Question

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?
○ 3
○ 4
○ 8
○ 16

Explanation:

Step1: Find remaining radioisotope mass

The total initial mass of the radioisotope (assuming no daughter initially) is the sum of remaining radioisotope and daughter isotope. Wait, actually, initial radioisotope is 1.28 g. Now daughter is 1.12 g, so remaining radioisotope is \( 1.28 - 1.12 = 0.16 \) g.

Step2: Use half - life formula

The formula for remaining quantity \( N = N_0(\frac{1}{2})^n \), where \( N_0 = 1.28 \) g, \( N = 0.16 \) g, and \( n \) is number of half - lives.
So \( 0.16=1.28\times(\frac{1}{2})^n \)
Divide both sides by 1.28: \( \frac{0.16}{1.28}=(\frac{1}{2})^n \)
\( \frac{1}{8}=(\frac{1}{2})^n \)
Since \( \frac{1}{8}=(\frac{1}{2})^3 \), so \( n = 3 \)

Answer:

3