The half-life of Bi^{210} is 5 days. What ti | vedclass

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(C) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$

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$15$

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(C) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T$ is the half-life.Given that $(7/8)^{th}$ of the sample decays,the remaining amount $N$ is $N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.Substituting the values into the formula: $\frac{1}{8}N_0 = N_0 \left( \frac{1}{2} \right)^{t/5}$.This simplifies to $\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{t/5}$.Comparing the exponents,we get $3 = t/5$,which implies $t = 15 \ days$.

The half-life of $Bi^{210}$ is $5 \ days$. What time is taken by $(7/8)^{th}$ part of the sample to decay?

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