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NEET PHYSICSNUCLEIEasy

Question

The number of \beta particles emitted by a radioactive substance is twice the number of \alpha particles emitted by it. The resulting daughter is an:

A

isobar of parent

B

isomer of parent

C

isotone of parent

D

isotope of parent

Step-by-Step Solution

  1. Decay Rules:
  • In Alpha (α\alpha) decay, the nucleus emits a helium nucleus (24He{}_{2}^{4}\mathrm{He}). The atomic number (ZZ) decreases by 2 and the mass number (AA) decreases by 4 .
  • In Beta (β\beta^-) decay, the nucleus emits an electron (10e{}_{-1}^{0}e). The atomic number (ZZ) increases by 1 and the mass number (AA) remains unchanged .
  1. Calculation:
  • Let the number of α\alpha particles emitted be nn. The change in ZZ due to α\alpha emission is n×(2)=2nn \times (-2) = -2n.
  • The number of β\beta particles emitted is given as twice the number of α\alpha particles, i.e., 2n2n. The change in ZZ due to β\beta emission is 2n×(+1)=+2n2n \times (+1) = +2n.
  • Net change in atomic number (ZZ): ΔZ=2n+2n=0\Delta Z = -2n + 2n = 0.
  1. Conclusion: The final daughter nucleus has the same atomic number (ZZ) as the parent nucleus but a different mass number (since AA decreases by 4n4n). Atoms with the same atomic number but different mass numbers are defined as isotopes .

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.

PHYSICSNUCLEInumberparticlesemittedradioactivesubstance

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