Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A radioisotope X with a half-life 1.4 × 10^9 yr decays into 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:

1.96 × 10^9 yr

3.92 × 10^9 yr

4.20 × 10^9 yr

8.40 × 10^9 yr

Step-by-Step Solution

Understand the Decay Process: The radioisotope X (parent) decays into stable isotope Y (daughter). The total number of nuclei is conserved. If we assume the rock initially contained only X, then the initial amount of X ( $N_0$ ) is equal to the sum of the varying amounts of X and Y present at time $t$ . $N_0 = N_X + N_Y$
Analyze the Ratio: The observed ratio of X to Y is given as $1:7$ . This means for every 1 nucleus of X remaining, there are 7 nuclei of Y formed.

Fraction of X remaining ( $N/N_0$ ) = $\frac{N_X}{N_X + N_Y} = \frac{1}{1 + 7} = \frac{1}{8}$ .

Calculate Number of Half-Lives ( $n$ ): Using the radioactive decay formula $N = N_0 \left(\frac{1}{2}\right)^n$ : $\frac{1}{8} = \left(\frac{1}{2}\right)^n$ $\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n \implies n = 3$ So, 3 half-lives have passed.
Calculate Age ( $t$ ): $t = n \times T_{1/2}$ $t = 3 \times (1.4 \times 10^9 \text{ yr})$ $t = 4.2 \times 10^9 \text{ yr}$

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