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?\n1\n5\n5.1\n5.44
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 equal to the amount of cobalt - 59, $A$). So $N_0=N + A$. Here, $N = 0.17$ g and $A=5.27$ g, then $N_0=0.17+5.27 = 5.44$ g.
Step2: Use the radioactive - decay formula for half - lives
The radioactive - decay formula is $N = N_0\times(\frac{1}{2})^n$, where $n$ is the number of half - lives. We know $N = 0.17$ g and $N_0 = 5.44$ g. Substitute these values into the formula: $0.17=5.44\times(\frac{1}{2})^n$. Then $(\frac{1}{2})^n=\frac{0.17}{5.44}=\frac{1}{32}$.
Step3: Solve for the number of half - lives
Since $\frac{1}{32}=(\frac{1}{2})^5$, then $n = 5$.
Answer:
5