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 (a) `l_H gt l_(Li) and |E_H| gt |E_(Li)|` (b) `l_H = l_(Li) and |E_H| lt |E_(Li)|`
(C ) `l_H = l_(Li) and |E_H| gt |E_(Li)|` (d) `l_H lt l_(Li) and |E_H| lt|E_(Li)|`

Explanation

To solve the problem, we need to analyze the properties of a hydrogen atom and a lithium ion (Li²⁺) in their second excited state. We will determine the relationships between their electronic angular momenta and energies.

### Step 1: Identify the quantum states
Both the hydrogen atom and the lithium ion are in the second excited state. The second excited state corresponds to the principal quantum number \( n = 3 \) (since the ground state is \( n = 1 \), the first excited state is \( n = 2 \), and the second excited state is \( n = 3 \)).

### Step 2: Calculate the electronic angular momentum
The electronic angular momentum \( L \) for an electron in an atom is given by the formula:
\[ L = n \cdot \hbar \]
where \( n \) is the principal quantum number and \( \hbar \) is the reduced Planck's constant.

For both hydrogen and Li²⁺:
- \( n_H = 3 \)
- \( n_{Li} = 3 \)

Thus, the angular momentum for both atoms is:
\[ L_H = 3 \cdot \hbar \]
\[ L_{Li} = 3 \cdot \hbar \]

This implies:
\[ L_H = L_{Li} \]

### Step 3: Calculate the energies
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

For hydrogen (\( Z_H = 1 \)):
\[ E_H = -\frac{1^2 \cdot 13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

For lithium ion (\( Z_{Li} = 3 \)):
\[ E_{Li} = -\frac{3^2 \cdot 13.6 \, \text{eV}}{3^2} = -\frac{9 \cdot 13.6 \, \text{eV}}{9} = -13.6 \, \text{eV} \]

### Step 4: Compare the energies
Now we can compare the magnitudes of the energies:
\[ |E_H| \approx 1.51 \, \text{eV} \quad \text{and} \quad |E_{Li}| = 13.6 \, \text{eV} \]

Thus, we find:
\[ |E_H| < |E_{Li}| \]

### Conclusion
From the analysis, we have:
- \( L_H = L_{Li} \)
- \( |E_H| < |E_{Li}| \)

Therefore, the correct answer is:
**(b) \( l_H = l_{Li} \) and \( |E_H| < |E_{Li}| \)**.