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A radioactive isotope X with a half-life of $1.37 \times {10^9}$ years decays to Y which is stable. A sample of rock from the moon was found to contain both the elements X and Y which were in the ratio of 1 : 7. The age of the rock is

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$4.11 \times {10^9}\,\,years$

$3.85 \times {10^9}\,\,years$

$9.59 \times {10^9}\,\,years$

$1.96 \times {10^8}\,\,years$

answer is C.

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Detailed Solution

If in the rock there is no Y element intially, then the time taken by element X to reduce to $\frac{1}{8} th$ the initial value will be equal to

\frac{1}{8} = {\left( {\frac{1}{2}} \right)^n}

or n =3

Therefore, from the beginning three half life time is spent. Hence the age of the rock is = 3x1.37x10⁹= 4.11x10⁹ years

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