To determine the correct nuclide that completes the nuclear reaction:
[tex]\[
{ }^{64} \text{Cu} \rightarrow+\beta+\text{ ? }
\][/tex]
we need to understand the process of beta decay. In beta decay, a neutron in the nucleus of the atom is converted into a proton, with the emission of a beta particle (an electron, [tex]\(\beta\)[/tex]) and an antineutrino.
1. Change in Atomic Number: During beta decay, the atomic number of the element increases by 1, because a neutron is converted to a proton. Therefore, we add 1 to the atomic number of Copper ([tex]\(^{64}\text{Cu}\)[/tex]), which has an atomic number of 29 (Cu is element number 29 in the periodic table).
[tex]\[
\text{New Atomic Number} = 29 + 1 = 30
\][/tex]
2. Change in Mass Number: The mass number (A) remains the same because a neutron is converted to a proton, and the total number of nucleons does not change. Copper-64 has a mass number of 64.
[tex]\[
\text{Mass Number} = 64
\][/tex]
3. Identifying the Resulting Element: The element with atomic number 30 is Zinc (Zn). Therefore, the resultant nuclide is Zinc-64 ([tex]\(^{64}\text{Zn}\)[/tex]).
So, the nuclide that completes the reaction
[tex]\[
{ }^{64} \text{Cu} \rightarrow \beta + { }^{64} \text{Zn}
\][/tex]
is Zinc-64 ([tex]\(^{64}\text{Zn}\)[/tex]).
From the given options, the correct choice is:
- C: [tex]\(^{64} \text{Zn}\)[/tex]
Therefore, the answer is:
[tex]\[
\boxed{C}
\][/tex]
