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Step-by-Step Solution
Step 1: Understand the Ionization Energy Formula
For a hydrogen-like atom, the energy required to remove an electron from the ground state is given by
$E = 13.6 \times \dfrac{Z^2}{n^2} \text{ eV}$,
where:
$Z$ is the atomic number of the ion/atom.
$n$ is the principal quantum number (for the ground state, $n=1$).
Step 2: Calculate Energy for the Singly Ionized Helium Atom
Helium has atomic number $Z = 2$. For a singly ionized helium ion (He⁺), we use $Z=2$ and $n=1$:
$E_{1} = 13.6 \times \dfrac{2^2}{1^2} = 13.6 \times 4 = 54.4 \text{ eV}.$
This is the energy required to remove the remaining electron from the singly ionized helium ion (He⁺).
Step 3: Relate the Energies Using the Given Ratio
Let $E_{2}$ be the energy required to remove an electron from the neutral helium atom (He). According to the problem statement:
$E_{1} = 2.2\,E_{2}.$
Substituting $E_{1} = 54.4\text{ eV}$, we get
$54.4 = 2.2\,E_{2} \quad \Longrightarrow \quad E_{2} = \dfrac{54.4}{2.2} \approx 24.72\text{ eV}.$
Step 4: Find the Total Energy to Ionize Helium Completely
To ionize helium completely, we first remove one electron from the neutral atom (energy $E_{2}$), and then remove the second electron from the singly ionized ion (energy $E_{1}$). Thus, the total ionization energy is:
$E_{\text{total}} = E_{1} + E_{2} = 54.4 \text{ eV} + 24.72 \text{ eV} \approx 79.12\text{ eV}.$
This is closest to $79\text{ eV}$ as given in the options.
Final Answer: 79 eV
