A hydrogen atom and a `Li^(2+)` ion are both in the second excited state. If `l_H` and `l_(Li)` are their respective electronic angular momenta, and `E_H and E_(Li)` their respective energies, then

Options

A. `l_(H)gtl_(Li) and |E_(H)|gt|E_(Li)|`
B. `l_(H)=l_(Li) and |E_(H)|lt|E_(Li)|`
C. `l_(H)ltl_(Li) and |E_(H)|gt|E_(Li)|`
D. `l_(H)gtl_(Li) and |E_(H)|gtgt|E_(Li)|`

Answer

To solve the problem, we need to analyze the angular momentum and energy of a hydrogen atom and a lithium ion in the second excited state.

### Step-by-Step Solution:

1. **Identify the Quantum Numbers**:
- The second excited state corresponds to the principal quantum number \( n = 3 \) for both the hydrogen atom (H) and the lithium ion (\( Li^{2+} \)).

2. **Calculate Angular Momentum**:
- The angular momentum \( L \) of an electron in a hydrogen-like atom is given by the formula:
\[ L = n \frac{h}{2\pi} \]
- Since both atoms are in the same state (\( n = 3 \)), the angular momentum for both will be:
\[ L_H = L_{Li} = 3 \frac{h}{2\pi} \]
- Therefore, we conclude that:
\[ l_H = l_{Li} \]

3. **Calculate Energy**:
- The energy of an electron in a hydrogen-like atom is given by:
\[ E = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
- For hydrogen (\( Z = 1 \)):
\[ E_H = -\frac{13.6 \times 1^2}{3^2} = -\frac{13.6}{9} \text{ eV} \]
- For lithium (\( Z = 3 \)):
\[ E_{Li} = -\frac{13.6 \times 3^2}{3^2} = -\frac{13.6 \times 9}{9} = -13.6 \text{ eV} \]
- The magnitudes of the energies are:
\[ |E_H| = \frac{13.6}{9} \text{ eV} \quad \text{and} \quad |E_{Li}| = 13.6 \text{ eV} \]
- Since \( |E_{Li}| > |E_H| \), we conclude that:
\[ |E_{Li}| > |E_H| \]

4. **Final Relations**:
- From the above calculations, we have:
\[ l_H = l_{Li} \quad \text{and} \quad |E_{Li}| > |E_H| \]

### Conclusion:
The correct relation is:
- Angular momentum of hydrogen and lithium are equal: \( l_H = l_{Li} \)
- The energy of lithium is greater in magnitude than that of hydrogen: \( |E_{Li}| > |E_H| \)