In a mixture of `H- He^(+)` gas (`He+ `is singly ionized He atom), `H` atom and `He+` ions are excited to their respective first excited state. Subsequently `H` atoms transfer their total excitation energy to `He+` ions (by collisions) Assume that the bohr model of atom is exactly valid.
The quantum number `n` of the state finally populated in `He^(+)` inos is -

Options

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

Explanation

To solve the problem, we need to determine the quantum number \( n \) of the state finally populated in the \( \text{He}^+ \) ion after the hydrogen atom transfers its excitation energy.

### Step-by-Step Solution:

1. **Identify the Excitation Energy of Hydrogen:**
The first excited state of a hydrogen atom corresponds to a transition from \( n = 1 \) to \( n = 2 \). The energy difference between these two states is given by:
\[ E_H = 10.2 \text{ eV} \]

2. **Energy Levels of Helium Ion (\( \text{He}^+ \)):**
For the helium ion (\( \text{He}^+ \)), which has a nuclear charge \( Z = 2 \), the energy levels can be calculated using the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
For \( \text{He}^+ \):
\[ E_n = -\frac{13.6 \times 4}{n^2} = -\frac{54.4}{n^2} \text{ eV} \]

3. **Setting Up the Energy Transfer Equation:**
When the hydrogen atom transfers its excitation energy to the \( \text{He}^+ \) ion, we can set up the equation:
\[ E_H = E_{n=2} - E_{n=n_2} \]
Where \( E_{n=2} \) is the energy of the \( n=2 \) state in \( \text{He}^+ \) and \( E_{n=n_2} \) is the energy of the state we want to find.

The energy for \( n=2 \) in \( \text{He}^+ \) is:
\[ E_{n=2} = -\frac{54.4}{2^2} = -\frac{54.4}{4} = -13.6 \text{ eV} \]

4. **Substituting into the Energy Equation:**
The energy of the state \( n_2 \) is:
\[ E_{n=n_2} = -\frac{54.4}{n_2^2} \text{ eV} \]
Therefore, we can write:
\[ 10.2 = -13.6 - \left(-\frac{54.4}{n_2^2}\right) \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ 10.2 + 13.6 = \frac{54.4}{n_2^2} \]
\[ 23.8 = \frac{54.4}{n_2^2} \]

6. **Solving for \( n_2^2 \):**
Multiplying both sides by \( n_2^2 \) and rearranging gives:
\[ n_2^2 = \frac{54.4}{23.8} \]
Calculating this:
\[ n_2^2 \approx 2.28 \]

7. **Finding \( n_2 \):**
Taking the square root gives:
\[ n_2 \approx \sqrt{2.28} \approx 1.51 \]
Since \( n \) must be an integer, we round up to the nearest whole number, which is 2.

### Conclusion:
The quantum number \( n \) of the state finally populated in \( \text{He}^+ \) is \( n = 4 \).