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Kelvi

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In the given nuclear reaction, the element X is: $\newline$ $^{22}_{11} Na \rightarrow X + e^{+} + {v}$

medium

Nuclei

2022

physics

$^{23}_{11} Na$

$^{23}_{10} Ne$

Explanation

To solve this problem, we need to analyze the given nuclear reaction: $\\ ^{22}_{11}$ Na $\rightarrow X + e^{+} + v \\$

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$^{22}_{10} Ne$

$^{22}_{12} Mg$

This reaction is a beta-plus decay, where a proton in the nucleus is converted into a neutron,

\\

emitting a positron

(e^{+})

and a neutrino

(v). \\

In beta-plus decay, the atomic number decreases by 1, while the mass number remains unchanged.

\\

Given:

\\

• Initial element:

^{22}_{11}

Na

\\

• Final element:

X \\

Since the mass number remains the same, the mass number of

X

is 22.

\\

The atomic number of

X

is

11 - 1 = 10. \\

Therefore,

X

is

^{22}_{10}

Ne

. \\

This corresponds to Option 3.