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

For hydrogen hydrogen like atoms 
 $E_n=\frac{-13.6z^2}{n^2}eV/atom$
For hydrogen atom  $E_1=-13.6eV(forn=1)$
 $(Z=1)E_2=-3.4eV(forn=2)$
 $\therefore \Delta E=E_2-E_1=-3.4-(-13.6)=10.2eV$
i.e., When hydrogen comes to ground state from its first excited state it will release 10.2 eV 
of energy.
$For H{e}^{+}ion E_1=-13.6\times 4eV=-54.4eV(forn=1)$

$$\begin{gathered} E_2=-13.6\times eV(forn=2) \\ {E}_3=-6.04eV(forn=3) \\ {E}_4=-3.4eV(forn=4) \end{gathered}$$

Here $H{e}^{+}$ ion is in the first excited state i.e..possessing energy  $-13.6eV$. After receiving 
energy  of  $+10.2eV$ from excited hydrogen atom on collision, the energy of $(-13.610.2)eV=-3.4eV.$ electron will be  Hence the quantum number of the state finally populated in $H{e}^{+}$ ions , n=4.