Suggested languages for you:

Americas

Europe

Q12PE

Expert-verified
Found in: Page 1112

### College Physics (Urone)

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.

See the step by step solution

## Step 1: Definition of ionisation energy

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

## Step 2: Given information and formula to be used

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.

## Step 3: Determine the value of n for a hydrogen atom if 0.850eV of energy can ionize it.

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.

## Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.