Q: Which state of the hydrogen atom has energy \( E = -\dfrac{13.60}{9} \text{ eV} \) and angular momentum \( L = \sqrt{6} \hbar \)?
🔍 Step 1: Use the Energy Level Formula
The energy levels of the hydrogen atom are given by:
E_n = -\dfrac{13.6 \, \text{eV}}{n^2}
We are given:
E = -\dfrac{13.6}{9} \, \text{eV}
So, equating the two expressions:
-\dfrac{13.6}{n^2} = -\dfrac{13.6}{9} \Rightarrow n^2 = 9 \Rightarrow \boxed{n = 3}
This tells us the principal quantum number is n = 3.
🔹 Step 2: Use the Angular Momentum Formula
The angular momentum of an electron in a hydrogen atom is given by:
L = \sqrt{l(l + 1)} \hbar
We are given:
L = \sqrt{6} \hbar
Equating both sides:
\sqrt{l(l + 1)} \hbar = \sqrt{6} \hbar \Rightarrow l(l + 1) = 6
Testing values of l:
- l = 1 → 1(1+1) = 2 ❌
- l = 2 → 2(2+1) = 6 ✅
Therefore, l = 2 corresponds to a d orbital.
✅ Final Answer
The hydrogen atom state is characterized by:
- n = 3
- l = 2
Hence, the correct state is 3d.