Question

Two radioactive nuclides $A$ and $B$ have half-lives of $t$ and $2t$ respectively. If we start an experiment with one mole of each of them, the ratio of number of moles remaining after a time interval of $6t$, will be,

• A. $1:1$
• B. $1:2$
• C. $1:6$
• D. $1:8$

Answer 1

The correct option is D $1:8$

We know that, $t=nT$, $M=\frac{M_0}{2^n}$
For $A$ :-
$t_A=n_A{T}_A$
$6t=n_{A}t$
$\therefore n_A=6$
And, $M_A=\frac{M_0}{2^{\left(n_A\right)}}=\frac{M_0}{2^6}$
For B:-
$t_B=n_B{T}_B$
$6t=n_B\left(2t\right)$
$\therefore n_B=3$
And, $M_B=\frac{M_0}{2^{\left(n_B\right)}}=\frac{M_0}{2^3}$

$\therefore \frac{M_A}{M_B}=\frac{\left(\frac{M_0}{2^{\left(n_A\right)}}\right)}{\left(\frac{M_0}{2^{\left(n_B\right)}}\right)}=\frac{2^3}{2^6}=\frac{1}{2^3}$

$\therefore M_A:M_B=1:8$

Hence, $\left(D\right)$ is the correct answer.