Final Populated Quantum State in Helium Ion After Collision

To determine the quantum number $n$ of the state finally populated in the $\text{He}^+$ ions, we analyze the energy levels according to the Bohr model:

The energy of an electron in the $n$-th state of a hydrogen-like atom or ion of atomic number $Z$ is given by:
$E_n = -13.6 \, \frac{Z^2}{n^2} \text{ eV}$

1. Excitation Energy of the Hydrogen Atom ($\text{H}$, $Z = 1$):

   • The ground state corresponds to $n = 1$, and its energy is:
     $E_1(\text{H}) = -13.6 \text{ eV}$
   • The first excited state corresponds to $n = 2$, and its energy is:
     $E_2(\text{H}) = -\frac{13.6}{2^2} = -3.4 \text{ eV}$
   • The excitation energy of the $\text{H}$ atom in its first excited state is:
     $\Delta E_{\text{H}} = E_2(\text{H}) - E_1(\text{H}) = -3.4 - (-13.6) = 10.2 \text{ eV}$

2. Initial Energy of the Singly Ionized Helium Ion ($\text{He}^+$, $Z = 2$):

   • The ions are initially excited to their first excited state ($n = 2$).
   • The energy of this initial state is:
     $E_2(\text{He}^+) = -13.6 \times \frac{2^2}{2^2} = -13.6 \text{ eV}$

3. Final Energy State of the $\text{He}^+$ Ion:

   • The $\text{H}$ atoms transfer their total excitation energy ($\Delta E_{\text{H}} = 10.2 \text{ eV}$) to the $\text{He}^+$ ions through collisions.
   • Therefore, the final energy of the $\text{He}^+$ ion becomes:
     $E_n(\text{He}^+) = E_2(\text{He}^+) + \Delta E_{\text{H}} = -13.6 \text{ eV} + 10.2 \text{ eV} = -3.4 \text{ eV}$

4. Determining the Quantum Number $n$:

   • Using the energy formula for $\text{He}^+$:
     $E_n(\text{He}^+) = -13.6 \times \frac{2^2}{n^2} = -\frac{54.4}{n^2} \text{ eV}$
   • Equating this to the final energy:
     $-\frac{54.4}{n^2} = -3.4$
     $n^2 = \frac{54.4}{3.4} = 16 \implies n = 4$

Hence, the quantum number of the state finally populated in the $\text{He}^+$ ions is $4$.

Correct Option: (C)