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×10th Jan 2020 @ 8 min read
The Bohr atomic model or Rutherford-Bohr model was an earlier atomic model developed by Niels Bohr in 1913. Bohr was a Danish Nobel laureate. The model supplanted the Rutherford model. It was an improvement to its predecessor Rutherford's atomic theory. The model was effective in explaining many limitations of its predecessors, especially the hydrogen spectrum. Bohr refined Rutherford's model with the help of the old quantum theory.
Although the Bohr model is inaccurate, it is very helpful for understanding the quantized nature of subatomic particles, particularly electrons. Its postulates are described below.
Note: There are two ways in which we can represent the orbits of an atom: n = 1, 2, 3… and K, L, M… shell.
Note: If notice, the radius of an atom is inversely proportional to the atomic number. Thus, the radius decreases with an increase in the atomic number—which is a general trend as we move from left to right in the modern periodic table.
Note: eV in the above expression stands for the electronvolt, a unit of energy. 1 eV ≈ 1.602 × 10−19 J.
Bohr's model holds good for the hydrogen atom. Hydrogen atom is the simplest atom with one proton and one electron. The model is also applicable to ions similar to hydrogen, e.g. He+, Li2+, Be3+, which have only one electron.
The radius of the hydrogen atom (Z = 1):
When n = 1, r1 = 52.9 pm. This is the first stationary orbit, and its value is also called the Bohr radius. For n = 2, 3…, r = 211.6 pm, 476.1 pm…
The velocity of the hydrogen atom:
For n = 1, 2, 3…, v = 2.187 m s−1, 1.094 m s−1, 0.547 m s−1… Therefore, the velocity of an electron in the first orbit is the highest.
The energy of the hydrogen atom:
The energy of the first orbit (n = 1) is −13.6 eV. For n = 2, 3…, −3.40 eV, −1.51 eV… Although the energy increases with n, its absolute value decreases.
An electron in the hydrogen atom is mostly found in its ground state, the lowest energy state. The electron moves to a higher energy state, an excited state, when it absorbs energy.
The energy of the electron is negative; it indicates we have to supply energy to release an electron from an orbit. When n approaches infinity, the energy tends to zero. Hence, no energy is required and it is called the free electron.
As explained earlier, an electron can jump from an orbit to another. In this process, it emits energy if it descends or absorbs energy if it ascends. This is similar to a staircase. Every stair represents a stationary orbit. When an electron climbs down stairs, its energy is reduced and the difference in the energies is emitted. Similarly, when external energy is supplied, it absorbs and climbs up stairs. The figure below depicts the same.
We can even calculate the difference in the energy levels in a transition of an electron.
Let Ei and Ef be the energies in the initial and final states.
The difference between both is ∆E.
We also know, ∆E = hν.
The frequency of an electromagnetic radiation is . Here, c and λ are the speed of light (≈ 3 × 108 m s−1) and the wavelength.
Here, ν̄ is the wavenumber, which is equal to the reciprocal of the wavelength, and RH is the Rydberg constant of hydrogen and its value is given below.
One of the achievements of Bohr's model was that it could explain the hydrogen spectra.
When a white light is projected on a sample of hydrogen, electrons absorb the radiation and transit to an excited state. Since electrons can only absorb the light of certain wavelengths, most of the light remains unabsorbed. The outcoming light will be deficient in these absorbed wavelengths. This can be seen in the absorption spectrum of hydrogen (see the figure below).
The emission spectrum of hydrogen is similar to that of the absorption spectrum. Here, the light coming from an excited sample of hydrogen is used. The wavelengths emitted are the same in the absorption spectrum. See the figure below; it is self-explanatory.
None of the earlier models to Bohr's was able to explain these spectral lines. In fact, we can even calculate each wavelength using the equation below.
Let's say an electron jumps from n = 2 to n = 3.
The wavelength from the above equation is 656.47 nm. The value matches that in the spectral diagrams. In the same manner, we can determine the wavelengths for ni = 2 and nf = 4, 5, 6, 7…∞; the table below represents all of them.
ni | nf | λ (nm) | Colour |
---|---|---|---|
2 | 3 | 656.47 | Red |
2 | 4 | 486.27 | Cyan |
2 | 5 | 434.17 | Violet |
2 | 6 | 410.29 | Violet |
2 | 7 | 397.12 | Violet |
2 | 8 | 389.01 | near UV |
2 | 9 | 383.65 | near UV |
2 | ∞ | 364.70 | In UV |
The above series of the wavelengths, aka the Balmer series, reciprocates the hydrogen spectral lines.
Although the Bohr model was successful in explaining the hydrogen spectra, it had several limitations and was replaced by the quantum mechanic model.
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