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Ungrouped Data: Know Formulas, Definition, & Applications
December 11, 2024Rutherford’s Atom Model was undoubtedly a breakthrough in atomic studies. However, it was not wholly correct. The great Danish physicist Niels Bohr (1885–1962) made immediate use of Rutherford’s planetary model of the atom. Bohr became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on the planetary model of the atom. For decades, many questions had been asked about atomic characteristics. It needed slight modifications that were made by a legendary scientist named Neil’s Bohr.
Bohr recognised that the lines in an atomic spectrum are related to the arrangement of electrons in that atom. However, the concept of the nucleus as enunciated by the Rutherford Atomic Model was retained. Bohr postulated that Planck’s Quantum theory applies to the electron that revolves around the nucleus. On this basis, Bohr proposed this theory to explain the structure of an atom.
To overcome the drawback of Rutherford’s atom model and explain the line spectrum of hydrogen, Neil’s Bohr, a Danish physicist in 1913, proposed a new atom model based on Planck’s quantum theory. This new model is called Bohr’s model of an atom.
Bohr’s model is based on applying the quantum theory of radiation for revolving electrons around the nucleus. In this model, the concept of the atomic nucleus as proposed by Rutherford’s planetary model is retained.
The main postulates of Bohr’s model of atom are as follows:
The expressions for radius and energy are explained below:
Consider an electron of mass ‘\(\rm{m}\)’ and charge ‘\(\rm{-e}\)’ revolving around the nucleus of charge ‘\(\rm{+Ze}\)’ in a circular orbit of radius ‘\(\rm{r}\)’. Let \(\rm{v}\) be the tangential velocity of the electron.
As per the coulomb’s law, the electrostatic force of attraction between the moving electron and the nucleus is \({\text{F}} = – \frac{{{\text{KZ}}{{\text{e}}^2}}}{{{{\text{r}}^2}}}\). The coulombic force of attraction tends to pull the electron towards the nucleus.In a stationary orbit
\({\text{F}} = – \frac{{{\text{KZ}}{{\text{e}}^2}}}{{{{\text{r}}^2}}}\)
Where; \(\rm{K} =\) constant \( = \frac{1}{{4{\text{π}}{{{\varepsilon }}_0}}} = 9 \times {10^9}\,{\text{N}}{{\text{m}}^2}/{{\text{C}}^2}\)
And the centripetal force \({\text{F=}}\frac{{{\text{m}}{{\text{v}}^{\text{2}}}}}{{\text{r}}}\)
Here \(\rm{m}\) is the mass of the electron, and \(\rm{r}\) is the radius of the orbit.
Hence, \(\frac{{{\text{m}}{{\text{v}}^{\text{2}}}}}{{\text{r}}}{\text{=}}\frac{{{\text{KZ}}{{\text{e}}^{\text{2}}}}}{{{{\text{r}}^{\text{2}}}}}\)
Or, \({{\text{v}}^2}{\text{=}}\frac{{{\text{KZ}}{{\text{e}}^{\text{2}}}}}{{{\text{mr}}}}\) …..(1)
As per Bohr’s quantum condition, \({\text{mvr=n}}\frac{{\text{h}}}{{{\text{2 π}}}}\)
\(\therefore \,{\text{v}} = \frac{{{\text{nh}}}}{{2{\text{π mr}}}}\)
\({{\text{v}}^2} = \frac{{{{\text{n}}^2}{{\text{h}}^2}}}{{4{{\text{π }}^2}{{\text{m}}^2}{{\text{r}}^2}}}\) …….(2)
From equation (1) and (2)
\({\text{r}} = \frac{{{{\text{n}}^2}{{\text{h}}^2}}}{{4{{\text{π }}^2}{\text{mKZ}}{{\text{e}}^2}}}\)
on putting the value of \(e,\,h,\,m\)
\({\text{r}} = 0.529 \times \frac{{{{\text{n}}^2}}}{{\text{Z}}}{{\text{A}}^ \circ }\)
We know that, for hydrogen, \(\rm{Z} =1\).
Hence, the radius of the hydrogen atom is \({\text{r}} = 0.529 \times {{\text{n}}^2}{{\text{A}}^ \circ }\)
The total energy revolving in orbit is obtained by summing up its kinetic and potential energy.
