# Rydberg Formula: Understanding the Energy Spectrum of the Hydrogen Atom

The Rydberg formula describes the wavelengths \( \lambda\) of lines in the spectrum of the hydrogen atom. However, it also describes the discrete energy levels \( W\) of the hydrogen atom:

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.### Exercise: Transition from \( n = 2 \) to \( n = 3 \) in the Hydrogen Atom

In the Bohr atomic model, consider the transition from the energy level \( n = 2 \) to \( n = 3 \).

- How much
**energy**(in electron volts) does the electron need to move from the state \( n = 2 \) to a higher state \( n = 3 \)? - What
**light frequency**\( f \) can realize this transition? - What
**wavelength**\( \lambda \) corresponds to this frequency? - What
**wavenumber**\( k \) corresponds to this frequency?

#### Solution to Exercise #1.1

A transition of the electron from the energy level \( n = 2 \) to \( n = 3 \) corresponds to the so-called \( H_{\alpha} \) line in the hydrogen atom in the Bohr atomic model.

The binding energy of the electron in the state \( n = 2 \) is: $$ W_2 = \frac{-13.6 \, \text{eV}}{n^2} = \frac{-13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV} $$ Analogously for the binding energy in the state \( n = 3 \).

Accordingly, the electron needs the following amount of energy for the transition from \( n = 2 \) to \( n = 3 \): $$W_2 - W_3 = -3.4 \, \text{eV} - (-1.5 \, \text{eV}) = -1.9 \, \text{eV} $$

#### Solution to Task #1.2

To calculate the frequency of the photon capable of ensuring this transition, use the following relationship: $$ W = h \, f $$

The necessary energy \( W = |-1.9 \, \text{eV}| \) for the transition was calculated in part (a) of the task. So the required frequency \( f \) of the photon is: $$ f = \frac{E}{h} = \frac{ 1.9 \, \text{eV} }{ 6.6 \cdot10^{-34} \, \text{Js} } = \frac{ 1.9 \, \text{V} \cdot 1.6 \cdot 10^{-19} \, \text{As} }{ 6.6 \cdot 10^{-34} \, \text{Js} } = 4.6 \cdot 10^{14} \, \frac{1}{\text s} = 460 \, \text{THz} $$

#### Solution to Exercise #1.3

Each frequency \( f \) of a photon can also be assigned a wavelength \( \lambda \). The frequency and wavelength are linked by the speed of light \( c \): $$ \lambda = \frac{c}{f} = \frac{ 3 \cdot 10^8 \, \frac{\text m}{\text s} }{ 4.6 \cdot 10^{14} \, \frac{1}{\text s} } = 6.5 \cdot 10^{-7} \, \text{m} = 650 \cdot 10^{-9} \, \text{m} = 650 \, \text{nm}$$

#### Solution to Exercise #1.4

To illustrate how many times the photon with a wavelength \( \lambda = 650 \, \text{nm} \) (see task part 1.3) performs an oscillation per meter, you calculate the wavenumber: $$k = \frac{1}{\lambda} = \frac{1}{650 \cdot 10^{-9} \, \text{m} } = 1.5 \cdot 10^6 \, \frac{1}{\text m} $$