To determine the number of protons, neutrons, and electrons in the ion [tex]\(103 \text{Rh}^{6+}\)[/tex], we need to consider the following information:
1. Number of Protons (p⁺):
The atomic number of Rhodium (Rh) is 45. This means that a neutral atom of Rhodium contains 45 protons. Since changing the charge of the ion does not affect the number of protons, the ion [tex]\(103 \text{Rh}^{6+}\)[/tex] also contains 45 protons.
2. Number of Neutrons (n⁰ or n):
We are given that this isotope of Rhodium has a mass number of 103. The mass number (A) is the sum of protons (Z) and neutrons (N):
[tex]\[
A = Z + N
\][/tex]
For [tex]\(\text{Rh}\)[/tex], [tex]\(A = 103\)[/tex] and [tex]\(Z = 45\)[/tex]:
[tex]\[
103 = 45 + N
\][/tex]
Solving for neutrons (N):
[tex]\[
N = 103 - 45 = 58
\][/tex]
Thus, the number of neutrons is 58.
3. Number of Electrons (e⁻):
In a neutral atom, the number of electrons is equal to the number of protons, which is 45 for Rhodium. However, this ion has a [tex]\(6+\)[/tex] charge, indicating that it has lost 6 electrons. Therefore:
[tex]\[
\text{Number of electrons} = 45 - 6 = 39
\][/tex]
Therefore, the number of protons, neutrons, and electrons in the ion [tex]\(103 \text{Rh}^{6+}\)[/tex] is:
- Protons: 45
- Neutrons: 58
- Electrons: 39
The last option,
- [tex]\(45 \, p^+, 58 \, n^{\circ}, 39 \, e\)[/tex], is the correct one.
