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Rydberg Formula: Understanding the Energy Spectrum of the Hydrogen Atom

Important Formula

Formula: Rydberg energy formula for H atom

$$W ~=~ \frac{ 13.6 \, \mathrm{eV} }{ n^2 }$$ $$W ~=~ \frac{ 13.6 \, \mathrm{eV} }{ n^2 }$$ $$n ~=~ \sqrt{ \frac{ 13.6 \, \mathrm{eV} }{ W } }$$

Rearrange formula

What do the formula symbols mean?

Binding energy

$$ W $$ Unit $$ \mathrm{J} = \mathrm{Nm} = \frac{ \mathrm{kg} \, \mathrm{m^2} }{ \mathrm{s}^2 } $$

Energy required to knock an electron, which is in the $ n $-th state, out of the hydrogen atom. For example, the binding energy of the electron in the ground state is $ n = 1 $: $ 13.6 \, \mathrm{eV} $.

Quantum number

$$ n $$ Unit $$ - $$

Principal quantum number is an integer indicating an energy level of the H atom. The electron in the H atom can occupy this energy state, which is characterized by $n$.

Here is:

$ n = 1 $ is the ground state.
$ n = 2 $ is the first excited state.
$ n = 3 $ is the second excited state.
and so on...

Table of contents

Important Formula
Exercises with Solutions

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:

$$ \begin{align} W ~=~ \frac{ 13.6 \, \mathrm{eV} }{ n^2 } \end{align} $$

Hydrogen atom - term diagram (energy levels)

$$ \begin{align} \lambda ~=~ \frac{1}{R \, \left( \frac{1}{n^2} - \frac{1}{m^2} \right)} \end{align} $$

$$ \begin{align} f ~=~ R_{\text f} \, \left( \frac{1}{n^2} ~-~ \frac{1}{m^2} \right) \end{align} $$

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: $$k = \frac{1}{\lambda} = \frac{1}{650 \cdot 10^{-9} \, \text{m} } = 1.5 \cdot 10^6 \, \frac{1}{\text m} $$