A singly ionized helium atom in an excited state (n = 4) emits a photon of energy 2.6 eV. Given that the ground state energy of hydrogen atom is – 13.6 eV, the energy (\[{{E}_{f}}\]) and quantum number (n) of the resulting state are respectively,

\[A.\,{{E}_{f}}=-13.6\,eV,\,n=1\]
\[B.\,{{E}_{f}}=-6.0\,eV,\,n=3\]
\[C.\,{{E}_{f}}=-6.0\,eV,\,n=2\]
\[D.\,{{E}_{f}}=-13.6\,eV,\,n=2\]

Hint: The formula used to calculate the energy of the excited atom (mainly singly ionized atom) is taken from the Bohr’s theory of the single electron species. For only iso-electronic (single electron species) species like \[H,\,H{{e}^{+}},L{{i}^{2+}}\] this model is applicable.

Formula used:
\[E=-13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}\]

Complete answer:
From given, we have the data,
The excited state of a singly ionized helium atom, n = 4
The energy of the photon emitted by the atom, \[{{E}_{p}}=2.6\,eV\]
To determine: The energy of the resulting state, \[{{E}_{f}}\]= ?
The quantum number of the resulting state, n = ?

The formula used to calculate the energy of the excited atom is,
\[E=-13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}\]
Where Z is the atomic number and n is the quantum number.

A single ionized helium atom has an atomic number of \[Z=2\].
Substitute the values in the above formula to find the amount of energy.
The energy of the fourth state Helium ion is given as,
\[\begin{align}
  & E_{4}^{He+}=-13.6\times \dfrac{{{2}^{2}}}{{{4}^{2}}} \\
 & \implies E_{4}^{He+}=-3.4\,eV \\
\end{align}\]

As the energy of the emitted photon is \[{{E}_{p}}=2.6\,eV\], thus, the energy after emitting the photon is given as,
\[\begin{align}
  & {{E}_{f}}=E_{4}^{He+}-{{E}_{p}} \\
 & \implies {{E}_{f}}=-3.4-2.6 \\
 & \implies {{E}_{f}}=-6\,eV \\
\end{align}\]
Therefore, the energy of the resulting state is -6 eV.

Now consider the quantum number formula,
\[n={{\left( \dfrac{-13.6\times {{Z}^{2}}}{{{E}_{n}}} \right)}^{\dfrac{1}{2}}}\]
Substitute the value of the atomic number and the obtained value of the energy in the above equation to find the quantum number.
\[\begin{align}
  & n=\sqrt{\dfrac{-13.6\times {{2}^{2}}}{-6}} \\
 & \implies n=\sqrt{9} \\
 & \implies n=3 \\
\end{align}\]
Therefore, the quantum number of the resulting state is 3.

As the energy of the atom in the resulting state is – 6 eV and the quantum number of the resulting state is 3.

So, the correct answer is “Option B”.

Note:
The things to be on your figure tips for further information on solving these types of problems are: Bohr’s atomic model is valid only for the species like hydrogen, that is, single-electron species, as the model discards the inter electron attraction force.