# Degeneracy

This problem was asked in JEE 2015. It states:
If the degeneracy of hydrogen atom in second excited state ($n=3$) is 9 then find the degeneracy of hydride anion i.e. $\ce{H^-}$ ion in second excited state.

Note: electronic spin is not to be considered here.

Note by Dipanjan Chowdhury
3 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $ ... $ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Before I answer this, let me ask you a question that will be key to understanding what's happening here:
Why is the degeneracy of hydrogen atom 9 when n = 3?

- 3 years, 3 months ago

I think because of the one 3s three 3p and five 3d orbitals which are not excited as have no electron and as a whole exhibit degeneracy

- 3 years, 3 months ago

Okay, good! The fact that hydrogen has only one electron means that orbitals with the same value of n have the same energies. n = 3 has 9 orbitals, as you said, so 9 is the degeneracy.
But for H-, there are two electrons. The orbitals' energies now depend on l (the azimuthal quantum number that differentiates between s,p, d, f and so on.) like, $1s < 2s < 2p < 3s < 3p ...$
Now the ground state of H- is $1s^2$.
The 1st excited state is $1s^1 2s^1$.
The second excited state is $1s^1 2s^0 2p^1$ (Compare with H atom for which the 2nd excited state is $3s^1$.)
But there are 3 p orbitals for the electron to 'choose' between when making the transition from first excited state to second excited state.
So the degeneracy is 3. Hope this helps.

- 3 years, 3 months ago

Absolutely clear, Sir. Thank You . :D I was unaware of this sort of splitting.

- 3 years, 3 months ago

Sir isn't the ground state of an atom 1s¹

- 2 years, 2 months ago

Yes .

- 2 years, 2 months ago

1s$^2$ is the ground state electronic configuration. So, 1s$^1$ 2s$^1$ is the first excited state and 1s$^1$ 2p$^1$ is the second excited state. Now in 2p, all 3 orientations are degenerate so the degeneracy is 3.

- 2 years ago

3

- 1 year, 1 month ago

3

- 12 months ago