A radioactive isotope has a t_{1/2} of 10 da | vedclass

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(D) The half-life period $t_{1/2} = 10 \ days$.Total time elapsed $t = 40 \ days$.The number of half-lives $n = \frac{t}{t_{1/2}} = \frac{40}{10} = 4$

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$2000$

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(D) The half-life period $t_{1/2} = 10 \ days$.Total time elapsed $t = 40 \ days$.The number of half-lives $n = \frac{t}{t_{1/2}} = \frac{40}{10} = 4$.The formula for radioactive decay is $N = N_0 	imes (\frac{1}{2})^n$,where $N$ is the remaining amount and $N_0$ is the initial amount.Given $N = 125 \ g$ and $n = 4$,we have $125 = N_0 	imes (\frac{1}{2})^4$.$125 = N_0 	imes \frac{1}{16}$.$N_0 = 125 	imes 16 = 2000 \ g$.

$A$ radioactive isotope has a $t_{1/2}$ of $10 \ days$. If today $125 \ g$ of it is left,what was its weight $40 \ days$ earlier?

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