A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is n for a hydrogen atom if 0.850 eV of energy can ionize it?

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.