Taking the Bohr radius `a_(0) = 53` pm, the radius of `Li^(++)` ion in its gnround state, on the basis of Bohr's model, will be about.

Options

A. 53 pm
B. 27 pm
C. 18 pm
D. 13 pm

Explanation

To find the radius of the \( \text{Li}^{++} \) ion in its ground state using Bohr's model, we can use the formula for the radius of an electron in a hydrogen-like atom:

\[ r_n = \frac{n^2 a_0}{Z} \]

where:
- \( r_n \) is the radius of the electron's orbit,
- \( n \) is the principal quantum number (for the ground state, \( n = 1 \)),
- \( a_0 \) is the Bohr radius (given as \( 53 \) pm),
- \( Z \) is the atomic number of the ion.

### Step-by-Step Solution:

1. **Identify the atomic number \( Z \)**:
For lithium (\( \text{Li} \)), the atomic number is \( 3 \). Since we are considering the \( \text{Li}^{++} \) ion, which has lost two electrons, the effective atomic number for the remaining electron is still \( 3 \).

2. **Determine the principal quantum number \( n \)**:
For the ground state, \( n = 1 \).

3. **Substitute the values into the formula**:
Using the formula \( r_n = \frac{n^2 a_0}{Z} \):
\[ r_1 = \frac{1^2 \cdot 53 \text{ pm}}{3} \]

4. **Calculate the radius**:
\[ r_1 = \frac{53 \text{ pm}}{3} = 17.67 \text{ pm} \]

Thus, the radius of the \( \text{Li}^{++} \) ion in its ground state is approximately \( 17.67 \) pm.

### Final Answer:
The radius of the \( \text{Li}^{++} \) ion in its ground state is about \( 17.67 \) pm.