The radus of hydrogen atom in the ground state `0.53 Å`. The radius of `Li^(2+)` ion (Atomic number = 3) in a similar state is

Options

A. 0.13
B. 0.06
C. 4.77
D. 2.12

Explanation

To find the radius of the Li²⁺ ion in the ground state, we can use the formula derived from the Bohr model of the atom. The radius of an electron orbit in a hydrogen-like atom is given by:

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

Where:
- \( R_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number (for ground state, \( n = 1 \)),
- \( Z \) is the atomic number,
- \( R_1 \) is the radius of the hydrogen atom in the ground state (which is given as \( 0.53 \, \text{Å} \)).

### Step-by-Step Solution:

1. **Identify the values:**
- For hydrogen (\( H \)):
- Atomic number \( Z_H = 1 \)
- Ground state radius \( R_1 = 0.53 \, \text{Å} \)
- For lithium ion (\( Li^{2+} \)):
- Atomic number \( Z_{Li} = 3 \)
- Ground state \( n = 1 \)

2. **Write the formula for the radius of Li²⁺:**
\[ R_1 (Li^{2+}) = \frac{n^2}{Z} R_1 (H) \]

3. **Substitute the known values into the formula:**
\[ R_1 (Li^{2+}) = \frac{1^2}{3} \times 0.53 \, \text{Å} \]

4. **Calculate the radius:**
\[ R_1 (Li^{2+}) = \frac{1}{3} \times 0.53 \, \text{Å} = 0.1767 \, \text{Å} \]

5. **Final Answer:**
The radius of the \( Li^{2+} \) ion in the ground state is approximately \( 0.1767 \, \text{Å} \).