What is the degeneracy of the level of the hydrogen atom that has the energy \[ - \dfrac{{{R_H}}}{9}\] ?
Answer
271.5k+ 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.
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