First slide

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

A

$1.96 \times 10^{9} years$
B

$3.92 \times 10^{9} years$
C

$8.4 MeV$
D

$8.40 \times 10^{9} years$

Solution

Number of Nuclei at t = 0
Number of Nuclei at time t
As per question
$\frac{N_{0} - x}{x} = \frac{1}{7}$
$7 N_{0} - 7 x = x or x = \frac{7}{8} N_{0}$
$∴ Remaining nuclei of isotope X$
$= N_{0} - x = N_{0} - \frac{7}{8} N_{0} = \frac{1}{8} N_{0} = {(\frac{1}{2})}^{3} N_{0}$
So three half lives would have been passed
$∴ t = n T_{1 / 2} = 3 \times 1.4 \times 10^{9} years = 4.2 \times 10^{9} years$
$Hence, the age of the rock is 4.2 \times 10^{9} years.$

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