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NEET PHYSICSNUCLEIEasy

Question

The rate of radioactive disintegration at an instant for a radioactive sample of half-life 2.2×109 s2.2 \times 10^9 \text{ s} is 1010 s110^{10} \text{ s}^{-1}. The number of radioactive atoms in that sample at that instant is:

A

3.7 × 10^{20}

B

3.17 × 10^{17}

C

3.17 × 10^{18}

D

3.17 × 10^{19}

Step-by-Step Solution

  1. Recall the Radioactive Decay Formulae: The rate of disintegration (RR or Activity) is given by R=λNR = \lambda N, where λ\lambda is the decay constant and NN is the number of radioactive atoms. The decay constant is related to the half-life (T1/2T_{1/2}) by the formula: λ=0.693T1/2\lambda = \frac{0.693}{T_{1/2}} .
  2. Combine Equations: Substituting λ\lambda into the rate equation: R=(0.693T1/2)N    N=RT1/20.693R = \left( \frac{0.693}{T_{1/2}} \right) N \implies N = \frac{R \cdot T_{1/2}}{0.693}
  3. Substitute Values: R=1010 s1R = 10^{10} \text{ s}^{-1} T1/2=2.2×109 sT_{1/2} = 2.2 \times 10^9 \text{ s} N=1010×2.2×1090.693N = \frac{10^{10} \times 2.2 \times 10^9}{0.693}
  4. Calculate: N=2.2×10190.6933.1746×1019N = \frac{2.2 \times 10^{19}}{0.693} \approx 3.1746 \times 10^{19}
  5. Conclusion: The number of radioactive atoms is approximately 3.17×10193.17 \times 10^{19}.

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.

PHYSICSNUCLEIradioactivedisintegrationinstantradioactivesample

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