The quantum number n of the state finally populated in He? ions is [2008]

Question

In a mixture of ${H–He}^{+}$ gas ( ${He}^{+}$ is a singly ionized He atom), H atoms and ${He}^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to ${He}^{+}$ ions (by collisions). Assume that the Bohr model of the atom is exactly valid. [2008]

Q. The quantum number $n$ of the state finally populated in ${He}^{+}$ ions is ________.

Smart Achievers · Accepted Answer

(3)

For hydrogen or hydrogen-like atoms,

$E_{n} = \frac{- 13.6 Z^{2}}{n^{2}} eV/atom$

For hydrogen atom $E_{1} = - 13.6 eV$ $(for n = 1)$

$(Z = 1),$ $E_{2} = - 3.4 eV$ $(for n = 2)$

$∴$ $Δ E = E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV$

i.e., when hydrogen comes to the ground state from its first excited state, it releases 10.2 eV of energy.

For ${He}^{+}$ ion $E_{1} = - 13.6 \times 4 eV = - 54.4 eV$ $(for n = 1)$

$(Z = 2)$ $E_{2} = - 13.6 eV$ $(for n = 2)$

$E_{3} = - 6.04 eV$ $(for n = 3)$

$E_{4} = - 3.4 eV$ $(for n = 4)$

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

Solution

1 2

2 3

3 4

4 5

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