To determine the energy in the \(n^{\text{th}}\) orbit of a singly ionized helium atom, we'll use the concept of energy levels in atoms based on the Bohr model. For hydrogen-like atoms, the energy of an electron in the \(n^{\text{th}}\) orbit is given by the formula:
\(E_n = -\dfrac{13.6 \, \text{eV} \cdot Z^2}{n^2}\)
Where:
For a hydrogen atom (\(Z = 1\)), the energy at the \(n^{\text{th}}\) orbit is:
\(E_n^{\text{Hydrogen}} = -\dfrac{13.6 \, \text{eV}}{n^2}\)
Now, consider a singly ionized helium atom (\(Z = 2\)). The energy in the \(n^{\text{th}}\) orbit for this atom is:
\(E_n^{\text{He}^+} = -\dfrac{13.6 \, \text{eV} \cdot (2)^2}{n^2} = -\dfrac{54.4 \, \text{eV}}{n^2}\)
Thus, the energy of the \(n^{\text{th}}\) orbit in a singly ionized helium atom is four times that of the \(n^{\text{th}}\) orbit of a hydrogen atom. This is because the factor by which the energy increases depends on the square of the atomic number \(Z\), which is 4 for helium as \(Z = 2\).
Therefore, the energy in the \(n^{\text{th}}\) orbit of a singly ionized helium atom is \(4E_n\):
Let's briefly address why other options are incorrect:
Hence, the correct answer reflects the increased binding energy due to a more substantial nuclear charge in helium compared to hydrogen.