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What is the degeneracy of the level of hydrogen atom that has energy$(\dfrac{-{{R}_{H}}}{9})$?
(A) 16
(B) 9
(C) 4
(D) 1

seo-qna
Last updated date: 17th May 2024
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Answer
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Hint:Hydrogen atom is a uni-electronic system. It contains only one electron and one proton. The repulsive forces due to electrons are absent in hydrogen atoms. Degeneracy of level means that the orbitals are of equal energy in a particular sub-shell.

Complete step by step solution:
Let’s see the answer to the given question:
We know that the energy is inversely proportional to the square of the level of the shell in which the electron is present.
That is, $E\propto -\dfrac{1}{{{n}^{2}}}$
$\Rightarrow \,\,E=-\dfrac{{{R}_{H}}}{{{n}^{2}}}$…..equation 1
Now, it is given in the question that:
$E=-\dfrac{{{R}_{H}}}{9}$
$E=-\dfrac{{{R}_{H}}}{{{3}^{2}}}$…..equation 2
On comparing equation 1 and equation 2
We get, n=3
Therefore, the electron in in the third level or shell of the hydrogen atom
Now, we know that the azimuthal quantum number ‘l’ gives the number of subshells and the magnetic quantum number ‘m’ gives the number of orbitals present in a shell.
Now, for n=3
l = 0 and m = 0
l = 1 and m = +1, 0, -1
l = 2 and m = -2, -1, 0, +1, +2
So, the total number of degenerate orbitals = 1+3+5 = 9
Hence, the answer of the given question is option (B).

Note: Degeneracy of orbitals means that the orbitals are of equal energy. Such orbitals are called degenerate orbitals. In hydrogen the level of energy degeneracy is as follows:
1s, 2s = 2p, 3s = 3p = 3d, 4s = 4p = 4d = 4f,…