Question

The ionisation energy of hydrogen atom is \[13.6eV\], the ionisation energy of helium atom would be
A.\[13.6eV\]
B.$27.2eVC$
C.$6.8eV$
D.$54.4eV$

Answer 1

Hint: The ionisation energy is the energy required in order to remove an electron from the isolated gaseous atom.
In a periodic table, as we go from left to right the value of ionisation energy increases as the nuclear charge increases and one electron is added to the outermost orbital with each consecutive atom.

Complete step-by-step answer: An atom is made up of a nucleus which contains protons as well as neutrons. These two species are collectively known as nucleons. We know that the nucleus is surrounded by a number of negatively charged particles called electrons. These electrons move inside the electron clouds or orbitals. These electrons could be removed from a specific atom, but it will require some energy, because there are some attractive forces which are acting between the electrons which are negatively charged particles and the nucleus which has proton, having positive charge. So, we need sufficient energy to break this attractive bond between the electrons and the nucleus.
Ionisation energy of an element is the energy required to remove the most loosely bound electron from an isolated gaseous atom. Now, in the given question we can see that the ionisation energy of the hydrogen atom is given as \[13.6eV\], and we know that an atom of hydrogen contains only one electron. So, from this we know that the ionised hydrogen atom is nothing but a proton. So, the question asks us the ionisation energy of the helium atom.
Determination of the ionisation energy of helium atoms can be done by following these simple steps. At first determine the atomic number of the species, which is $2$ in our case. Now, in the next step we need to calculate the number of electrons which would be removed, which is $1$. Now, we will use the formula
$13.6\times \dfrac{{{Z}^{2}}}{{{n}^{2}}}=I.E$
Here, \[13.6eV\] is the ionisation energy of the hydrogen atom which is provided to us. ‘Z’ is the atomic number of the atom, ‘n’ is the number of electrons under consideration, and I.E refers to the ionisation energy of helium. Now, after putting the values in the formula, we get,
$I.E=13.6\times \dfrac{{{2}^{2}}}{{{1}^{2}}}=54.4eV$
Now, as we can see the ionisation energy of the helium comes out to be $54.4eV$ which matches the option D.

So, the correct option would be D.

Note: The standard ionisation energy in terms of electron volts is \[13.6eV\] which is the ionisation energy of the hydrogen.
We can calculate the ionisation energy of the species or atoms having more than one atom by using this value of ionisation energy and atomic number of the atom.