To determine the type of nuclear reaction from the equation [tex]\({ }_{30}^{90} Y \rightarrow{ }_{40}^{90} Zr + { }_{-1}^0 e\)[/tex], let's break it down step by step.
1. Identify the Participants:
- The reactant is an atom of yttrium ([tex]\(Y\)[/tex]) with an atomic number [tex]\(30\)[/tex] and an atomic mass [tex]\(90\)[/tex].
- The products are an atom of zirconium ([tex]\(Zr\)[/tex]) with an atomic number [tex]\(40\)[/tex] and an atomic mass [tex]\(90\)[/tex], and an electron ([tex]\(e\)[/tex]) with an atomic mass [tex]\(0\)[/tex] and an atomic number [tex]\(-1\)[/tex].
2. Analyze the Change:
- Atomic Number Change: The atomic number of yttrium (30) increases to 40, which corresponds to zirconium.
- Atomic Mass Change: The atomic mass remains the same at 90.
3. Electron Emission:
- The equation shows the emission of an electron, with an atomic number [tex]\(-1\)[/tex] and an atomic mass [tex]\(0\)[/tex].
4. Type of Decay:
- The emission of an electron (also called a beta particle) is characteristic of beta decay. In beta decay, a neutron in the nucleus is transformed into a proton with the release of an electron (beta particle).
- The atomic number increases by 1 (from 30 to 40) because a neutron is converted into a proton.
- The atomic mass remains unchanged because the number of nucleons (protons and neutrons) remains the same.
Based on this analysis, we conclude that the nuclear reaction [tex]\({ }_{30}^{90} Y \rightarrow { }_{40}^{90} Zr + { }_{-1}^0 e\)[/tex] is a Beta decay.
Therefore, the correct answer is:
[tex]\[ \text{B. Beta decay} \][/tex]
