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)|`

Options

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"|`

Answer

To solve the problem, we will analyze the angular momentum and energy of a hydrogen atom and a lithium ion \( \text{Li}^{2+} \) in their second excited states.

### Step 1: Determine the quantum numbers for the second excited state
The second excited state corresponds to \( n = 3 \) for both the hydrogen atom and the lithium ion.

### Step 2: Calculate the angular momentum
According to Bohr's model, the angular momentum \( l \) of an electron in orbit is given by:
\[ l = n \frac{h}{2\pi} \]
For both the hydrogen atom and the lithium ion in the second excited state (\( n = 3 \)):
\[ l_H = 3 \frac{h}{2\pi} \]
\[ l_{Li} = 3 \frac{h}{2\pi} \]
Thus, we find:
\[ l_H = l_{Li} \]

### Step 3: Calculate the energy of the hydrogen atom and lithium ion
The energy of an electron in a hydrogen-like atom is given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
For hydrogen (\( Z = 1 \)):
\[ E_H = -\frac{13.6 \cdot 1^2}{3^2} = -\frac{13.6}{9} \text{ eV} \approx -1.51 \text{ eV} \]
For lithium (\( Z = 3 \)):
\[ E_{Li} = -\frac{13.6 \cdot 3^2}{3^2} = -\frac{13.6 \cdot 9}{9} = -13.6 \text{ eV} \]

### Step 4: Compare the magnitudes of the energies
Now we compare the absolute values of the energies:
\[ |E_H| \approx 1.51 \text{ eV} \]
\[ |E_{Li}| = 13.6 \text{ eV} \]
Thus, we find:
\[ |E_H| < |E_{Li}| \]

### Conclusion
From the calculations, we have:
- \( l_H = l_{Li} \)
- \( |E_H| < |E_{Li}| \)

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

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