Question

Radon-222 has a half-life of \(3.8\) days. If one starts with \(0.064 \mathrm{~kg}\) of Radon-222, the quantity of Radon-222 left aft er 19 days will be
(A) \(0.002 \mathrm{~kg}\)
(B) \(0.062 \mathrm{~kg}\)
(C) \(0.032 \mathrm{~kg}\)
(D) \(0.024 \mathrm{~kg}\)

Answer 1

Ans(A)

\(\frac{19}{3.8}=5,\left(\frac{N}{N_{0}}\right)=\left(\frac{1}{2}\right)^{5} \Rightarrow N \quad 0.064 \times \frac{1}{32}=0.002 \mathrm{~kg}\)

Answer 2

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Answer 3

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