Question

Two substances A and B are present such that [A]=4[B] and half -life of A is 5 minute and of B is 15 minute . If they start decaying at the same time following first order, how much time later will the concentration of both of them would be same?

• A. 15 Minute
• B. 10 Minute
• C. 5 Minute
• D. 12 Minute

Answer 1

The correct option is A

15 Minute

Amount of $A$ left in $n_1$ halves $=\frac{\left[A_0\right]}{2^{\prime 1}}$

Amount of $B$ left in $n_2$ halves $=\frac{\left[B_0\right]}{2^2}$

Also if $\frac{\left[A_0\right]}{2^7}=\frac{\left[B_0\right]}{2^{\ast }2}$ When $A$ decay to $n_2$ halves and $B$ decay to $n_2$ halves
$\because \left[A_0\right]=4\left[B_0\right]$

$\therefore 4=\frac{2^{n_1}}{2^{n_2}}=(2{)}^{n_1-n_2}$

Or $(n_1-n_2)=2$

$\therefore n_2=n_1-2$

Now, $T=n_1\times t_{1/2}A$ and $T=n_2\times t_{1/2}B$

$\frac{n_1\times t_{1/2}A}{n_2\times t_{1/2}B}=1$

Or $\frac{n_1\times 5}{n_2\times 15}=1$

Or $\frac{n_1}{n_2}=3$
$\therefore$ By Eqs. $\left(i\right)$ and $\left(ii\right)$ $n_2=3,n_2=1$
Thus, $T=3\times 5=15$ minute