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Answer :
Final answer:
After 66.9 years, which corresponds to 3 half-lives of Lead-20 with each half-life being 22.3 years, 10 mg of the original 80 mg samples remains.
Explanation:
The question involves the concept of half-life, which is the amount of time it takes for half of a radioactive substance to decay. Lead-20 has a half-life of 22.3 years. To determine how much of the original 80 mg of Lead-20 will be left after 66.9 years, we calculate the number of half-lives that have passed, which is 66.9 years ÷ 22.3 years/half-life = 3 half-lives.
With each half-life, the amount of substance remaining is halved. So after:
One half-life (22.3 years), there would be 50% left: 80 mg / 2 = 40 mg.
Two half-lives (44.6 years), we halve it again: 40 mg / 2 = 20 mg.
Three half-lives (66.9 years), halve it once more: 20 mg / 2 = 10 mg.
The answer is that after 66.9 years, 10 mg of the original 80 mg of Lead-20 will be left.
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