Question

Two radioactive isotopes $A$ and $B$ of atomic mass $X$ and $Y$ are mixed in equal amount by mass. After $20$ days, their mass ratio is found to be $1:4$. Half life of '$A$' is $1$ day. What will be the half life of $B$?

• A. $1.11$ day
• B. $0.6237$ day
• C. $0.11\frac{X}{Y}$ day
• D. $1.11\frac{Y}{X}$ day

Answer 1

Answer: (A)

$1.11$ day

Let $1$ $g$ of both $A$ and $B$ are taken initially. $W_A$ and $W_B$ are the amounts left after $20$ days.
${\lambda }_A=\frac{2.303}{20}\log \frac{1}{W_A}$
${\lambda }_B=\frac{2.303}{20}\log \frac{1}{W_B}$
${\lambda }_A-{\lambda }_B=\frac{2.303}{20}\log \frac{W_B}{W_A}$
$=\frac{2.303}{20}\log 4=0.0693$
${\lambda }_B={\lambda }_A-0.0693=\frac{0.693}{t_{1/2}}-0.0693$
$=\frac{0.693}{1}-0.0693=0.6237$
$t_{1/2}B=\frac{0.693}{0.6237}=1.11$ day