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In the given nuclear reaction, the element X is : $^{22}_{11}Na\rightarrow X+e^++\nu$
A
$^{22}_{12}Mg$
B
$^{22}_{11}Na$
C
$^{22}_{10}Ne$
D
$^{22}_{10}Ne$
Detailed Solution
$^{22}_{11}Na \rightarrow X + e^+ + \nu$ is $\beta^+$ decay: mass number stays the same and atomic number decreases by 1.
$^{22}_{11}Na \rightarrow {}^{22}_{10}Ne + e^+ + \nu$
So X is $^{22}_{10}Ne$.