A $H{e^ + }$ ion is in its first excited state. Determine its ionization energy.

Hint According to Bohr’s Theory, angular momentum of an electron is $mvr = \dfrac{{nh}}{{2\pi }}$(where m is mass of the electron, v is the velocity, n is the orbit in which electron is and r is the radius of the nth orbit). Here n provides the centrifugal force, which is the force of attraction between positive and negative charge. And therefore, it can move in a circular orbit. So, the total energy (E) of an electron in nth orbit is $ - 13.6 \times \dfrac{{{z^2}}}{{{n^2}}}eV.$

Complete step by step answer
$H{e^ + }$has two electrons, as its atomic number is 2. Now if it loses one electron, it leaves with one electron which is the same as a hydrogen atom.
According to the question $H{e^ + }$ is in its first excited state.

Now according to Bohr’s Theory, the total energy (E) of an electron in nth orbit is $ - 13.6 \times \dfrac{{{z^2}}}{{{n^2}}}eV$
Here z is the atomic number.
For Helium ion, the total energy (E) of an electron in 2nd orbit is $ = - 13.6 \times \dfrac{{{2^2}}}{{{2^2}}} = - 13.6eV$
We have to find the ionization energy. As we know Ionization energy is the amount of energy required to remove an electron from an isolated atom or molecule. Always this amount is positive.

As total energy (E) of $H{e^ + }$in its first excited state is $ - 13.6eV$, ionization energy will be $13.6eV.$

Note We have to remember that the ionization energy is always positive as it requires energy to remove an electron. Loss of an electron from an atom requires energy input. Hence it is an endothermic process and the value is$ + ve$ for endothermic processes. Ionization enthalpies are always positive.