Understanding the Problem

We are comparing the degeneracy of the second excited state (n=3) for two systems: the Hydrogen atom (H) and the Hydrogen anion (H⁻). For the Hydrogen atom, it is given that the degeneracy is 9 (ignoring electron spin). We need to find the degeneracy for H⁻ in its second excited state.

Key Concepts

Degeneracy refers to the number of distinct quantum states that have the same energy. For a hydrogen-like atom, the energy depends only on the principal quantum number $n$ and is given by:

$E_n=-\frac{Z{}^{2}R}{n^2}$

where $Z$ is the atomic number and $R$ is the Rydberg constant.

The total degeneracy for a given energy level $n$ is the sum of the degeneracies from all possible orbital angular momentum states. For each $n$, the possible values of the orbital quantum number $l$ are $l=0,1,2,...,n-1$. For each $l$, there are $(2l+1)$ values of the magnetic quantum number $m_l$.

Therefore, the orbital degeneracy for a shell $n$ is:

$g_{n,\text{orbital}}=\sum _{l=0}^{n-1}(2l+1)=n^2$

For the Hydrogen atom (H, Z=1), the second excited state is n=3. Ignoring electron spin, the degeneracy is indeed $3^2=9$.

Step-by-Step Solution for H⁻

Step 1: Identify the System

The Hydrogen anion, H⁻, is a two-electron system. Its energy level structure is not hydrogen-like. The presence of a second electron means the electrons interact with each other, and the simple formula $E_n=-\frac{Z{}^{2}R}{n^2}$ no longer applies. The energy now depends on both the principal quantum number $n$ and the orbital angular momentum $l$.

Step 2: Determine the Second Excited State Configuration

We find the ground and excited state configurations for H⁻ by filling orbitals according to the Aufbau principle.

• Ground State (1s)²: Both electrons occupy the 1s orbital. This is the only possible configuration for the lowest energy.
• First Excited State (1s)¹(2s)¹: One electron is promoted from the 1s to the 2s orbital.
• Second Excited State (1s)¹(2p)¹: One electron is promoted from the 1s to the 2p orbital. This is higher in energy than the (1s)(2s) configuration.

Therefore, the second excited state of H⁻ has the electron configuration 1s¹2p¹.

Step 3: Calculate the Degeneracy of the 1s¹2p¹ Configuration

We must now find how many distinct quantum states exist for this configuration. We do this by considering the possible orbital and spin states of the two electrons.

A. Orbital Degeneracy:

• The 1s electron has $l=0$. The degeneracy for an s-orbital is $2l+1=1$ ($m_l=0$).
• The 2p electron has $l=1$. The degeneracy for a p-orbital is $2l+1=3$ ($m_l=-1,0,1$).

The total number of possible orbital configurations is the product of the individual degeneracies: $1\times 3=3$.

B. Spin Degeneracy:

Each electron has a spin of $1/2$ with $m_s=\pm 1/2$. For two electrons, the number of possible spin microstates is $2\times 2=4$. These four states form a spin triplet (degeneracy=3, symmetric) and a spin singlet (degeneracy=1, antisymmetric).

C. Total Degeneracy (Orbital + Spin):

The total number of microstates is the product of orbital and spin configurations: $3\times 4=12$.

However, because electrons are identical fermions, the total wavefunction (orbital × spin) must be antisymmetric under the exchange of the two electrons. This Pauli exclusion principle restricts the allowed combinations of orbital and spin states.

For the 1s¹2p¹ configuration, the two electrons are in different orbitals (non-equivalent electrons). This means the Pauli principle does not forbid any of the 12 possible combinations; it only dictates the symmetry of the total wavefunction. All 12 states are physically allowed and have very similar energies.

Step 4: Final Answer

The degeneracy of the second excited state of H⁻ is 12.

Related Topics & Formulae

Related Topics:

• Hydrogen-like Atoms
• Multi-electron Atoms
• Pauli Exclusion Principle
• Electron Configurations
• Term Symbols (for classifying states)

Key Formulae and Theory:

• Hydrogen Atom Energy: $E_n=-\frac{Z{}^{2}R}{n^2}$
• Orbital Degeneracy: $g_{\text{orbital}}=n^2$
• Spin Degeneracy for one electron: $g_{\text{spin}}=2$
• Non-Equivalent Electrons: When two electrons are in different orbitals (e.g., 1s and 2p), the Pauli exclusion principle does not limit the number of allowed states. The total degeneracy is simply the product of the orbital and spin degeneracies.