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A 2,000g of C-14 is left to decay radioactively. The half-life of Carbon-14 is approximately 5,700 years. What percentage of the sample will remain after 17,100 years?

User DanCouper
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2 Answers

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1/8 when using usatestprep

User Algamest
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Answer is: 12,6% (1/8) percentage of the sample will remain.
c₀ - initial amount of C-14.
c - amount of C-14 remaining at time.
t = 5700
y.
First calculate the radioactive decay rate constant λ:
λ = 0,693 ÷ t = 0,693 ÷ 5700
y = 0,000121 1/y = 1,21·10⁻⁴ y.
c = c₀·e∧-λ·t.
c = 2000 · e∧-(0,000121 1/y · 17100 y).
c = 252 g.
ω = 252 g ÷ 2000 g = 0,126 = 12,6%.


User Hmlasnk
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