A hydrogen atom makes a transition from `n=2` to `n=1` and emits a photon. This photon strikes a doubly ionized lithium atom `(Z=3)` in excited state and completely removes the orbiting electron. The least quantum number for the excited stated of the ion for the process is:

Options

A. 2
B. 4
C. 5
D. 3

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the energy of the photon emitted during the transition in the hydrogen atom.
The energy levels of the hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

For the transition from \( n = 2 \) to \( n = 1 \):
- Energy at \( n = 1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]
- Energy at \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]

The energy of the emitted photon is:
\[ E_{\text{photon}} = E_1 - E_2 = -13.6 \, \text{eV} - (-3.4 \, \text{eV}) = 10.2 \, \text{eV} \]

### Step 2: Determine the energy required to remove an electron from the doubly ionized lithium atom.
For a doubly ionized lithium atom (Li\(^{2+}\)), the energy required to remove an electron from the excited state is given by:
\[ E = 13.6 \, \text{eV} \cdot \frac{Z^2}{n^2} \]
where \( Z = 3 \) for lithium.

### Step 3: Set up the inequality for the energy of the emitted photon and the ionization energy.
We need the energy of the photon to be greater than or equal to the ionization energy:
\[ 10.2 \, \text{eV} \geq 13.6 \, \text{eV} \cdot \frac{3^2}{n^2} \]

### Step 4: Rearrange the inequality to solve for \( n^2 \).
Rearranging gives:
\[ n^2 \geq 13.6 \cdot \frac{9}{10.2} \]
Calculating the right side:
\[ n^2 \geq \frac{122.4}{10.2} \approx 12 \]

### Step 5: Take the square root to find \( n \).
Taking the square root:
\[ n \geq \sqrt{12} \approx 3.46 \]

Since \( n \) must be a whole number, the least integer value for \( n \) that satisfies this condition is:
\[ n = 4 \]

### Final Answer
Thus, the least quantum number for the excited state of the ion for the process is:
\[ \boxed{4} \]

---