Question

Half-life of a radioactive substances $A$ and $B$ are as $t_{1/2}A=2\times t_{1/2}B$. If $N_A^0$ and $N_B^0$ represents initial number of atoms of $A$ and $B$ respectively and after 3 half-life of $A$, the number of atoms of $A$ and $B$ left undecayed becomes same, the initial ratio of $N_A^0$ and $N_B^0$ is equal to:

• A. $\frac{1}{2}$
• B. $\frac{1}{8}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{8}$

$N_A^0=N_A{e}^{{\lambda }_A.t_A}$
$N_B^0=N_B{e}^{{\lambda }_B.t_B}$
$\frac{N_A^0}{N_B^0}=\frac{N_A}{N_B}{e}^{[{\lambda }_A.t_A-{\lambda }_B.t_B]}$
If $t=t_{1/2}A\times 3then\frac{N_A}{N_B}=1$
$\therefore \frac{N_A^0}{N_B^0}=e^{[\frac{0.693}{t_{1/2A}}\times 3\times t_{1/2}A-\frac{0.693}{t_{1/2B}}\times 3\times t_{1/2}A]}$
$=e^{[0.693\times 3-\frac{0.693}{t_{1/2}A}\times 2\times 3\times t_{1/2}A]}$
$=e^{0.693[3-6]}=e^{-2.079}$
$=0.125=\frac{1}{8}$