Question

$A$ and $B$ are two radioactive substances whose half-lives are $1$ and $2$ years respectively. Initially $10g$ of $A$ and $1g$ of $B$ is taken. The time (approximate) after which they will have same quantity remaining is

• A. $6.62year$
• B. $5year$
• C. $3.2year$
• D. $7year$

Answer 1

Answer: (A)

$6.62year$

$N=N_0{\left(\frac{1}{2}\right)}^{t_{1/2}}$
$\Rightarrow N_A=10{\left(\frac{1}{2}\right)}^{t/1}$ and $N_8=1{\left(\frac{1}{2}\right)}^{t/2}$
Given, $N_A=N_B$
$\Rightarrow 10{\left(\frac{1}{2}\right)}^t={\left(\frac{1}{2}\right)}^{t/2}\Rightarrow 10={\left(\frac{1}{2}\right)}^{-t/2}$
$\Rightarrow 10=2^{t/2}$
Taking log on both the sides
${\log }_{10}=\frac{t}{2}\times 10{\log }_{10}2\Rightarrow 1=\frac{t}{2}\times 0.3010$.

t = $\frac{2}{0.301}=6.62years$