}} \approx \frac{-1.38629}{-0.693147} \\&=2 \end{aligned}[/tex].
