An orbit electron in the ground state of hydrogen has an angular momentum `L_(1)`, and an orbital electron in the first orbit in the ground state of lithium (dounle ionised positively) has an angular momentum `L_(2)`. Then :

Options

A. `L_(1) = L_(2)`
B. `L_(1) = 3 L_(2)`
C. `L_(2) = 3L_(1)`
D. `L_(2) = 9 L_(1)`

Explanation

To solve the problem, we need to determine the angular momentum \( L_1 \) of an electron in the ground state of hydrogen and \( L_2 \) of an electron in the first orbit of lithium (double ionized). We will use Bohr's model of the atom to find these angular momenta.

### Step-by-Step Solution:

1. **Understanding Angular Momentum in Bohr's Model**:
According to Bohr's model, the angular momentum \( L \) of an electron in an orbit is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number and \( h \) is Planck's constant.

2. **Calculating \( L_1 \) for Hydrogen**:
For hydrogen, in the ground state, the principal quantum number \( n \) is 1. Therefore, we can calculate \( L_1 \):
\[ L_1 = 1 \cdot \frac{h}{2\pi} = \frac{h}{2\pi} \]

3. **Calculating \( L_2 \) for Lithium (Li\(^{++}\))**:
Lithium double ionized (Li\(^{++}\)) means it has lost two electrons and has one electron left. In the ground state, this remaining electron also occupies the first orbit, so \( n = 1 \) for Li\(^{++}\) as well. Thus, we can calculate \( L_2 \):
\[ L_2 = 1 \cdot \frac{h}{2\pi} = \frac{h}{2\pi} \]

4. **Comparing \( L_1 \) and \( L_2 \)**:
Now, we can compare the two angular momenta:
\[ L_1 = \frac{h}{2\pi} \quad \text{and} \quad L_2 = \frac{h}{2\pi} \]
Therefore, we conclude that:
\[ L_1 = L_2 \]

### Final Result:
The relationship between the angular momenta is:
\[ L_1 = L_2 \]