As per Bohr model , the minimum energy (in eV) required to remove electron from the ground state of doubly ioinized `Li` alom `(Z = 3)` is

Options

A. 1.51
B. 13.6
C. 40.8
D. 122.4

Explanation

To find the minimum energy required to remove an electron from the ground state of a doubly ionized lithium atom (Li²⁺), we can use the Bohr model of the atom. Here’s the step-by-step solution:

### Step 1: Understand the Bohr Model Formula
The energy levels of hydrogen-like atoms are given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where:
- \(E_n\) is the energy of the nth level,
- \(Z\) is the atomic number,
- \(n\) is the principal quantum number.

### Step 2: Identify the Values for Li²⁺
For doubly ionized lithium (Li²⁺):
- The atomic number \(Z = 3\) (since lithium has 3 protons).
- The ground state corresponds to \(n = 1\).

### Step 3: Substitute the Values into the Formula
Now, substituting \(Z\) and \(n\) into the formula:
\[ E_1 = -\frac{13.6 \times 3^2}{1^2} \]

### Step 4: Calculate the Energy
Calculating \(3^2\):
\[ 3^2 = 9 \]
Now substituting this back into the energy formula:
\[ E_1 = -\frac{13.6 \times 9}{1} = -122.4 \, \text{eV} \]

### Step 5: Determine the Ionization Energy
The minimum energy required to remove the electron from the ground state (ionization energy) is equal to the absolute value of the ground state energy:
\[ \text{Ionization Energy} = |E_1| = 122.4 \, \text{eV} \]

### Final Answer
Thus, the minimum energy required to remove an electron from the ground state of a doubly ionized lithium atom is:
\[ \boxed{122.4 \, \text{eV}} \]

---