Consider atoms `H, He^(+), Li^(++)` in their ground states. If `L_(1), L_(2)` and `L_(3)` are magnitude of angular momentum of their electrons about the nucleus respectively then:

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

Options

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

Explanation

To solve the problem, we need to analyze the angular momentum of the electrons in the ground states of the hydrogen atom (H), the singly ionized helium atom (He⁺), and the doubly ionized lithium atom (Li²⁺).

### Step-by-Step Solution:

1. **Identify the Principal Quantum Numbers (n)**:
- In the ground state, the principal quantum number \( n \) for hydrogen (H) is 1.
- For helium (He⁺), which has one electron in the ground state, \( n \) is also 1.
- For lithium (Li²⁺), which has two electrons but only one is in the ground state (the outermost electron), \( n \) is still 1.

2. **Angular Momentum Formula**:
- The magnitude of angular momentum \( L \) of an electron in an atom is given by the formula:
\[ L = n \frac{h}{2\pi} \]
- Here, \( h \) is Planck's constant and \( n \) is the principal quantum number.

3. **Calculate Angular Momentum for Each Atom**:
- For hydrogen (H):
\[ L_1 = 1 \cdot \frac{h}{2\pi} = \frac{h}{2\pi} \]
- For helium (He⁺):
\[ L_2 = 1 \cdot \frac{h}{2\pi} = \frac{h}{2\pi} \]
- For lithium (Li²⁺):
\[ L_3 = 1 \cdot \frac{h}{2\pi} = \frac{h}{2\pi} \]

4. **Compare the Angular Momenta**:
- We find that:
\[ L_1 = L_2 = L_3 = \frac{h}{2\pi} \]

5. **Conclusion**:
- Since the angular momenta of all three atoms are equal, we conclude that:
\[ L_1 = L_2 = L_3 \]

### Final Answer:
The correct option is **A. \( L_1 = L_2 = L_3 \)**.