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(D) Let $N_0$ be the initial number of atoms of $A$. After $n$ half-lives,the number of atoms of $A$ remaining is $N_A = N_0 \left(\frac{1}{2}\right)^

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Between $4.5$ to $6$

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(D) Let $N_0$ be the initial number of atoms of $A$. After $n$ half-lives,the number of atoms of $A$ remaining is $N_A = N_0 \left(\frac{1}{2}ight)^n$.Since the element $A$ converts into $B$,the number of atoms of $B$ formed is $N_B = N_0 - N_A = N_0 \left(1 - \left(\frac{1}{2}ight)^night)$.The ratio of atoms of $A$ to $B$ is given as $\frac{N_A}{N_B} = \frac{1}{8}$.Substituting the expressions: $\frac{N_0 (1/2)^n}{N_0 (1 - (1/2)^n)} = \frac{1}{8}$.This simplifies to $\frac{(1/2)^n}{1 - (1/2)^n} = \frac{1}{8}$.Let $x = (1/2)^n$. Then $\frac{x}{1-x} = \frac{1}{8} \implies 8x = 1 - x \implies 9x = 1 \implies x = \frac{1}{9}$.Since $(1/2)^3 = 1/8$ and $(1/2)^4 = 1/16$,and $1/16 Given the half-life $T_{1/2} = 1.5 \ hrs$,the time $t = n 	imes T_{1/2}$.Since $3 < n < 4$,the time $t$ is between $3 	imes 1.5 = 4.5 \ hrs$ and $4 	imes 1.5 = 6 \ hrs$.

$A$ radioactive element $A$ converts into another stable element $B$. The half-life of $A$ is $1.5 \ hrs$. After time $t$,the ratio of the number of atoms of $A$ to $B$ is found to be $1:8$. Then $t$ in hours is:

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