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JEE Main 2010
Exam
JEE Main
Year
2010
Subject
Physics
Chapter
Nuclear Physics

JEE Main 2010 · Physics · Nuclear Physics

A radioactive nucleus (initial mass number A and atomic number Z)...

A radioactive nucleus (initial mass number AA and atomic number Z) emits 3α3 \alpha-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Options

  1. AA−Z−8Z−4\frac{A-Z-8}{Z-4}
  2. BA−Z−4Z−8\frac{A-Z-4}{Z-8}
  3. CA−Z−12Z−4\frac{A-Z-12}{Z-4}
  4. DA−Z−4Z−2\frac{A-Z-4}{Z-2}

Correct answer

B. A−Z−4Z−8\frac{A-Z-4}{Z-8}

Step-by-step solution

In positive beta decay a proton is transformed into a neutron and a positron is emitted. \mathrmp^{+} \longrightarrow \mathrm{n}^0+\mathrme^{+} no. of neutrons initially was A−ZA-Z no. of neutrons after decay (A−Z)−3×2(A-Z)-3 \times 2 (due to alpha particles) +2×1+2 \times 1 (due to positive beta decay) The no. of proton will reduce by 8 . [as 3×23 \times 2 (due to alpha particles) +2+2 (due to positive beta decay)] Hence atomic number reduces by 8 .
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