Not considering the electronic spin, the degeneracy of the second excited state `(n = 3)` of H atom is 9, while the degeneracy of the second excited state of `H^+` is

Explanation

To solve the problem of finding the degeneracy of the second excited state of the hydrogen ion \( H^+ \), we will follow these steps:

### Step 1: Understand Degeneracy
Degeneracy refers to the number of different quantum states that have the same energy level. For a hydrogen atom, the degeneracy of a given energy level can be calculated based on the principal quantum number \( n \).

### Step 2: Determine the Degeneracy of Hydrogen Atom
For the hydrogen atom, the degeneracy of the energy level can be calculated using the formula:
\[ \text{Degeneracy} = n^2 \]
For the second excited state, where \( n = 3 \):
\[ \text{Degeneracy} = 3^2 = 9 \]
This means that the second excited state of the hydrogen atom has a degeneracy of 9.

### Step 3: Analyze the Hydrogen Ion \( H^+ \)
The hydrogen ion \( H^+ \) is formed when one electron is removed from the hydrogen atom. Therefore, \( H^+ \) has no electrons and only consists of a single proton.

### Step 4: Determine the Energy Levels of \( H^+ \)
Since \( H^+ \) has no electrons, it does not have any energy levels or states associated with electron orbitals. Thus, it does not have any degeneracy in the traditional sense used for hydrogen atoms.

### Step 5: Conclusion
The degeneracy of the second excited state of \( H^+ \) is 0 because there are no electrons to occupy any energy levels.

### Final Answer
The degeneracy of the second excited state of \( H^+ \) is **0**.

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