A sample contains 16, g of a radioactive mate | vedclass

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(A) The initial amount of radioactive material is $N_0 = 16\, g$.The half-life of the material is $T_{1/2} = 2\, days$.The total time elapsed is $t =

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Less than $1\, mg$

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(A) The initial amount of radioactive material is $N_0 = 16\, g$.The half-life of the material is $T_{1/2} = 2\, days$.The total time elapsed is $t = 32\, days$.The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{32}{2} = 16$.The amount of radioactive material remaining $N$ is given by the formula $N = N_0 	imes (\frac{1}{2})^n$.Substituting the values, we get $N = 16 	imes (\frac{1}{2})^{16} = 2^4 	imes (\frac{1}{2})^{16} = (\frac{1}{2})^{12}$.Since $(\frac{1}{2})^{10} = \frac{1}{1024} \approx 10^{-3} = 1\, mg$, then $(\frac{1}{2})^{12} = \frac{1}{4096}\, g \approx 0.244\, mg$.Therefore, the remaining amount is less than $1\, mg$.

$A$ sample contains $16\, g$ of a radioactive material, the half-life of which is $2\, days$. After $32\, days$, the amount of radioactive material left in the sample is:

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