Question

Two samples $X$ and $Y$ contain equal amount of radioactive substances. If ${\left(\frac{1}{16}\right)}^{\text{th }}$ of the sample $X$ and ${\left(\frac{1}{256}\right)}^{\text{th }}$ of the sample $Y$, remain after $8$ hours, then the ratio of half periods of $X$ and $Y$ is

• A. 2 : 1
• B. 1 : 2
• C. 1 : 4
• D. 4 : 1

Answer 1

Answer: (A)

As $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$
where, Number of half lives, $n=\frac{t}{T}$
$T$ is the half life period.
For $X$ sample
$\frac{1}{16}={\left(\frac{1}{2}\right)}^{8/T_X}$ or
${\left(\frac{1}{2}\right)}^4={\left(\frac{1}{2}\right)}^{8/T_X}$ or
$4=\frac{8}{T_X}\dots$(i)
For $Y$ sample
$\left(\frac{1}{256}\right)={\left(\frac{1}{2}\right)}^{8/T_Y}$ or
${\left(\frac{1}{2}\right)}^8={\left(\frac{1}{2}\right)}^{8/T_Y}$ or
$8=\frac{8}{T_Y}\dots$(ii)
Divide (i) by (ii) we get
$\frac{4}{8}=\frac{8}{T_X}\times \frac{T_Y}{8}$
$\Rightarrow \frac{1}{2}=\frac{T_Y}{T_X}$ or
$\frac{T_X}{T_Y}=\frac{2}{1}$