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NEET PHYSICSNUCLEIMedium

Question

The half-life of a radioactive sample undergoing α\alpha-decay is 1.4×1017 s1.4 \times 10^{17}\text{ s}. If the number of nuclei in the sample is 2.0×10212.0 \times 10^{21}, the activity of the sample is nearly equal to:

A

104 Bq10^4\text{ Bq}

B

105 Bq10^5\text{ Bq}

C

106 Bq10^6\text{ Bq}

D

103 Bq10^3\text{ Bq}

Step-by-Step Solution

Radioactive decay follows first-order kinetics . The activity (AA) of a sample is determined by the product of the decay constant (λ\lambda) and the number of nuclei (NN), such that A=λNA = \lambda N. The relationship between the decay constant and half-life (T1/2T_{1/2}) is given by λ=ln2T1/20.693T1/2\lambda = \frac{\ln 2}{T_{1/2}} \approx \frac{0.693}{T_{1/2}} .

Given data:

  • T1/2=1.4×1017 sT_{1/2} = 1.4 \times 10^{17}\text{ s}
  • N=2.0×1021N = 2.0 \times 10^{21}

Calculation:

  1. First, find λ\lambda: λ=0.6931.4×1017 s1\lambda = \frac{0.693}{1.4 \times 10^{17}}\text{ s}^{-1}.
  2. Then, find AA: A=(0.6931.4×1017)×(2.0×1021)A = \left(\frac{0.693}{1.4 \times 10^{17}}\right) \times (2.0 \times 10^{21}).
  3. Simplifying the powers of ten and the coefficients: A=(0.6931.4)×2.0×10(2117)=(0.6930.7)×1040.99×104 BqA = \left(\frac{0.693}{1.4}\right) \times 2.0 \times 10^{(21-17)} = \left(\frac{0.693}{0.7}\right) \times 10^4 \approx 0.99 \times 10^4\text{ Bq}.

The activity is nearly equal to 104 Bq10^4\text{ Bq}.

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.

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