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# Lesson 6.3: Line Spectra and The Bohr Model

← Lesson 6.2: Lesson 6.5: →

The Bohr Model: The Bohr Model →

The model of the atom we have been working with since lesson 2 is one that has a positive nucleus with electrons moving about that nucleus in some sort of orbit. A problem with this model is that according to classical physics an electron moving in orbit should lose energy and soon collapse into the nucleus. Since his does not happen, where have we gone wrong in our thinking?

We saw in the previous lesson that energy is quantized. Bohr applied this notion to the electron orbiting the nucleus of a hydrogen atom. He suggested that the electrons in an atom could only exist in certain allowed orbits, and in those orbits they violated the rules of classical physics – they did not lose energy as the moved about the nucleus. Given certain assumptions, he derived two equations. The first allowed him to calculate the radii of the allowed orbits, and the second their energy.

Although you do not need to concern yourself with the details, the following video gives a clear presentation of how that first equation is derived. Based on the assumption that the angular momentum of an electron orbiting a proton in a hydrogen is quantized, he was able to arrive at an equation that can be used to calculate the allowed radii of the electron. When the angular momentum can only have integer values of h/2π, the radii are given by r = r1n2 where r1 is the closet allowed radius of 5.3 x 10-11 m.

Here is a representation of the first two allowed orbits for Bohr's hydrogen atom. The first radius, r1, equals 5.3 x 10-11 m. The second allowed orbit, r2, has a radius four times as large equal to 21.1 x 10-11 m. The third orbit would be 9 times as great etc.

Step by step calculations are shown in the following video.

Here is a table showing the first six allowed orbits.

allowed orbit, n radius, r (m)
1 0.529 x 10-10
2 2.12 x 10-10
3 4.76 x 10-10
4 8.46 x 10-10
5 13.2 x 10-10
6 19.0 x 10-10

Here is a diagram of the same data.

Based on the same assumptions the energy associated with each of these allowed orbits can be calculated. This video works out the math if you want to see it.

These allowed orbits were called energy shells. The allowed orbits have different energies, and for the hydrogen atom can be calculated with the equation $$E_n = \frac{E_1}{n^2}$$ where E1 = -13.6 eV and n can be an integer starting with n = 1.

Step by step calculations are shown in the following video for finding the energy at variaous allowed orbits.

Here is a table showing the energy for first six allowed orbits.

allowed orbit, n energy, E (eV)
1 -13.6
2 -3.4
3 -1.51
4 -0.85
5 -0.54
6 -0.37

Here is a diagram of the same data in eV and then in Joules.

Bohr's model was able to accurately predict the size of the hydrogen atom and the energy needed to ioinize the atom. However, what made it even more useful is that it gave a meaningful interpretation of the emission spectrum of hydrogen.

Consider the diagram below showing the light emitted by a sample of heated hydrogen.

How can we make sense of this?

After absorbing energy imagine an electron existing in one of the allowed Bohr orbits – say the 3rd orbit. If it lost energy and ended up in the second orbit, it would give up a discrete amount of energy - a chunk – the difference in energy between the 2nd and the 3rd orbit, between n = 2 and n = 3.

Other transitions would also be possible. Those that involve differences in energy that correspond to light in the visible region of the electromagnetic spectrum could be represented as follows.

Do these predicted transitions match up with the experimental the lines shown in the line spectrum of hydrogen? Yes.

Example 1: Consider the transition from n = 3 to n = 2. The difference in energy is \begin{align*} -1.51 \, eV - ( -3.4 \, eV) & = 1.89 \, eV \\\\ 1.89 \, eV \times \frac{1.6 \times 10^{-19} \, J}{1 \, eV} & = 3.02 \times 10^{-19} \, J \\\\ \end{align*} The frequency of light with this much energy per photon is \begin{align*} E & = h\nu\\\\ \nu & = \frac{3.02 \times 10^{-19} \, J}{h} \\\\ \nu & = 4.56 \times 10^{14} \, Hz\\\\ \end{align*} We can now find the wavelength \begin{align*} c & = \lambda\nu\\\\ \lambda & = \frac{c}{\nu} \\\\ \lambda & = 6.57 \times 10^{-7} \, m \\\\ \end{align*} Which apart from rounding corresponds to the experimental value.

More examples are worked out in the videos below.