
What is the degeneracy of the level of the hydrogen atom that has the energy \[ - \dfrac{{{R_H}}}{9}\] ?
Answer
221.4k+ views
Hint: From quantum mechanics, it can be understood 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 categorized 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:
\[\dfrac{{ - {R_H}}}{{{n^2}}}\] where n is the level of the energy state.
Complete step by step answer:
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 \[\dfrac{{ - {R_H}}}{{{n^2}}}\] .
But we know that, energy can be obtained using the formula:
\[E{\text{ }} = \;\dfrac{{ - {R_H}}}{{{n^2}}} = \dfrac{{ - {R_H}}}{9}\]
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
Therefore, 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.
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. The quantities derived from Schrodinger’s wave equations are represented by the letters l, m and n. These letters stand for the quantities mentioned as Orbital angular momentum quantum number, Magnetic quantum number and Principal quantum number respectively.
Formula used:
\[\dfrac{{ - {R_H}}}{{{n^2}}}\] where n is the level of the energy state.
Complete step by step answer:
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 \[\dfrac{{ - {R_H}}}{{{n^2}}}\] .
But we know that, energy can be obtained using the formula:
\[E{\text{ }} = \;\dfrac{{ - {R_H}}}{{{n^2}}} = \dfrac{{ - {R_H}}}{9}\]
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
Therefore, 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.
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. The quantities derived from Schrodinger’s wave equations are represented by the letters l, m and n. These letters stand for the quantities mentioned as Orbital angular momentum quantum number, Magnetic quantum number and Principal quantum number respectively.
Recently Updated Pages
The hybridization and shape of NH2 ion are a sp2 and class 11 chemistry JEE_Main

What is the pH of 001 M solution of HCl a 1 b 10 c class 11 chemistry JEE_Main

Aromatization of nhexane gives A Benzene B Toluene class 11 chemistry JEE_Main

Show how you will synthesise i 1Phenylethanol from class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

Which of the following Compounds does not exhibit tautomerism class 11 chemistry JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Other Pages
NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

NCERT Solutions ForClass 11 Chemistry Chapter Chapter 5 Thermodynamics

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

