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I Multiple Choice Questions · Q13

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

(a) A−Z−4Z−2\dfrac{A-Z-4}{Z-2}
(b) A−Z−2Z−6\dfrac{A-Z-2}{Z-6}
(c) A−Z−4Z−6\dfrac{A-Z-4}{Z-6}
(d) A−Z−12Z−4\dfrac{A-Z-12}{Z-4}
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Step 1. Each alpha decay reduces mass number AA by 4 and atomic number ZZ by 2. Two alpha decays: A→A−8A\to A-8, Z→Z−4Z\to Z-4.

Step 2. Each positron (β+\beta^+) decay leaves AA unchanged but reduces ZZ by 1 (a proton converts to a neutron). Two positron decays: Z→(Z−4)−2=Z−6Z\to (Z-4)-2=Z-6, with AA still A−8A-8.

Step 3. So the final nucleus has Af=A−8A_f=A-8 and Zf=Z−6Z_f=Z-6. Its number of neutrons is Nf=Af−Zf=(A−8)−(Z−6)=A−Z−2N_f=A_f-Z_f=(A-8)-(Z-6)=A-Z-2.

Step 4. The required ratio of neutrons to protons in the final nucleus is …

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