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Question: Not considering the electron spin, the degeneracy of the second excited state (n = 3) of H atom is 9...

Not considering the electron spin, the degeneracy of the second excited state (n = 3) of H atom is 9. What is the degeneracy of second excited state of H−H^- ?

A

9

B

3

C

6

D

5

Answer

6

Explanation

Solution

The question asks for the degeneracy of the second excited state of the H- ion, without considering electron spin. We are given that for the H atom, the degeneracy of the second excited state (n=3) is 9.

First, let's understand the term "degeneracy" in the context of the H atom. For a hydrogen atom (a single-electron system), the energy levels depend only on the principal quantum number 'n'. The degeneracy of an energy level 'n' (without considering electron spin) is n2n^2. For n=3, the degeneracy is 32=93^2=9. These 9 states correspond to:

  • 3s orbital: 1 state (ml=0m_l=0)
  • 3p orbitals: 3 states (ml=−1,0,1m_l=-1, 0, 1)
  • 3d orbitals: 5 states (ml=−2,−1,0,1,2m_l=-2, -1, 0, 1, 2)

Total = 1 + 3 + 5 = 9. This confirms that for the H atom, "not considering electron spin" means we count the number of distinct spatial wavefunctions (orbitals) for a given energy level.

Now, let's consider the H- ion. H- is a two-electron system (isoelectronic with He atom). For multi-electron systems, electron-electron repulsion lifts the degeneracy of orbitals with the same 'n' but different 'l' (e.g., 2s and 2p are not degenerate). The energy levels are also split by spin-spin interactions (exchange energy), leading to distinct singlet and triplet states.

We need to find the "second excited state" of H- and its degeneracy, "not considering electron spin". This implies we should count the number of distinct spatial wavefunctions for that energy level.

Let's list the electron configurations and their corresponding terms (states) in increasing order of energy:

  1. Ground state: Configuration: 1s21s^2 This configuration gives rise to only one term: 1S^1S. The orbital angular momentum L = 0. Degeneracy (not considering spin) = 2L+1=2(0)+1=12L+1 = 2(0)+1 = 1.

  2. First excited configurations: These involve promoting one electron from 1s to the next available higher energy orbitals (2s or 2p). In multi-electron atoms, 2s is lower in energy than 2p. So, the lowest energy excited configuration is 1s12s11s^1 2s^1. This configuration gives rise to two terms: 3S^3S (triplet) and 1S^1S (singlet). According to Hund's rule, the term with higher spin multiplicity is lower in energy. So, 3S^3S is lower than 1S^1S.

    • First excited state: 1s12s11s^1 2s^1, 3S^3S term. L = 0. Degeneracy (not considering spin) = 2L+1=2(0)+1=12L+1 = 2(0)+1 = 1.
    • Next state: 1s12s11s^1 2s^1, 1S^1S term. L = 0. Degeneracy (not considering spin) = 2L+1=2(0)+1=12L+1 = 2(0)+1 = 1.

  3. Second excited configurations: The next higher energy configuration is 1s12p11s^1 2p^1. This configuration gives rise to two terms: 3P^3P (triplet) and 1P^1P (singlet). Again, 3P^3P is lower in energy than 1P^1P.

    • Next state: 1s12p11s^1 2p^1, 3P^3P term. L = 1. Degeneracy (not considering spin) = 2L+1=2(1)+1=32L+1 = 2(1)+1 = 3.
    • Next state: 1s12p11s^1 2p^1, 1P^1P term. L = 1. Degeneracy (not considering spin) = 2L+1=2(1)+1=32L+1 = 2(1)+1 = 3.

Let's list the states in increasing order of energy and their degeneracies (not considering spin):

  1. Ground state: 1s21s^2, 1S^1S. Degeneracy = 1.
  2. First excited state: 1s12s11s^1 2s^1, 3S^3S. Degeneracy = 1.
  3. Second excited state: 1s12s11s^1 2s^1, 1S^1S. Degeneracy = 1.
  4. Third excited state: 1s12p11s^1 2p^1, 3P^3P. Degeneracy = 3.

