Certainly! Let's analyze the nuclear reaction given:
[tex]\[
{ }_7^{13} N \rightarrow{ }_6^{13} C +{ }_{+1}^0 e + \nu_e
\][/tex]
First, let's identify the particles involved in the reaction:
- [tex]\(_7^{13}\text{N}\)[/tex] is a nitrogen nucleus with a mass number of 13 and an atomic number of 7.
- [tex]\(_6^{13}\text{C}\)[/tex] is a carbon nucleus with a mass number of 13 and an atomic number of 6.
- [tex]\({ }_{+1}^0 e\)[/tex] represents a positron, which is a particle with the same mass as an electron but a positive charge.
- [tex]\(\nu_e\)[/tex] represents a neutrino, specifically an antineutrino in this context.
Given the characteristics of the particles and the change happening to the nucleus:
- The nitrogen nucleus [tex]\(_7^{13}\text{N}\)[/tex] is converting into a carbon nucleus [tex]\(_6^{13}\text{C}\)[/tex].
- A positron ([tex]\({ }_{+1}^0 e\)[/tex]) is emitted, indicating that a proton is being converted into a neutron.
- A neutrino ([tex]\(\nu_e\)[/tex]) is also emitted along with the positron.
The type of decay in which a proton within the nucleus is converted into a neutron, accompanied by the emission of a positron and a neutrino, is known as beta plus decay or positron emission.
Therefore, the correct answer is:
(C) beta plus decay
