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NEET PHYSICSNUCLEIMedium

Question

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:

A

1.96 × 10^9 yr

B

3.92 × 10^9 yr

C

4.20 × 10^9 yr

D

8.40 × 10^9 yr

Step-by-Step Solution

  1. 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 (N0N_0) is equal to the sum of the varying amounts of X and Y present at time tt. N0=NX+NYN_0 = N_X + N_Y
  2. Analyze the Ratio: The observed ratio of X to Y is given as 1:71:7. This means for every 1 nucleus of X remaining, there are 7 nuclei of Y formed.
  • Fraction of X remaining (N/N0N/N_0) = NXNX+NY=11+7=18\frac{N_X}{N_X + N_Y} = \frac{1}{1 + 7} = \frac{1}{8}.
  1. Calculate Number of Half-Lives (nn): Using the radioactive decay formula N=N0(12)nN = N_0 \left(\frac{1}{2}\right)^n : 18=(12)n\frac{1}{8} = \left(\frac{1}{2}\right)^n (12)3=(12)n    n=3\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n \implies n = 3 So, 3 half-lives have passed.
  2. Calculate Age (tt): t=n×T1/2t = n \times T_{1/2} t=3×(1.4×109 yr)t = 3 \times (1.4 \times 10^9 \text{ yr}) t=4.2×109 yrt = 4.2 \times 10^9 \text{ yr}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from NUCLEI. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSNUCLEIradioisotopehalflifedecaysstablesample

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