Find the radius of `Li^(++)` ions in its ground state assuming Bohr's model to be valid.

Options

A. `12xx10^(-10)m`
B. `12xx10^(-12)m`
C. `16xx10^(-10)m`
D. `18xx10^(-12)m`.

Answer

To find the radius of the `Li^(++)` ion in its ground state using Bohr's model, we can follow these steps:

### Step 1: Understand the Bohr's Model Formula
According to Bohr's model, the radius of the nth orbit of an electron in a hydrogen-like atom is given by the formula:
\[ r_n = \frac{0.529 \, n^2}{Z} \, \text{Å} \]
where:
- \( r_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number (1 for ground state),
- \( Z \) is the atomic number of the element.

### Step 2: Identify the Parameters for `Li^(++)`
For the lithium ion `Li^(++)`:
- The atomic number \( Z \) of lithium (Li) is 3.
- Since we are looking for the ground state, the principal quantum number \( n \) is 1.

### Step 3: Substitute the Values into the Formula
Now, we substitute \( n = 1 \) and \( Z = 3 \) into the formula:
\[ r_1 = \frac{0.529 \times 1^2}{3} \, \text{Å} \]
\[ r_1 = \frac{0.529}{3} \, \text{Å} \]

### Step 4: Calculate the Radius
Calculating the above expression:
\[ r_1 = 0.1763 \, \text{Å} \]
To convert angstroms to meters, we use the conversion \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):
\[ r_1 = 0.1763 \times 10^{-10} \, \text{m} = 1.763 \times 10^{-11} \, \text{m} \]

### Step 5: Final Result
Thus, the radius of the `Li^(++)` ion in its ground state is approximately:
\[ r \approx 1.76 \times 10^{-11} \, \text{m} \]