Question

question 2 of 28
the half - life of cobalt - 60 is 5.3 years. if you have a sample that contains 10.0 g of the isotope, how much will be left after 15.9 years?
a. 5.0 g
b. 0.625 g
c. 1.25 g
d. 2.5 g

Explanation:

Step1: Calculate number of half - lives

The formula to calculate the number of half - lives $n$ is $n=\frac{t}{t_{1/2}}$, where $t$ is the time elapsed and $t_{1/2}$ is the half - life. Given $t = 15.9$ years and $t_{1/2}=5.3$ years.
$n=\frac{15.9}{5.3}=3$

Step2: Calculate remaining amount

The formula for the remaining amount $N$ of a radioactive substance is $N = N_0\times(\frac{1}{2})^n$, where $N_0$ is the initial amount and $n$ is the number of half - lives. Here, $N_0 = 10.0$ g and $n = 3$.
$N=10.0\times(\frac{1}{2})^3=10.0\times\frac{1}{8}=1.25$ g

Answer:

C. 1.25 g