Question: 16: In the given nuclear reaction, the element X is
{ }_{11}^{22} \mathrm{Na} \rightarrow X+e^{+}+v(1) { }_{10}^{22} \mathrm{Ne}
(2) { }_{12}^{22} \mathrm{Mg}
(3) { }_{11}^{23} \mathrm{Na}
(4) { }_{10}^{23} \mathrm{Ne}
Answer: Option (1)
Explanation:
The given nuclear reaction shows the emission of a positron e^{+} and
a neutrino v. This type of decay is known as positron emission or \beta^{+} decay.
In positron emission, a proton inside the nucleus converts into a neutron according to the process
p \rightarrow n + e^{+} + v.
As a result of this conversion, the atomic number of the nucleus decreases by 1,
while the mass number remains unchanged.
The given nucleus is { }_{11}^{22} \mathrm{Na}.
After positron emission, the new atomic number becomes Z = 11 - 1 = 10,
and the mass number remains A = 22.
The element with atomic number 10 and mass number 22 is neon.
Thus, the daughter nucleus formed is { }_{10}^{22} \mathrm{Ne}.
Hence, the correct answer is option (1).