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

Options

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

Explanation

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

### Step 1: Understand the formula for the radius in Bohr's model
The radius of an electron in the nth orbit of a hydrogen-like atom is given by the formula:
\[ r_n = \frac{a_0 n^2}{Z} \]
where:
- \( r_n \) is the radius of the nth orbit,
- \( a_0 \) is the Bohr radius,
- \( n \) is the principal quantum number (1 for ground state),
- \( Z \) is the atomic number of the ion.

### Step 2: Identify the values for lithium ion
For \( \text{Li}^{++} \):
- The atomic number \( Z \) of lithium (Li) is 3.
- In the ground state, the principal quantum number \( n = 1 \).
- The Bohr radius \( a_0 = 53 \) pm (picometers).

### Step 3: Substitute the values into the formula
Now, substituting the values into the formula:
\[ r_1 = \frac{a_0 \cdot n^2}{Z} = \frac{53 \, \text{pm} \cdot 1^2}{3} \]

### Step 4: Calculate the radius
Calculating the above expression:
\[ r_1 = \frac{53 \, \text{pm}}{3} \approx 17.67 \, \text{pm} \]

### Step 5: Round the answer
Rounding to two decimal places, we get:
\[ r_1 \approx 18 \, \text{pm} \]

### Final Answer
The radius of the \( \text{Li}^{++} \) ion in its ground state is approximately **18 pm**.

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