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

Explanation

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 formula for the radius in Bohr's model
The radius of the nth orbit in Bohr's model 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 (given as 53 pm),
- \( n \) is the principal quantum number (for ground state, \( n = 1 \)),
- \( Z \) is the atomic number of the ion.

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

### Step 3: Substitute the values into the formula
Now, we can substitute these values into the formula:

\[ r_1 = \frac{53 \, \text{pm} \cdot (1)^2}{3} \]

### Step 4: Calculate the radius
Now, we perform the calculation:

\[ r_1 = \frac{53 \, \text{pm}}{3} = 17.67 \, \text{pm} \]

Rounding this to two decimal places gives us:

\[ r_1 \approx 18 \, \text{pm} \]

### Conclusion
The radius of the `Li^(++)` ion in its ground state is approximately **18 pm**.

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