91Ó°ÊÓ

The energy that is enough to separate the electron in the last shell of the atom also called valance electron is the ionization energy.

Consider the hydrogen atom has 4 states.

Consider the formula for the ionization energy as follows:

\[{E_n} = \frac{{13.6\;{\rm{eV}}}}{{{n^2}}}\]

Here, \[{E_n}\] is the ionization energy and nis the energy level.

Substitute the values and solve for the ionization state.

\[\begin{array}{l}{E_n} = 0.85\;{\rm{eV}}\\{E_n} = \frac{{13.6\;{\rm{eV}}}}{{{n^2}}}\\0.85eV = \frac{{13.6\;{\rm{eV}}}}{{{n^2}}}\\{n^2} = \frac{{13.6\;{\rm{eV}}}}{{0.85\;{\rm{eV}}}}\end{array}\]

Substitute the values and solve as:

\[\begin{array}{l}{n^2} = 16\\n = \sqrt {16} \\n = \pm 4\end{array}\]

Also, n can never have a negative value.

Therefore, the n value is 4.