The degencracy of 1st excited state of `H` atom is ______ (Ignore efffect of spin)

Options

A. 2
B. 3
C. 4
D. 8

Answer

To find the degeneracy of the first excited state of the hydrogen atom, we can follow these steps:

### Step 1: Identify the Principal Quantum Number
The ground state of the hydrogen atom corresponds to the principal quantum number \( n = 1 \). The first excited state corresponds to \( n = 2 \).

**Hint:** The principal quantum number \( n \) indicates the energy level of an electron in an atom.

### Step 2: Determine the Subshells for \( n = 2 \)
For \( n = 2 \), the possible subshells are:
- 2s
- 2p

**Hint:** Each principal quantum number \( n \) can have subshells that are designated by letters (s, p, d, f, etc.).

### Step 3: Count the Orbitals in Each Subshell
- The 2s subshell has **1 orbital**.
- The 2p subshell has **3 orbitals** (2px, 2py, 2pz).

**Hint:** Each type of subshell has a specific number of orbitals: s has 1, p has 3, d has 5, and f has 7.

### Step 4: Calculate the Total Number of Degenerate Orbitals
The total number of orbitals in the first excited state (n = 2) is:
- 1 (from 2s) + 3 (from 2p) = 4 orbitals.

**Hint:** Degeneracy refers to the number of orbitals that have the same energy level.

### Step 5: Conclusion
Since we are ignoring the effect of spin, all these orbitals are considered to be of equal energy. Therefore, the degeneracy of the first excited state of the hydrogen atom is **4**.

**Final Answer:** The degeneracy of the first excited state of the hydrogen atom is **4**.

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