Radon-222 has a half-life of 3.8 days. If one | vedclass

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(A) The half-life of Radon-$222$ is $T_{1/2} = 3.8 	ext{ days}$.Total time elapsed is $t = 19 	ext{ days}$.The number of half-lives $n$ is given by

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$0.002$

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(A) The half-life of Radon-$222$ is $T_{1/2} = 3.8 	ext{ days}$.Total time elapsed is $t = 19 	ext{ days}$.The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{19}{3.8} = 5$.The remaining quantity $N$ is given by the formula $N = N_0 \left(\frac{1}{2}ight)^n$, where $N_0$ is the initial amount.Given $N_0 = 0.064 \ kg$ and $n = 5$, we have:$N = 0.064 	imes \left(\frac{1}{2}ight)^5$$N = 0.064 	imes \frac{1}{32}$$N = 0.002 \ kg$.Thus, the quantity of Radon-$222$ left after $19$ days is $0.002 \ kg$.

Radon-$222$ has a half-life of $3.8$ days. If one starts with $0.064 \ kg$ of radon-$222$, the quantity of radon-$222$ left after $19$ days will be (in $kg$)

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