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May 15, 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 electrons move around the nucleus with definite velocity in a certain fixed closed circular path called orbits (or) shells. These shells are numbered as one, two, three, four or termed as \(\rm {K, L, M, N}\) from the nucleus.
- Out of the infinite number of possible circular orbits around the nucleus, an electron can revolve only in those orbits whose angular momentum is a multiple integral factor \(\frac{ {\text{h}}}{ { {\text{2 π}}}}\), where \(\rm{h}\) is Planck’s constant. It is Bohr’s quantum condition for angular momentum.

\( {\text{mvr=n}}\frac{ {\text{h}}}{ { {\text{2 π}}}}\)

Where \(\rm{m} =\) mass of the electron, \(\rm{v} =\) velocity of the electron, \(\rm{r} =\) radius of the orbit and ‘\(\rm{n}\)’ is the integral number like \(1,\,2,\,3,\,4 ….\) is called principal quantum number. - Each stationary state is associated with a definite amount of energy. Also, long as an electron is revolving in orbit, it neither loses nor gains energy. Hence these orbits are called stationary states or stable orbits.
- The energy of an electron changes when it moves from one orbit to another. Outer orbits have higher energies, while inner orbits have lower energies. The energy is absorbed when an electron moves from inner orbit to outer orbit. The energy is emitted when the electron jumps from one outer orbit \(\rm{E}_2\) to the inner orbit \(\rm{E}_1\).

Thus, the energy emitted or absorbed is in the form of quanta. \( { {\text{E}}_2} – { {\text{E}}_1} = {\text{∆ E}} = {\text{hv}}.\)

Here \(\rm{E}_1\) and \(\rm{E}_2\) are the energies of the lower and higher allowed energy states. - The centrifugal force of the revolving electron in a stationary orbit is balanced by the electrostatic attraction between the electron and the nucleus.

- Bohr’s model explains the stability of the atom. The electron revolves in a stationary orbit, does not lose energy, and remains in orbit forever.
- Bohr’s theory successfully explains the atomic spectrum of hydrogen.
- The theory explains the hydrogen spectrum and the spectra of one electron species such as \(\rm{He}^+\), \(\rm{Li}^{2+}\) and \(\rm{Be}^{3+}\), etc.
- The experimentally determined frequencies of spectral lines are in close agreement with those calculated by Bohr’s theory.
- The value of the Rydberg constant for hydrogen calculated from Bohr’s equation tallies with the value determined experimentally.

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’s theory fails to explain the line spectra of multielectron atoms.
- It does not explain the splitting of spectral lines into a group of finer lines under the influence of magnetic field (Zeeman effect) and electric field (Stark effect).
- Inability to explain the three-dimensional model of the atom. According to Bohr’s atom model, the electrons move along certain circular paths in one plane. Thus, it gives a flat model of the atom. However, now, it is well established that the atom is three-dimensional and not flat, as Bohr suggested.
- Inability to explain the shapes of molecules. Now it is well known that in covalent molecules, the bonds have directional characteristics (i.e., atoms are linked to each other in particular directions), and hence they possess definite shapes. Bohr’s model is unable to explain it.
- Inability to explain de Broglie concept of the dual nature of matter and Heisenberg’s uncertainty principle.
- Inability to explain the concept of elliptical orbits. Bohr’s model assumes that electrons revolve around the nucleus in circular orbits. However. Sommerfeld explained the splitting of spectral lines of hydrogen by assuming that some electron orbits are elliptical.
- Inability to explain the reason for the assumption of angular momentum. The assumption that angular momentum is an integral multiple of \( {\text{h/2 π}}\) was made without any explanation.
- Inability to consider electron spin energy. Bohr’s theory assumes that the total energy of the electrons is the sum of potential and kinetic energies. Now, it is known that electron spins about its axis and total energy should also include spin energy.

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:*

- The electrons occupy certain orbits around the nucleus.
- As long as the electron revolves in a particular orbit, the electron neither gains nor loses energy. Therefore, these orbits are called stationary orbits or main energy states and the electrons are said to be in stationary energy states.
- The energy associated with the energy levels increases with the increase in the value of \(\rm{n}\).

*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:*

- Bohr’s theory fails to explain the spectra of multielectron atoms.
- It does not explain the splitting of spectral lines into a group of finer lines under the influence of magnetic field (Zeeman effect) and electric field (Stark effect).
- The theory could not explain why the angular momentum of an electron is quantised.
- The theory failed to explain the splitting of spectral lines when observed with a high-resolution power spectrometer.
- Bohr’s model of electronic structure could not account for the ability of atoms to form molecules through chemical bonds.

*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:

- Bohr’s model explains the stability of the atom. The electron revolves in a stationary orbit, does not lose energy, and remains in orbit forever.
- Bohr’s theory successfully explains the atomic spectrum of hydrogen.
- The experimentally determined frequencies of spectral lines are in close agreement with those calculated by Bohr’s theory.
- The value of the Rydberg constant for hydrogen calculated from Bohr’s equation tallies with the value determined experimentally.

*We hope this article on Bohr’s Model of the Hydrogen Atom has helped you. If you have any queries, drop a comment below, and we will get back to you.*