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NEET PHYSICSEasy

In the given nuclear reaction, the element X is: 1122Na→X+e++ν{}^{22}_{11}\mathrm{Na} \rightarrow X + e^+ + \nu

A

1222Mg{}^{22}_{12}\mathrm{Mg}

B

1123Na{}^{23}_{11}\mathrm{Na}

C

1023Ne{}^{23}_{10}\mathrm{Ne}

D

1022Ne{}^{22}_{10}\mathrm{Ne}

Step-by-Step Solution

  1. Identify the Decay Process: The reaction shows the emission of a positron (e+e^+) and a neutrino (ν\nu). This is characteristic of β+\beta^+ decay (positron emission) .
  2. Apply Conservation Laws:
  • Mass Number (AA): The total mass number is conserved. The positron and neutrino have zero mass number. 22=AX+0⇒AX=2222 = A_X + 0 \Rightarrow A_X = 22
  • Atomic Number (ZZ): The total electric charge (atomic number) is conserved. The positron has a charge of +1+1. 11=ZX+1⇒ZX=1011 = Z_X + 1 \Rightarrow Z_X = 10
  1. Identify the Element: The element with atomic number Z=10Z=10 is Neon (Ne). Thus, the daughter nucleus XX is 1022Ne{}^{22}_{10}\mathrm{Ne}.
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