Question

The one electron species having ionisation energy of 54.4eV is:
(a) $H{{e}^{+}}$
(b) H
(c) $B{{e}^{2+}}$
(d) $B{{e}^{3+}}$

Answer 1

We know that ionisation energy is the energy required to remove an electron from a valence shell of an atom. This loss of an electron happens in the ground state of chemical species. The equation for ionisation energy according to Bohr’s theory is $-{{Z}^{2}}{{E}_{1}}$, where Z is the atomic number and ${{E}_{1}}$ is energy 13.6eV.

Complete step by step solution:
We know according to Bohr’s theory
Ionisation energy = $-{{Z}^{2}}{{E}_{1}}$
Here Z is the atomic number and
${{E}_{1}}=13.6eV$
It is given that ionisation energy is 5.4eV.
Therefore,
$13.6eV{{Z}^{2}}=54.4eV$
${{Z}^{2}}=\frac{54.4eV}{13.6eV}=4$
${{Z}^{2}}=4={{2}^{2}}$
${{Z}^{{}}}={{2}^{{}}}$
Since, the atomic number of helium is 2, $H{{e}^{+}}$ is the correct answer and the option is (a).

Additional Information:
-Bohr proposed that electrons cannot radiate energy as they orbit a nucleus. But they exist in states of constant energy which he called as stationary states.
-So, this says that electrons orbit a fixed distance from the nucleus. Bohr’s work was based on emission spectra of hydrogen which is also referred to as planetary model of the atom.
-Bohr’s theory mainly explains the inner workings of hydrogen atoms. When an electron is in one orbit, it has a fixed energy. The ground state, the energy is the lowest and the electron is in the orbit closest to the nucleus.
-Electrons are not allowed to occupy any spaces between the orbits.
-Bohr’s model is an analogy to the rungs of a ladder. As you move up a ladder, you can occupy only specific rungs and the spaces cannot be occupied. Moving up the ladder will increase the potential energy, and moving down the ladder decreases your energy.

Note: The formula of ionisation energy is obtained from Bohr’s atomic model of hydrogen. It shows the relation between the atomic number and potential.