A certain radioactive substance reduces to 25 | vedclass

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

Vedclass · Accepted Answer

$8$

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(B) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} ight)^{\frac{t}{T}}$, 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 the substance reduces to $25\%$ of its initial value, we have $N = 0.25 N_0 = \frac{N_0}{4}$.Substituting the values into the formula: $\frac{N_0}{4} = N_0 \left( \frac{1}{2} ight)^{\frac{16}{T}}$.This simplifies to $\left( \frac{1}{2} ight)^2 = \left( \frac{1}{2} ight)^{\frac{16}{T}}$.Equating the exponents: $2 = \frac{16}{T}$.Therefore, $T = \frac{16}{2} = 8 	ext{ days}$.

$A$ certain radioactive substance reduces to $25\%$ of its initial value in $16$ days. Its half-life is ........ $days$.

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