a meteorite contains 0.17 g of nickel - 59, a radioisotope that decays to form cobalt - 59. the meteorite…
a meteorite contains 0.17 g of nickel - 59, a radioisotope that decays to form cobalt - 59. the meteorite also contains 5.27 g of cobalt - 59. how many nickel - 59 half - lives have passed since the meteorite formed?\no 1\no 5\no 5.1\no 5.44
Answer
Explanation:
Step1: Calculate initial amount of nickel - 59
The initial amount of nickel - 59 ($N_0$) is the sum of the remaining nickel - 59 ($N$) and the amount that has decayed (which is now cobalt - 59). So $N_0=0.17 + 5.27=5.44$ g and $N = 0.17$ g.
Step2: Use radioactive decay formula
The radioactive decay formula is $N = N_0\times\left(\frac{1}{2}\right)^n$, where $n$ is the number of half - lives. Rearranging for $n$ gives $n=\frac{\log\left(\frac{N_0}{N}\right)}{\log(2)}$. Substitute $N_0 = 5.44$ g and $N = 0.17$ g into the formula: $\frac{N_0}{N}=\frac{5.44}{0.17}=32$. Then $n=\frac{\log(32)}{\log(2)}$. Since $\log(32)=\log(2^5) = 5\log(2)$, $n = 5$.
Answer:
5