Gadolinium-153,which is used to detect osteop | vedclass

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(C) The half-life $(t_{1/2})$ of $Gd-153$ is $242$ days. The total time $(t)$ is $2$ years,which is $2 	imes 365 = 730$ days. The number of half-live

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$12.0$

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(C) The half-life $(t_{1/2})$ of $Gd-153$ is $242$ days. The total time $(t)$ is $2$ years,which is $2 	imes 365 = 730$ days. The number of half-lives $(n)$ is calculated as $n = \frac{t}{t_{1/2}} = \frac{730}{242} \approx 3.016$. Since $n \approx 3$,the remaining amount of the substance is given by the formula: $	ext{Remaining percentage} = \frac{100}{2^n} = \frac{100}{2^3} = \frac{100}{8} = 12.5\%$. The value closest to $12.5\%$ among the given options is $12.0\%$.

Gadolinium-$153$,which is used to detect osteoporosis (porous bones),has a half-life of $242$ days. Which value is closest to the percentage of $Gd-153$ left in a patient's system after $2$ years? .......... $\%$

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