Question

A hydrogen atom makes a transition from n=2 to n=1 and emits a photon. This photon strikes a doubly ionized lithium atom (z=3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is
A. 3
B. 2
C. 5
D. 4

Answer 1

We are given a hydrogen atom that transits from n = 2 to n = 1 and emits a photon. This emitted photon strikes a doubly ionized lithium atom which is in an excited state and completely removes an electron. We have the equation to calculate energy of hydrogen like species. We know that for the photon to remove an electron from the orbit its energy should be more than the energy required to remove the electron from the orbit.
Formula used:
$E=-13.6\times \dfrac{{{z}^{2}}}{{{n}^{2}}}eV$

Complete answer:
In the question it is said that during the transition of the hydrogen atom from n = 2 to n = 1 it emits a photon. This photon strikes doubly ionized lithium which is in an excited state and removes the orbiting electron.
We know that energy of an electron in hydrogen like species is given by the formula,
$E=-13.6\times \dfrac{{{z}^{2}}}{{{n}^{2}}}eV$, were ‘E’ is the energy, ‘z’ is the atomic number and ‘n’ is the orbit.
Here it is said that the hydrogen atom transits from n = 2 to n =1. Therefore the change in energy due to the transition will be,
$\Delta E={{E}_{2}}-{{E}_{1}}$
We know that atomic number of hydrogen is 1, then
$\Rightarrow {{E}_{2}}=-13.6\times \dfrac{{{1}^{2}}}{{{2}^{2}}}$
$\Rightarrow {{E}_{2}}=-13.6\times \dfrac{1}{4}$
And we also have,
${{E}_{1}}=-13.6\times \dfrac{{{1}^{2}}}{{{1}^{2}}}$
$\Rightarrow {{E}_{1}}=-13.6$
Therefore we get the change in energy as,
$\Rightarrow \Delta E=\dfrac{-13.6}{4}-\left( -13.6 \right)$
$\Rightarrow \Delta E=10.2eV$
This is the energy of the photon which is emitted during the transition of the hydrogen atom.
It is said that this photon strikes a doubly ionized lithium atom in an excited state and it completely removes an electron from the orbit.
Since it is a doubly ionized atom we can say that it is a hydrogen like species and we also know that for the photon to remove the electron completely, the energy of the electron should be greater than the energy required for removing the electron from the orbit to infinity.
We know that $\Delta E$ is the energy of the photon and the energy required for removing the electron from ${{n}^{th}}$ orbit to infinity will be,
${{E}_{\infty }}-{{E}_{n}}$
Since it is hydrogen like species, we can write this as
$\Rightarrow \left( -13.6\times \dfrac{{{z}^{2}}}{{{\infty }^{2}}} \right)-\left( -13.6\times \dfrac{{{z}^{2}}}{{{n}^{2}}} \right)$
We know that for lithium z = 3, therefore,
$\Rightarrow \left( -13.6\times \dfrac{{{3}^{2}}}{{{\infty }^{2}}} \right)-\left( -13.6\times \dfrac{{{3}^{2}}}{{{n}^{2}}} \right)$
$\Rightarrow 0-\left( -13.6\times \dfrac{9}{{{n}^{2}}} \right)$
$\Rightarrow 13.6\times \dfrac{9}{{{n}^{2}}}eV$
Since, for the photon to remove the electron completely, the energy of electron should be greater that the energy required for removing the electron from the orbit to infinity we can write,
$\Delta E>{{E}_{\infty }}-{{E}_{n}}$
From earlier calculations we have,
$\Delta E=10.2eV$
${{E}_{\infty }}-{{E}_{n}}=13.6\times \dfrac{9}{{{n}^{2}}}eV$
Therefore we get,
$\Rightarrow 10.2eV>13.6\times \dfrac{9}{{{n}^{2}}}eV$
$\Rightarrow 10.2\times {{n}^{2}}>13.6\times 9$
$\Rightarrow {{n}^{2}}>\dfrac{13.6\times 9}{10.2}$
$\Rightarrow {{n}^{2}}>12$
$\Rightarrow n>\sqrt{12}$
From this we get $n=3.5$. By rounding this to the nearest possible integer we get $n=4$.
Therefore the least quantum number for the excited state is 4.

So, the correct answer is “Option D”.

Note:
Here we say that the doubly ionized lithium atom is hydrogen like atom.
We know that ionization is the process where one or more electrons are added or removed from an atom.
We know that lithium has 3 electrons in it. So when we say that it is doubly ionized, two electrons will be removed from the atom. Therefore there will be only one electron left in the atom and hence it is hydrogen like species.