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13.Nuclei
hard
$A$ and $B$ are two radioactive substances whose half lives are $1$ and $2$ years respectively. Initially $10\, g$ of $A$ and $1\,g$ of $B$ is taken. The time (approximate) after which they will have same quantity remaining is ........... $years$
A
$6.62$
B
$5$
C
$3.2$
D
$7$
Solution
$N=N_{0}\left(\frac{1}{2}\right)^{t_{1 / 2}}$
$\Rightarrow N_{A}=10\left(\frac{1}{2}\right)^{t / 1}$ 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
$\log _{10}=\frac{t}{2} \times 10 \log _{10} 2 \Rightarrow 1=\frac{t}{2} \times 0.3010$
$\mathrm{t}=\frac{2}{0.301}=6.62$ years
Standard 12
Physics