This interpretation leads to a degeneracy of 1 for the second excited state, which is not among the options (9, 3, 6, 5). This suggests that the "second excited state" might refer to the entire configuration, or there's a different interpretation of "not considering electron spin" for degeneracy in multi-electron systems.

A common convention in such problems, particularly in the context of atomic spectroscopy for multi-electron atoms, is that terms arising from the same electron configuration (e.g., 1S^1S and 3S^3S from 1s12s11s^1 2s^1) are distinct energy levels. When asked for degeneracy "not considering electron spin", it usually refers to the orbital degeneracy (2L+12L+1) of a specific term. However, the problem statement's example for H atom (n=3, degeneracy 9) sums the degeneracies of all orbitals (3s,3p,3d3s, 3p, 3d) for that principal quantum number.

Let's reconsider the "second excited state" of H-. The ground state is 1s21s^2. The first excited configuration is 1s12s11s^1 2s^1. The second excited configuration is 1s12p11s^1 2p^1.

If the "second excited state" refers to all states arising from the 1s12p11s^1 2p^1 configuration, and "not considering electron spin" means we sum the orbital degeneracies of all terms that can arise from this configuration (ignoring the spin multiplicity factor 2S+12S+1), then: For 1s12p11s^1 2p^1:

  • It gives rise to 3P^3P (L=1, S=1) and 1P^1P (L=1, S=0) terms.
  • The orbital degeneracy for both P terms is 2L+1=2(1)+1=32L+1 = 2(1)+1 = 3.

If we consider these two terms as distinct energy levels, and we are asked for the degeneracy of the "second excited state", which is the 1S^1S from 1s12s11s^1 2s^1, the answer is 1.

However, if the question implies that all terms from a particular configuration are considered part of a single "excited state" for degeneracy counting, and we sum their orbital degeneracies:

  • For 1s12s11s^1 2s^1 configuration: Terms are 3S^3S and 1S^1S. Both have L=0. Orbital degeneracy = 1. Sum = 1+1 = 2.
  • For 1s12p11s^1 2p^1 configuration: Terms are 3P^3P and 1P^1P. Both have L=1. Orbital degeneracy = 3. Sum = 3+3 = 6.

Let's try to list states by "configuration" and sum the orbital degeneracies of all terms arising from that configuration.

  1. Ground configuration: 1s21s^2. Term: 1S^1S. Orbital degeneracy: 1.
  2. First excited configuration: 1s12s11s^1 2s^1. Terms: 3S,1S^3S, ^1S. Orbital degeneracy for each is 1. Total orbital degeneracy from this configuration = 1 + 1 = 2.
  3. Second excited configuration: 1s12p11s^1 2p^1. Terms: 3P,1P^3P, ^1P. Orbital degeneracy for each is 3. Total orbital degeneracy from this configuration = 3 + 3 = 6.

This interpretation yields 6 for the "degeneracy of the second excited state" (meaning the second excited configuration). This matches one of the options. This approach is consistent with how the degeneracy for H atom (n=3) is calculated by summing orbital degeneracies (1+3+5=91+3+5=9). In H- (a multi-electron system), the different L values (s, p, d) are no longer degenerate, but the singlet and triplet states for a given L are also distinct energy levels. If we interpret "not considering electron spin" as summing the orbital degeneracies (2L+12L+1) of all possible terms arising from a configuration, then 6 is the answer.

Final check of the interpretation: The H atom example (n=3, degeneracy 9) means all 3s,3p,3d3s, 3p, 3d orbitals are degenerate. Their individual orbital degeneracies are 1, 3, 5, which sum to 9. For H-, the electron-electron interaction removes the 'l' degeneracy. So, 2s2s is lower than 2p2p. The first excited configuration is 1s12s11s^1 2s^1. From this, we get 3S^3S and 1S^1S. If we sum their orbital degeneracies, it is 1+1=21+1=2. The second excited configuration is 1s12p11s^1 2p^1. From this, we get 3P^3P and 1P^1P. The orbital degeneracy for each is 3. Summing them (as per the H atom example of summing orbital degeneracies for 3s,3p,3d3s, 3p, 3d) gives 3+3=63+3=6.

This interpretation seems the most plausible given the options and the phrasing for the H atom example.