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

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answer is 4.

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Detailed Solution

For hydrogen hydrogen like atoms
$E_{n} = \frac{- 13.6 z^{2}}{n^{2}} e V / a t o m$
For hydrogen atom $E_{1} = - 13.6 e V (f o r n = 1)$
$(Z = 1) E_{2} = - 3.4 e V (f o r n = 2)$
$∴ Δ E = E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 e V$
i.e., When hydrogen comes to ground state from its first excited state it will release 10.2 eV
of energy.
$F o r H e^{+} i o n E_{1} = - 13.6 \times 4 e V = - 54.4 e V (f o r n = 1)$

$E_{2} = - 13.6 \times e V (f o r n = 2) E_{3} = - 6.04 e V (f o r n = 3) E_{4} = - 3.4 e V (f o r n = 4)$

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