Quantum Analysis of Hydrogen Atom in n = 2 State

Quantum Description of a Hydrogen Atom in State n = 2

Question:

A hydrogen atom is in a state with principal quantum number n = 2.

What are the energy, possible orbital angular momentum quantum numbers (ℓ), magnetic quantum numbers (m), and the total degeneracy of this energy level?

1. Energy of the State

The energy levels of a hydrogen atom are given by:

E_n = −13.6 eV / n²

For n = 2:

E₂ = −13.6 eV / 4 = −3.4 eV

2. Allowed Orbital Angular Momentum Quantum Numbers (ℓ)

The allowed values of the orbital angular momentum quantum number ℓ for a given n range from 0 to n−1. Thus, for n = 2:

ℓ = 0, 1

These correspond to the 2s (ℓ = 0) and 2p (ℓ = 1) orbitals.

3. Magnetic Quantum Numbers (m)

For each value of ℓ, the magnetic quantum number m takes values from −ℓ to +ℓ in integer steps:

For ℓ = 0: m = 0 → one orbital (2s)
For ℓ = 1: m = −1, 0, +1 → three orbitals (2p)

4. Degeneracy of the Level

The total number of distinct quantum states (degeneracy) for a given principal quantum number n is given by the total combinations of ℓ and m values:

1 state from ℓ = 0 (2s)
3 states from ℓ = 1 (2p)

Degeneracy (g_n=2) = 1 + 3 = 4

🔍 Summary:

Energy (E₂): −3.4 eV
Allowed ℓ values: 0 (2s), 1 (2p)
Allowed m values:
- ℓ = 0 → m = 0
- ℓ = 1 → m = −1, 0, +1
Total Degenerate States: 4

learnlyfly