Nuclei - Result Question 62

63. A radio isotope ’ $X$ ’ with a half life $1.4 \times 10^{9}$ 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: [2014]

(a) $1.96 \times 10^{9}$ years

(b) $3.92 \times 10^{9}$ years

(c) $4.20 \times 10^{9}$ years

(d) $8.40 \times 10^{9}$ years

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Answer:

Correct Answer: 63. (c)

Solution:

$ \begin{aligned} & \text{ (c) } As \frac{N_x}{N_y}=\frac{1}{7} \text{ (Given) } \\ & \Rightarrow \frac{N_x}{N_x+N_y}=\frac{1}{8}=(\frac{1}{2})^{3} \end{aligned} $

Therefore, age of the rock $t=3 T _{1 / 2}=3 \times 1.4 \times 10^{9} yrs=4.2 \times 10^{9} yrs$.



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