Question

A stable species $C$ can be obtained through unstable nuclides $A$ and $B$. The half life period for conversion $A\to C$ is $T$ and that for conversion $B\to C$ is $2T$. Initially the sample contains $N$ nuclides of $A,N$ nuclides of $B$ and no nuclides for $C$. After $t=4T$, the nuclides of species $C$ will be equal to

• A. $\frac{27N}{32}$
• B. $\frac{9N}{4}$
• C. $\frac{27N}{16}$
• D. $\frac{81N}{16}$

Answer 1

The correct option is C $\frac{27N}{16}$

At $t=4T$, $A$ undergoes $4$ half-lives and $B$ undergoes $2$ half lives.
$\therefore N_A=\frac{N}{2^4};N_B=\frac{N}{2^2}$
$N_A=\frac{N}{16};N_B=\frac{N}{4}$

Number of nuclides of $C$ obtained from$A$
$\Rightarrow N_{CA}=N-\frac{N}{16}$
$=\frac{15N}{16}$
and
Number of nuclides of $C$ obtained from$B$
$N_{CB}=N-\frac{N}{4}$
$N_{CB}=\frac{3N}{4}$
Total nuclides of $C$ are,
$N_C=\frac{15N}{16}+\frac{3N}{4}$
$N_C=\frac{27N}{16}$