Step-by-step Solution:

1. Understand the Bohr Model Energy Formula The energy of an electron in the n-th orbit for a hydrogen-like atom with atomic number Z is given by the Bohr model formula: $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$ where $n$ is the principal quantum number.

2. Calculate the Excitation Energy of the Hydrogen Atom For a Hydrogen atom (H), the atomic number is $Z=1$.

   • The ground state energy (n=1) is: $E_{1,H} = -13.6 \frac{1^2}{1^2} = -13.6 \text{ eV}$
   • The first excited state (n=2) is the state to which the H atoms are initially excited. Its energy is: $E_{2,H} = -13.6 \frac{1^2}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$
   • The total excitation energy of the H atom is the energy difference between the first excited state and the ground state. This is the amount of energy that the H atom will transfer when it de-excites to its ground state. $\Delta E_H = E_{2,H} - E_{1,H} = (-3.4 \text{ eV}) - (-13.6 \text{ eV}) = 10.2 \text{ eV}$

3. Determine the Initial Energy of the He$^+$ ion For a singly ionized Helium ion (He$^+$), the atomic number is $Z=2$.

   • The He$^+$ ions are also initially in their first excited state, which corresponds to $n=2$.
   • The energy of the He$^+$ ion in its first excited state is: $E_{initial, He^+} = E_{2,He^+} = -13.6 \frac{2^2}{2^2} = -13.6 \frac{4}{4} = -13.6 \text{ eV}$

4. Calculate the Final Energy of the He$^+$ ion after Collision The problem states that the H atom transfers its total excitation energy (10.2 eV) to the He$^+$ ion upon collision. The He$^+$ ion, which is already in its first excited state, absorbs this energy.

   • The final energy of the He$^+$ ion will be its initial energy plus the energy it absorbed: $E_{final, He^+} = E_{initial, He^+} + \Delta E_H$ $E_{final, He^+} = -13.6 \text{ eV} + 10.2 \text{ eV} = -3.4 \text{ eV}$

5. Find the Quantum Number (n) of the Final State of the He$^+$ ion Now we need to find the quantum number 'n' that corresponds to this final energy for the He$^+$ ion.

   • Using the Bohr energy formula for He$^+$ ($Z=2$): $E_{n,He^+} = -13.6 \frac{2^2}{n^2} = -\frac{54.4}{n^2} \text{ eV}$
   • We set this equal to the final energy we calculated: $E_{final, He^+} = E_{n,He^+}$ $-3.4 \text{ eV} = -\frac{54.4}{n^2} \text{ eV}$
   • Solving for $n^2$: $n^2 = \frac{-54.4}{-3.4} = \frac{54.4}{3.4} = 16$
   • Taking the square root to find n: $n = \sqrt{16} = 4$

Thus, the quantum number of the state finally populated in He$^+$ ions is 4.