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NEET PHYSICSNUCLEIEasy

Question

If a nucleus nXm_n X^m emits one α\alpha and two β\beta particles, the resulting nucleus will be:

A

nXm4_n X^{m-4}

B

n1Xm1_{n-1} X^{m-1}

C

nZm4_n Z^{m-4}

D

None of these

Step-by-Step Solution

The change in the nucleus can be calculated by applying the displacement laws for radioactive decay:

  1. α\alpha-decay: The emission of one \alpha particle (24He^4_2\text{He}) reduces the mass number (mm) by 4 and the atomic number (nn) by 2 . Result: n2Ym4_{n-2} Y^{m-4}.
  2. β\beta-decay: The emission of one \beta particle (10e^0_{-1}e) increases the atomic number by 1 while leaving the mass number unchanged. For two β\beta particles, the atomic number increases by 2×1=22 \times 1 = 2. Result: (n2)+2Xm4=nXm4_{ (n-2) + 2 } X^{m-4} = _n X^{m-4}.

Since the final atomic number (nn) is the same as the initial one, the chemical identity of the element remains the same (XX). Therefore, the resulting nucleus is an isotope of the parent element with a mass number of m4m-4.

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.

PHYSICSNUCLEInucleusparticlesresultingnucleus

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