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

Hint: 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.