The half-life of a radioactive isotope is 30 | vedclass

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(B) The half-life of the radioactive isotope is given as $T_{1/2} = 30 \,h$.Let the initial amount of the radioactive isotope be $N_0$.We want to find

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$90$

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(B) The half-life of the radioactive isotope is given as $T_{1/2} = 30 \,h$.Let the initial amount of the radioactive isotope be $N_0$.We want to find the time $t$ when the remaining amount $N$ is $12.5 \%$ of $N_0$.$N = 12.5 \% 	ext{ of } N_0 = \frac{12.5}{100} N_0 = \frac{1}{8} N_0$.Using the radioactive decay formula $N = N_0 \left( \frac{1}{2} ight)^n$, where $n = \frac{t}{T_{1/2}}$ is the number of half-lives:$\frac{1}{8} N_0 = N_0 \left( \frac{1}{2} ight)^n$$\frac{1}{8} = \left( \frac{1}{2} ight)^n$$\left( \frac{1}{2} ight)^3 = \left( \frac{1}{2} ight)^n$Therefore, $n = 3$.Since $n = \frac{t}{T_{1/2}}$, we have $t = n 	imes T_{1/2} = 3 	imes 30 \,h = 90 \,h$.

The half-life of a radioactive isotope is $30 \,h$. How long will it take to get reduced to $12.5 \%$ of its initial amount (in $\,h$)?

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