A radioactive element $A$ converts into another stable element $B$, half-life of $A$ is $1.5 \mathrm{hrs}$.…
A radioactive element $A$ converts into another stable element $B$, half-life of $A$ is $1.5 \mathrm{hrs}$. After time $t$ the ratio of atoms of $A$ and $B$ is found to be $1: 8$, then $t$ in hours is
$6$
$8$
Between 3 to 4.5
Between 4.5 to 6
Solution
Number of atoms of $A$ and $B$ are as shown
For $\mathrm{A} \rightarrow$
$N_0 \xrightarrow{T} \frac{N_0}{2} \xrightarrow{T} \frac{N_0}{4} \xrightarrow{T} \frac{N_0}{8} \stackrel{T}{\longrightarrow} \frac{N_0}{16} \ldots \ldots .$.
for $B \rightarrow$
$O \xrightarrow{T} \frac{N_0}{2} \stackrel{T}{\longrightarrow} \frac{3}{4} N_0 \xrightarrow{T} \frac{7}{8} N_0 \xrightarrow{T} \frac{15}{16} N_0 \ldots \ldots$
Ratio of atoms of $A$ and $B$
After one half-life, time period $=\frac{\frac{N_0}{2}}{\frac{N_0}{2}}=1$
After 2 half-life, time period $=\frac{\frac{N_0}{4}}{\frac{3 N_0}{4}}=\frac{1}{3}$
After 3 half-life, time period $=\frac{\frac{N_0}{8}}{\frac{7 N_0}{8}}=\frac{1}{7}$
After 4 half-life, time period $=\frac{\frac{N_0}{16}}{\frac{15 N_0}{16}}=\frac{1}{15}$
As $\frac{1}{7} < \frac{1}{8} < \frac{1}{15} ;$ ratio of $A: B$
becomes $\frac{1}{8}$ sometime between 3 half-lifes to
4 half -lifes. . So, ratio becomes $1: 8$ after $3 \times 1.5$ hrs but earlier than $4 \times 1.5$ hours.