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

Options

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

Explanation

To find the radius of the second Bohr orbit for the lithium ion \( Li^{2+} \), we can use the formula derived from Bohr's model of the atom. The radius of the \( n \)-th orbit in a hydrogen-like atom is given by:

\[ r_n = \frac{4 \pi \epsilon_0 \hbar^2 n^2}{m Z e^2} \]

Where:
- \( n \) is the principal quantum number (orbit number),
- \( Z \) is the atomic number,
- \( \hbar \) is the reduced Planck's constant,
- \( m \) is the mass of the electron,
- \( e \) is the charge of the electron,
- \( \epsilon_0 \) is the permittivity of free space.

### Step-by-Step Solution:

1. **Identify the values needed**:
- For \( Li^{2+} \):
- Atomic number \( Z = 3 \)
- Principal quantum number for the second orbit \( n = 2 \)

2. **Use the radius formula for hydrogen-like atoms**:
The radius for the hydrogen atom (where \( Z = 1 \)) is given as:
\[ r_1 = 0.529 \, \text{Å} \quad \text{(for } n = 1 \text{)} \]

3. **Calculate the radius for the second orbit**:
The radius for the \( n \)-th orbit in a hydrogen-like atom can be expressed as:
\[ r_n = r_1 \frac{n^2}{Z} \]
Substituting the known values:
\[ r_2 = 0.529 \, \text{Å} \times \frac{2^2}{3} \]

4. **Perform the calculations**:
\[ r_2 = 0.529 \, \text{Å} \times \frac{4}{3} = 0.529 \, \text{Å} \times 1.3333 \approx 0.705 \, \text{Å} \]

5. **Final Result**:
Therefore, the radius of the second Bohr orbit for \( Li^{2+} \) is approximately:
\[ r_2 \approx 0.705 \, \text{Å} \]