Question:
The degeneracy of a hydrogen atom whose energy equals \(-R_H/16\) is:
The degeneracy of a hydrogen atom whose energy equals \(-R_H/16\) is:
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For hydrogenic (one-electron) species like \(\text{H}\), \(\text{He}^+\), and \(\text{Li}^{2+}\), all subshells within a major shell are degenerate, meaning total degeneracy is simply \(n^2\). Note that if a problem explicitly mentions "including spin", you must multiply the result by \(2\), making it \(2n^2\).
Updated On: May 25, 2026
- \(8\)
- \(9\)
- \(16\)
- \(10\)
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The Correct Option is C
Solution and Explanation
Concept:
In quantum mechanics, degeneracy refers to the total number of distinct quantum states (orbitals) that share the exact same energy level. For a single-electron species like the hydrogen atom, the energy of a shell depends solely on its principal quantum number \(n\) according to Bohr's model:
\[
E_n = -\frac{R_H}{n^2}
\]
For a given principal quantum number \(n\), the total number of atomic orbitals (ignoring electron spin unless specified) is given by the mathematical summation of angular momentum states, which simplifies to \(n^2\).
Step 1: Determining the principal quantum number \(n\).
We are given that the energy of the hydrogen atom is: \[ E = -\frac{R_H}{16} \] Comparing this to the standard energy formula \(E_n = -\frac{R_H}{n^2}\): \[ n^2 = 16 \quad \Rightarrow \quad n = 4 \] This indicates that the electron resides in the fourth major energy shell (\(n = 4\)).
Step 2: Calculating the degeneracy of the shell.
The degeneracy (number of orbitals) for a single-electron system for a shell \(n\) is calculated using the formula: \[ \text{Degeneracy} = n^2 \] Substituting \(n = 4\) into the relation: \[ \text{Degeneracy} = 4^2 = 16 \] Alternatively, we can verify this by summing the subshells for \(n = 4\):
• For \(l = 0\) (\(s\)-subshell): \(1\) orbital (\(4s\))
• For \(l = 1\) (\(p\)-subshell): \(3\) orbitals (\(4p_x, 4p_y, 4p_z\))
• For \(l = 2\) (\(d\)-subshell): \(5\) orbitals (\(4d_{xy}, 4d_{yz}, 4d_{xz}, 4d_{x^2-y^2}, 4d_{z^2}\))
• For \(l = 3\) (\(f\)-subshell): \(7\) orbitals \[ \text{Total Orbitals} = 1 + 3 + 5 + 7 = 16 \]
We are given that the energy of the hydrogen atom is: \[ E = -\frac{R_H}{16} \] Comparing this to the standard energy formula \(E_n = -\frac{R_H}{n^2}\): \[ n^2 = 16 \quad \Rightarrow \quad n = 4 \] This indicates that the electron resides in the fourth major energy shell (\(n = 4\)).
Step 2: Calculating the degeneracy of the shell.
The degeneracy (number of orbitals) for a single-electron system for a shell \(n\) is calculated using the formula: \[ \text{Degeneracy} = n^2 \] Substituting \(n = 4\) into the relation: \[ \text{Degeneracy} = 4^2 = 16 \] Alternatively, we can verify this by summing the subshells for \(n = 4\):
• For \(l = 0\) (\(s\)-subshell): \(1\) orbital (\(4s\))
• For \(l = 1\) (\(p\)-subshell): \(3\) orbitals (\(4p_x, 4p_y, 4p_z\))
• For \(l = 2\) (\(d\)-subshell): \(5\) orbitals (\(4d_{xy}, 4d_{yz}, 4d_{xz}, 4d_{x^2-y^2}, 4d_{z^2}\))
• For \(l = 3\) (\(f\)-subshell): \(7\) orbitals \[ \text{Total Orbitals} = 1 + 3 + 5 + 7 = 16 \]
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