(B) For a first-order reaction,the amount remaining after $n$ half-lives is given by $N = N_0 \times (1/2)^n$. Given that the substance reduces to $1/

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(B) For a first-order reaction,the amount remaining after $n$ half-lives is given by $N = N_0 \times (1/2)^n$. Given that the substance reduces to $1/16$ of the original amount,we have $1/16 = (1/2)^n$. Since $1/16 = (1/2)^4$,it follows that $n = 4$ half-lives. The total time taken is $2 \, \text{hours} = 120 \, \text{min}$. Therefore,$4 \times t_{1/2} = 120 \, \text{min}$. $t_{1/2} = 120 / 4 = 30 \, \text{min}$.

Class 12 Chemistry Nuclear Chemistry Rate of decay and Half-life

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After $2 \, \text{hours}$,the amount of a certain radioactive substance reduces to $1/16^{th}$ of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is $...... \, \text{min}$.

KVPY 2011 All KVPY

A
$15$
B
$30$
C
$45$
D
$60$

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Rate of decay and Half-life

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After $2 \, \text{hours}$,the amount of a certain radioactive substance reduces to $1/16^{th}$ of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is $...... \, \text{min}$.

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