Radio activity

Question

A radioisotope X with a half life 1.4 × ${10}^{9}$ years decays to Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1: 7 . The age of the rock is

Easy

Solution

Number of Nuclei at t = 0

Number of Nuclei at time t

As per question

$\frac{{N}_{0}-x}{x}=\frac{1}{7}$

$7{N}_{0}-7x=x\text{or}x=\frac{7}{8}{N}_{0}$

$\therefore \text{Remaining nuclei of isotope}X$

$={N}_{0}-x={N}_{0}-\frac{7}{8}{N}_{0}=\frac{1}{8}{N}_{0}={\left(\frac{1}{2}\right)}^{3}{N}_{0}$

So three half lives would have been passed

$\therefore t=n{T}_{1/2}=3\times 1.4\times {10}^{9}\text{years}=4.2\times {10}^{9}\text{years}$

$\text{Hence, the age of the rock is}4.2\times {10}^{9}\text{years.}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Download Now