A radioactive material has an initial amount | vedclass

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(B) The relation between the remaining amount $N$ and initial amount $N_0$ after $n$ half-lives is given by: $N = N_0 \left( \frac{1}{2} \right)^n$ Gi

Vedclass · Accepted Answer

$30$

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(B) The relation between the remaining amount $N$ and initial amount $N_0$ after $n$ half-lives is given by: $N = N_0 \left( \frac{1}{2} \right)^n$ Given $N_0 = 16 \, gm$ and $N = 1 \, gm$: $1 = 16 \left( \frac{1}{2} \right)^n$ $\frac{1}{16} = \left( \frac{1}{2} \right)^n$ $\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^n$ So,the number of half-lives $n = 4$. We know that $n = \frac{t}{T_{1/2}}$,where $t$ is the total time and $T_{1/2}$ is the half-life. $4 = \frac{120}{T_{1/2}}$ $T_{1/2} = \frac{120}{4} = 30 \, days$.

$A$ radioactive material has an initial amount $16 \, gm$. After $120 \, days$ it reduces to $1 \, gm$. The half-life of the radioactive material is .......... $days$.

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