The degeneracy of hydrogen atom that has equal energy to \[\dfrac{{ - {R_H}}}{9}\] is:
[where \[{R_H}\] = Rydberg constant]
(A) 6
(B) 8
(C) 5
(D) 9

Hint: when understanding quantum mechanics, we understand that an energy level is said to be degenerate, if it is corresponding to two or more different measurable states of a given quantum system. On the other hand, the different states of a quantum system can be categorised as degenerate if they give the same value of energy upon measurement. The number of different states corresponding to a particular energy level is known as the degree of degeneracy of the level

Formula Used: E = \[\dfrac{{ - {R_H}}}{{{n^2}}}\]

Complete Step-by-Step Solution:
In order to understand the degeneracy of the hydrogen atom in the given energy state, we need to first understand the orbital to which the given electron belongs to, the state in which it is present and the number of degenerate orbitals present in that state.
The energy has been given to be equal to \[\dfrac{{ - {R_H}}}{9}\].
But we know that, energy can be obtained using the formula:
E = \[\dfrac{{ - {R_H}}}{{{n^2}}}\] = \[\dfrac{{ - {R_H}}}{9}\]
Hence the value of n = 3.
This means that the electron belongs to the third orbital. Now in the third orbital, we have 3s, 3p and 3d subshells present. Now the quantum numbers associated with each of these subshells can be identified as:
3s: l = 0; m = 0; hence 3s has 1 orbital
3p: l = 1; m = -1, 0, +1; hence 3p has 3 orbitals
3d: l = 2; m = -2, -1, 0, +1, +2; hence 3d has 5 orbitals
Thus the total number of degenerate orbitals present in the third shell are 1 + 3 + 5 = 9 degenerate orbitals. Hence the degeneracy of the given hydrogen atom is 9.

Hence, Option D is the correct option.

Note: From Schrodinger’s wave equations, we can derive certain quantities that describe the size, shape and orientation in space of the orbitals of the atoms. These quantities are known as quantum numbers.