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NEET PHYSICSNUCLEIEasy

Question

In a radioactive substance at t = 0, the number of atoms is 8×1048 \times 10^4. Its half-life period is 3 yr. The number of atoms equal to 1×1041 \times 10^4 will remain after an interval of:

A

9 yr

B

8 yr

C

6 yr

D

24 yr

Step-by-Step Solution

  1. Identify the Principle: Radioactive decay follows first-order kinetics. The number of undecayed nuclei NN remaining after nn half-lives is given by the formula N=N0(12)nN = N_0 (\frac{1}{2})^n, where N0N_0 is the initial number of nuclei .
  2. Analyze Given Data:
  • Initial number of atoms (N0N_0) = 8×1048 \times 10^4.
  • Final number of atoms (NN) = 1×1041 \times 10^4.
  • Half-life period (T1/2T_{1/2}) = 3 years.
  1. Calculate Number of Half-lives (nn):
  • NN0=1×1048×104=18\frac{N}{N_0} = \frac{1 \times 10^4}{8 \times 10^4} = \frac{1}{8}.
  • We know that 18=(12)3\frac{1}{8} = (\frac{1}{2})^3.
  • Therefore, the number of half-lives passed, n=3n = 3.
  1. Calculate Total Time (tt):
  • Total time t=n×T1/2t = n \times T_{1/2}.
  • t=3×3 years=9 yearst = 3 \times 3 \text{ years} = 9 \text{ years}.
  1. Conclusion: The atoms will reduce to the specified amount after 9 years .

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.

PHYSICSNUCLEIradioactivesubstancenumberhalflifeperiod

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