Chapter 28, Problem 34A

To determine

To calculate: Energies associated with a hydrogen atom’s energy levels of $E_2,E_3,E_4,E_5$ and $E_6$

Answer to Problem 34A

Energies associated with a hydrogen atom’s energy levels of:

  $E_2=\underset{\_}{-3.40\text{ eV}}$

  $E_3=\underset{\_}{-1.51\text{ eV}}$

  $E_4=\underset{\_}{-0.85\text{ eV}}$

  $E_5=\underset{\_}{-0.54\text{ eV}}$

  $E_6=\underset{\_}{-0.38\text{ eV}}$

Explanation of Solution

Given:

The following energy levels of hydrogen atom:

  $E_2,E_3,E_4,E_5$ and $E_6$

Formula used:

Energy of a hydrogen atom is given by,

  $E_n=-13.6\text{ eV }\times \frac{1}{n^2}$

Where $n$ is the principal quantum number.

Calculation:

The energies associated with the different energy levels of a hydrogen atom is calculated as:

  $E_2=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(2)}^2}=-3.40\text{ eV}$

  $E_3=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(3)}^2}=-1.51\text{ eV}$

  $E_4=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(4)}^2}=-0.85\text{ eV}$

  $E_5=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(5)}^2}=-0.54\text{ eV}$

  $E_6=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(6)}^2}=-0.38\text{ eV}$

Conclusion:

Energies associated with a hydrogen atom’s energy levels of:

  $E_2=\underset{\_}{-3.40\text{ eV}}$

  $E_3=\underset{\_}{-1.51\text{ eV}}$

  $E_4=\underset{\_}{-0.85\text{ eV}}$

  $E_5=\underset{\_}{-0.54\text{ eV}}$

  $E_6=\underset{\_}{-0.38\text{ eV}}$