Kinetic energy due to motion of an electron \( = \frac{1}{2}{\text{m}}{{\text{v}}^2}\) (\(\rm{m}\) is mass of electron and \(\rm{v}\) is its velocity)
Potential energy due to position \(= -\frac{{{\text{KZ}}{{\text{e}}^{\text{2}}}}}{{\text{r}}}\)
The total energy of the electron \(= \rm{K.E + P.E}\)
\( = \frac{1}{2}{\text{m}}{{\text{v}}^2} – \frac{{{\text{KZ}}{{\text{e}}^2}}}{{\text{r}}}\) ……(3)
We know that \( – \frac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}} = – \frac{{{\text{KZ}}{{\text{e}}^2}}}{{{{\text{r}}^2}}}\left({\therefore \frac{{{\text{KZ}}{{\text{e}}^2}}}{{\text{r}}} = {\text{m}}{{\text{v}}^2}} \right)\)
Substituting the value of \(\rm{mv}^2\) in equation (3), we get
Total energy \( = \frac{{{\text{KZ}}{{\text{e}}^2}}}{{2{\text{r}}}} – \frac{{{\text{KZ}}{{\text{e}}^2}}}{{\text{r}}} = – \frac{{{\text{KZ}}{{\text{e}}^2}}}{{2{\text{r}}}}\)
Now, substituting the value of \(\text{r}\) in the equation of T.E.
\({\text{E}} = – \frac{{{\text{KZ}}{{\text{e}}^2}}}{2} \times \frac{{4{{\text{π }}^2}{\text{Z}}{{\text{e}}^2}{\text{mK}}}}{{{{\text{n}}^2}{{\text{h}}^2}}} = – \frac{{2{{\text{π }}^2}{{\text{Z}}^2}{{\text{e}}^4}{\text{m}}{{\text{K}}^2}}}{{{{\text{n}}^2}{{\text{h}}^2}}}\)
The energy of electrons in \(\rm{n}^{\rm{th}}\) orbit, \({{\text{E}}_{\text{n}}}{= -}\frac{{{\text{2}}{{\text{π }}^{\text{2}}}{{\text{Z}}^{\text{2}}}{{\text{e}}^{\text{4}}}{\text{m}}{{\text{K}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}{{\text{h}}^{\text{2}}}}}\)
\({= -13}{.6 \times}\frac{{{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}\,{\text{eV/atom}}\)
\( = – 21.8 \times {10^{ – 19}} \times \frac{{{{\text{z}}^2}}}{{{{\text{n}}^2}}}\,{\text{J}}/{\text{atom}}\)
\( = – 313.6 \times \frac{{{{\text{Z}}^2}}}{{{{\text{n}}^2}}}\,{\text{Kcal}}/{\text{mole}}\)
We know that for hydrogen \(\rm{Z} =1\),
Thus, for a hydrogen atom \({{\text{E}}_{\text{H}}} = – 13.6 \times \frac{1}{{{{\text{n}}^2}}}\,{\text{eV}}/{\text{atom}}.\)
We know that
\({\text{mvr}} = {\text{n}}\frac{{\text{h}}}{{2{\text{π }}}}\); \({\text{v}} = \frac{{{\text{nh}}}}{{2\pi {\text{mr}}}}\)
By substituting for \(\rm{r}\), we are getting
\({\text{V}} = \frac{{{\text{nh}} \times 4{{\text{π }}^2}{\text{mKZ}}{{\text{e}}^2}}}{{2{\text{π m}}{{\text{n}}^2}{{\text{h}}^2}}}\)
\({\text{V}} = \frac{{2{\text{π KZ}}{{\text{e}}^2}}}{{{\text{nh}}}}\)
On putting the value of \(\rm{e},\,\rm{K}\) and \(\rm{h}\)
\({\text{V}} = 2.18 \times {10^6} \times \frac{{\text{Z}}}{{\text{n}}}\,{\text{m}}/{\text{sec}}\)
We know that for hydrogen \(\rm{Z} =1\)
Thus, for a hydrogen atom, \({\text{V}} = 2.18 \times {10^6} \times \frac{{\text{1}}}{{\text{n}}}\,{\text{m}}/{\text{sec}}\)
Energy expression for hydrogen-like ions \((\rm{He}^+,\,\rm{Li}^{2+}\),and \(\rm{Be}^{3+})\) is
\({{\text{E}}_{\text{n}}} = – \frac{{2{{\text{π }}^2}{{\text{Z}}^2}{{\text{e}}^4}{\text{m}}{{\text{K}}^2}}}{{{{\text{n}}^2}{{\text{h}}^2}}} = {{\text{E}}_{\text{H}}} \times {{\text{Z}}^2}\)
The radius and energy of an electron in the \(\rm{n}^{\rm{th}}\) orbit of a hydrogen atom can also be calculated in SI units.
