Question

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

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

Answer 1

Answer: (A)

2: 1

Answer 2

:

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 sample X

$\frac{1}{16} = \left( \frac{1}{2} \right)^{8/T_{X}}or\left( \frac{1}{2} \right)^{4} = \left( \frac{1}{2} \right)^{8/T_{X}}$

$\Rightarrow 4 = \frac{8}{T_{X}}$…… (i)

For sample Y

$\left( \frac{1}{256} \right) = \left( \frac{1}{2} \right)^{8/T_{Y}}$

Or $\left( \frac{1}{2} \right)^{8} = \left( \frac{1}{2} \right)^{8/T_{y}}$

$8 = \frac{8}{T_{Y}}$……….(ii)

Dividing (i) by (ii) we get

}{\frac{1}{2} = \frac{T_{Y}}{T_{X}}or\frac{T_{X}}{T_{Y}} = \frac{2}{1}}$$