1. ²²₁₂Mg 2. ²³₁₁Na 3. ²³₁₀Ne 4. ²²₁₀Ne

In the given nuclear reaction, the element X is :^22_11Na → X + e^+ +

1. \({}^{22}_{12}\text{Mg}\)

2. \({}^{23}_{11}\text{Na}\)

3. \({}^{23}_{10}\text{Ne}\)

4. \({}^{22}_{10}\text{Ne}\)

Doubtly Asked 30 Oct 2026 NEET UG 2022 neet Save WhatsApp Report

Best answer

Answer: 4. \({}^{22}_{10}\text{Ne}\)

The emission of a positron is beta-plus decay: a proton turns into a neutron, so the mass number stays the same and the atomic number drops by 1.

\({}^{22}_{11}\mathrm{Na}\to{}^{22}_{10}X+e^{+}+\nu\), and element with \(Z=10\) is neon.

So X is \({}^{22}_{10}\mathrm{Ne}\), option (4).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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