a sample originally contained 1.28 g of a radioisotope. it now contains 1.12 g of its daughter isotope. how…
a sample originally contained 1.28 g of a radioisotope. it now contains 1.12 g of its daughter isotope. how many half - lives have passed since the sample originally formed?\n3\n4\n8\n16
Answer
Explanation:
Step1: Calculate remaining radioisotope mass
The original mass of the radioisotope is $m_0 = 1.28$ g. The mass of the daughter - isotope is $m_d=1.12$ g. So the remaining mass of the radioisotope $m = m_0 - m_d=1.28 - 1.12=0.16$ g.
Step2: Use the radioactive - decay formula
The radioactive - decay formula is $m = m_0\times(\frac{1}{2})^n$, where $n$ is the number of half - lives. Substitute $m = 0.16$ g and $m_0 = 1.28$ g into the formula: $0.16=1.28\times(\frac{1}{2})^n$.
Step3: Solve for $n$
First, rewrite the equation as $\frac{0.16}{1.28}=(\frac{1}{2})^n$. Simplify the left - hand side: $\frac{0.16}{1.28}=\frac{1}{8}$. Since $\frac{1}{8}=(\frac{1}{2})^3$, then $n = 3$.
Answer:
3