A radioactive isotope has a half-life of 20 years. How many grams of a 6 gram sample will remain after 40 years?
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A
8g
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B
6g
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C
3g
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D
1.5g
1.5 grams will remain after 40 years.
Radioactive decay follows a predictable pattern based on half-life, which is the time required for half of a radioactive substance to decay. In this case, the isotope has a half-life of 20 years, and the initial mass of the sample is 6 grams.
After the first half-life (20 years), half of the original 6 grams decays, leaving 3 grams. After the second half-life (40 years total), half of the remaining 3 grams decays, leaving 1.5 grams. This means that after two half-lives, one-quarter of the original sample remains.
A. 8 g
Radioactive substances do not increase in mass over time. An increase would contradict the nature of radioactive decay.
B. 6 g
This would indicate that no decay occurred over 40 years, which is impossible given that two half-lives have passed.
C. 3 g
Three grams would be the amount remaining after only one half-life (20 years), not after two half-lives.
D. 1.5 g
This is correct. After two half-lives, the remaining mass is one-quarter of the original amount. One-quarter of 6 grams is 1.5 grams.
Conclusion
Over a period of 40 years, the radioactive sample undergoes two half-lives, reducing its mass by half twice. As a result, only one-quarter of the original 6-gram sample remains, which corresponds to 1.5 grams.

Topic Flashcards
Click to FlipThe definition of the time it takes for half of the atoms in a radioactive sample to decay.
Half-life.
The mathematical formula to calculate the remaining amount of a substance after n n half-lives, if the initial amount is N 0 N 0 .
N=N 0 ×( 2 1 ) n .
The fraction of a radioactive sample that remains after 3 half-lives have passed.
8 1 or 12.5%.
If a 20-gram sample has a half-life of 10 years, what mass remains after 30 years?
2.5 grams (30 years = 3 half-lives: 20 g → 10 g → 5 g → 2.5 g).
The constant that describes the rate of radioactive decay and is related to half-life by the equation T 1 / 2 = ln ( 2 ) λ T 1/2 = λ ln(2) .
Decay constant ( λ λ).