The radius of the second Bohr for `Li^(2+)` is

Options

A. `0.529 xx (4)/(3) Å`
B. `0.529 xx (2)/(3) Å`
C. `0.529 xx (4)/(9) Å`
D. `0.529 xx (2)/(9) Å`

Explanation

To find the radius of the second Bohr orbit for the ion \( \text{Li}^{2+} \), we can use the formula for the radius of the nth orbit in a hydrogen-like atom:

\[ R_n = R_1 \cdot \frac{n^2}{Z} \]

Where:
- \( R_n \) is the radius of the nth orbit,
- \( R_1 \) is the radius of the first orbit for hydrogen (which is \( 0.529 \) Å),
- \( n \) is the principal quantum number (for the second orbit, \( n = 2 \)),
- \( Z \) is the atomic number of the element.

### Step-by-Step Solution:

1. **Identify the atomic number (Z) of \( \text{Li}^{2+} \)**:
- Lithium (Li) has an atomic number of \( 3 \).

2. **Use the formula for the radius of the nth orbit**:
- We need to find \( R_2 \) (the radius of the second orbit), so we will set \( n = 2 \).

3. **Substitute the values into the formula**:
\[ R_2 = R_1 \cdot \frac{n^2}{Z} \]
\[ R_2 = 0.529 \, \text{Å} \cdot \frac{2^2}{3} \]

4. **Calculate \( n^2 \)**:
\[ n^2 = 2^2 = 4 \]

5. **Substitute \( n^2 \) into the formula**:
\[ R_2 = 0.529 \, \text{Å} \cdot \frac{4}{3} \]

6. **Perform the multiplication**:
\[ R_2 = 0.529 \, \text{Å} \cdot 1.3333 \approx 0.7053 \, \text{Å} \]

7. **Final result**:
The radius of the second Bohr orbit for \( \text{Li}^{2+} \) is approximately \( 0.7053 \, \text{Å} \).