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Q12PE

Expert-verifiedFound in: Page 1112

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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 value of n is 4.

**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 *n *is 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.

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