1. Identify the orbitals from the quantum numbers

Given:

• (A) $n=3,\ l=0,\ m=0 \Rightarrow 3s$
• (B) $n=4,\ l=0,\ m=0 \Rightarrow 4s$
• (C) $n=3,\ l=1,\ m=0 \Rightarrow 3p$
• (D) $n=3,\ l=2,\ m=1 \Rightarrow 3d$

Note: The magnetic quantum number $m$ does not affect energy in the absence of external field. So we only compare using $n$ and $l$.

2. Use the $(n+l)$ rule

For multi-electron atoms, orbital energy increases with increasing value of $(n+l)$. If two orbitals have same $(n+l)$ value, then the orbital with higher $n$ has higher energy.

Now calculate:

• For $3s$: $n+l = 3+0 = 3$
• For $4s$: $n+l = 4+0 = 4$
• For $3p$: $n+l = 3+1 = 4$
• For $3d$: $n+l = 3+2 = 5$

3. Arrange in decreasing order of energy

• Highest energy: $3d$ since $(n+l)=5$
• Next compare $4s$ and $3p$: both have $(n+l)=4$
  • Higher $n$ means higher energy, so $4s > 3p$
• Lowest: $3s$

Therefore, $3d > 4s > 3p > 3s$

So, $(D) > (B) > (C) > (A)$

4. Match with options

This corresponds to Option A.

5. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.