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# 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:

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## 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 $$n = 3$$.

1. How much energy (in electron volts) does the electron need to move from the state $$n = 2$$ to a higher state $$n = 3$$?
2. What light frequency $$f$$ can realize this transition?
3. What wavelength $$\lambda$$ corresponds to this frequency?
4. 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 $$n = 3$$: $$W_2 - W_3 = -3.4 \, \text{eV} - (-1.5 \, \text{eV}) = -1.9 \, \text{eV}$$

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: $$f = \frac{E}{h} = \frac{ 1.9 \, \text{eV} }{ 6.6 \cdot10^{-34} \, \text{Js} } = \frac{ 1.9 \, \text{V} \cdot 1.6 \cdot 10^{-19} \, \text{As} }{ 6.6 \cdot 10^{-34} \, \text{Js} } = 4.6 \cdot 10^{14} \, \frac{1}{\text s} = 460 \, \text{THz}$$
Each frequency $$f$$ of a photon can also be assigned a wavelength $$\lambda$$. The frequency and wavelength are linked by the speed of light $$c$$: $$\lambda = \frac{c}{f} = \frac{ 3 \cdot 10^8 \, \frac{\text m}{\text s} }{ 4.6 \cdot 10^{14} \, \frac{1}{\text s} } = 6.5 \cdot 10^{-7} \, \text{m} = 650 \cdot 10^{-9} \, \text{m} = 650 \, \text{nm}$$
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: $$k = \frac{1}{\lambda} = \frac{1}{650 \cdot 10^{-9} \, \text{m} } = 1.5 \cdot 10^6 \, \frac{1}{\text m}$$