\({{\text{E}}_{\text{n}}} = – \frac{{2{{\text{π }}^2}{{\text{e}}^4}{\text{m}}{{\text{K}}^2}}}{{{{\text{n}}^2}{{\text{h}}^2}}}\)
\({\text{r=}}\frac{{{{\text{n}}^{\text{2}}}{{\text{h}}^{\text{2}}}}}{{{\text{4}}{{\text{π }}^{\text{2}}}{\text{mK}}{{\text{e}}^{\text{2}}}}}\)
where, \({\text{K=}}\frac{{\text{1}}}{{{\text{4π }}{{\varepsilon }_{\text{0}}}}}\)
Where \(\rm{ε}_0\) being permittivity of air and is equal to \(8.854 \times {10^{ – 12}}\,{\text{Farad}\, \text{metre}}{{\text{}}^{ – 1}}\)
The energy of an electron can have only a certain discrete restricted value depending upon the value of \(\rm{n}\). It is called the quantisation of the energy of the electron. The principal quantum number determines the radius of the orbit and the electron’s energy in orbit.
It is important to note that the electron’s energy has a negative sign and is inversely proportional to the square of the principal quantum number. As the value of n increases, the radius of orbit increases and the absolute value of energy electron increases. When \(\rm{n = ∞}\), the energy becomes zero. It is the maximum energy that an electron can possess in an atom.
It corresponds to an ionised atom where the electron and the nucleus are infinitely separated. The electron in an atom is more stable than a electron. As the energy of the electron is taken as zero, the energy of an electron in an atom should be less than zero, i.e. negative. Hence the energy of an electron is always negative.
Bohr’s model of atom suffers from the following limitations.
Bohr Model of Hydrogen Atom Quantised: Niels Bohr introduced the atomic hydrogen model in 1913. He described it as a positively charged nucleus, comprised of protons and neutrons, surrounded by a negatively charged electron cloud. In the model, electrons orbit the nucleus in atomic shells.
Bohr model of hydrogen atom formula
The formula for Bohr’s model is \({\text{l=}}\frac{{{\text{nh}}}}{{{\text{2π }}}}\)
\(\rm{l}\) or \(\rm{mvr}=\) Angular momentum
\(\rm{n}=\) principal quantum number
\(\rm{h}=\) Planck’s constant.
The Bohr model of the atom was replaced by the Quantum Mechanics Model based upon the Schrodinger equation in the 1920s. The great success of the Bohr Model had been in explaining the spectra of hydrogen-like (single electron around a positive nucleus) atoms.
The Schrodinger equation replicated this explanation more sophisticatedly, and the Bohr analysis was considered obsolete. But the Schrodinger equation approach works only for a minimal number of models. Beyond this limited set, the Schrodinger equation approach gives no insights, whereas the Bohr Model does provide insights into diverse cases. In particular, the Schrodinger equation approach cannot be applied to the case, which considers the relativistic effects. On the other hand, the Bohr analysis can.
In this article, we learnt how Bohr overcome the drawback of Rutherford’s Atomic Model. The postulates, merits of Bohr’s theory. Radius and energy of hydrogen atoms, the velocity of an electron, Bohr’s theory concept of quantisation and Bohr’s Model’s limitations.
Q.1. What are the three principles of Bohr’s model?
Ans:
Q.2. How do you draw a Bohr model of hydrogen?
Ans: Bohr’s model of hydrogen is drawn as below.
Q.3. What are the limitations of the Bohr hydrogen model?
Ans:
Q.4. How do you find the Bohr model of an atom?
Ans: Electrons revolve around the nucleus in specified circular paths called orbits or shells. These orbits are numbered as \(1,\,2,\,3,\,4…\), etc., represented as \(\rm{K},\,\rm{L},\,\rm{M},\,\rm{N}….\) respectively and represented by the symbol ‘\(\rm{n}\)’. Electrons can jump from one orbit to another by emitting or absorbing energy. Each stationary state is associated with a definite amount of energy. Also, as long as an electron keeps revolving in orbit, it neither loses nor gains energy. Hence these orbits are called stationary states or stable orbits.
Q.5. What four assumptions did Bohr make about the electronic structure of the atom?
Ans: The four assumptions Bohr made are listed below:
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