If 75 % of a radioactive sample disintegrates | vedclass

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(C) Let $N_0$ be the initial amount of the radioactive sample.After $t = 16 	ext{ days}$, the remaining amount $N$ is:$N = N_0 - 75\% 	ext{ of } N_0

Vedclass · Accepted Answer

$8$

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(C) Let $N_0$ be the initial amount of the radioactive sample.After $t = 16 	ext{ days}$, the remaining amount $N$ is:$N = N_0 - 75\% 	ext{ of } N_0 = N_0 - 0.75 N_0 = 0.25 N_0 = \frac{N_0}{4}$.We know the radioactive decay formula is $N = N_0 \left( \frac{1}{2} ight)^n$, where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.Substituting the values:$\frac{N_0}{4} = N_0 \left( \frac{1}{2} ight)^n$$\frac{1}{4} = \left( \frac{1}{2} ight)^n$$\left( \frac{1}{2} ight)^2 = \left( \frac{1}{2} ight)^n$Thus, $n = 2$.Since $n = \frac{t}{T_{1/2}}$, we have $2 = \frac{16}{T_{1/2}}$.Therefore, $T_{1/2} = \frac{16}{2} = 8 	ext{ days}$.

If $75 \%$ of a radioactive sample disintegrates in $16 \text{ days}$, the half-life of the radioactive sample is (in $\text{ days}$)

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