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NEET PHYSICSEasy

In the given nuclear reaction, the element XX is 1122Na→X+e+^{22}_{11}\text{Na} \rightarrow X + e^+

1

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

2

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

3

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

4

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

Step-by-Step Solution

In a positron emission (e+e^+), the atomic number decreases by 1 and the mass number remains the same. 1122Na→1022Ne++10e^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^{0}_{+1}e. Thus, XX is 1022Ne^{22}_{10}\text{Ne}.

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