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A radio isotope 'X' with a half life $1.4\times10^9$ 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:
A
$1.96\times10^9$ years
B
$3.92\times10^9$ years
C
$4.20\times10^9$ years
D
$8.40\times10^9$ years
Detailed Solution
$\frac{N_x}{N_y} = \frac{1}{7} \Rightarrow \frac{N_x}{N_x + N_y} = \frac{1}{8} = \left(\frac{1}{2}\right)^3$
So $t = 3T_{1/2} = 3\times1.4\times10^9$ years $= 4.2\times10^9$ years
So $t = 3T_{1/2} = 3\times1.4\times10^9$ years $= 4.2\times10^9$ years