Bohr’s Model of Hydrogen Atom

Bohr gave the postulates as below:

(1) Electrons can only travel around the nucleus in certain circular orbits that are allowed.

Each orbit contains electrons at a fixed distance from the nucleus and defined energy. The orbits are given the letter n and numbered \(1, 2, 3,\) etc., as the distance from the nucleus increases (or \(\text {K}, \text {L}, \text {M},\) etc.).

(2) An electron does not emit (or lose) energy when revolving in these precise orbits.

As a result, the energy of an electron in each of these orbits remains constant, i.e. it does not lose or gain energy. As a result, the electron’s unique orbits in an atom are referred to as stationary energy levels or simply energy levels. 

(3) Only quantum or photon jumps allow an electron to shift from one energy level to another.

When an electron is in the lowest-energy orbit, it is said to be in the ground state (which is also the one nearest to the nucleus). When an electron is given energy, it obtains one quantum or photon of energy and leaps to a higher energy level. Because the electron possesses potential energy it is considered to be in an excited state.

\(\Delta {\rm{E}} = {{\rm{E}}_{{\rm{high }}}} – {{\rm{E}}_{{\rm{low }}}} = {\rm{hv}}…(1)\)

Where v is the frequency of a photon released or absorbed energy and h is Planck’s constant.

(4) An electron orbiting around the nucleus has an angular momentum (mvr) that is an integral multiple of Planck’s constant divided by \(2{\rm{\pi }}\).

Angular momentum \( = {\rm{mvr}} = {\rm{n}}\frac{{\rm{h}}}{{2{\rm{\pi }}}}…(2)\)

Where m signifies the electron’s mass, \(\text {v}\) defines its velocity, and \(\text {r}\) denotes the orbit’s radius; \(\text {n} = 1, 2, 3\), etc., and \(\text {h}\) denotes Planck’s constant.

There can’t possibly be a fractional number for \(\frac{{\rm{h}}}{{2{\rm{\pi }}}}\). As a result, the angular momentum is quantised. The atom’s Principal quantum number is the integer \(\text {n}\) in equation \((2)\) that can be used to define an orbit and a corresponding energy level.

Calculation of Radius of Orbits

Consider the rotation of a charge \(‘-\text {e}’\) electron around a charge \({\rm{Ze}}\) nucleus, where \(\text {Z}\) is the atomic number, and \(‘\text {e}’\) is the charge on a proton. Let \(\text {m}\) be the electron’s mass, \(\text {r}\) be the orbit’s radius, and \(\text {v}\) be the rotating electron’s tangential velocity.

Coulomb’s law describes the electrical attraction between the nucleus and the electron,

\( = \frac{{{\rm{KZ}}{{\rm{e}}^2}}}{{{{\rm{r}}^2}}}\)

The centrifugal force which is acting on the electron is given as:

\(\frac{{{\rm{KZ}}{{\rm{e}}^2}}}{{{{\rm{r}}^2}}} = \frac{{{\rm{m}}{{\rm{v}}^2}}}{{\rm{r}}}\)

To keep the electron in orbit, Bohr assumed that these two opposing forces must be perfectly balanced. Thus,

\(\frac{{{\rm{KZ}}{{\rm{e}}^{\rm{2}}}}}{{{{\rm{r}}^{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{m}}{{\rm{v}}^{\rm{2}}}}}{{\rm{r}}}\)

or,

\(\frac{{{\rm{K}}{{\rm{Z}}_{{{\rm{e}}^{\rm{2}}}}}}}{{\rm{r}}} = {\rm{m}}{{\rm{v}}^2}\)

The angular momentum of a rotating electron is provided by the following expression according to one of Bohr’s postulates.

\(\frac{{{\rm{K}}{{\rm{Z}}_{{{\rm{e}}^{\rm{2}}}}}}}{{\rm{r}}}{\rm{ = m}}{{\rm{v}}^{\rm{2}}}\)

Substituting the value of \({\rm{v}}\) in equation \(\left( {\rm{i}} \right),\)

\(\frac{{{\rm{KZ}}{{\rm{e}}^{\rm{2}}}}}{{\rm{r}}}{\rm{ = m}}{\left( {\frac{{{\rm{nh}}}}{{{\rm{2\Pi mr}}}}} \right)^{\rm{2}}}\)

Or we can write:

\({\rm{r = }}\frac{{{{\rm{n}}^{\rm{2}}}{{\rm{h}}^{\rm{2}}}}}{{{\rm{4}}{{\rm{\Pi }}^{\rm{2}}}{\rm{mZK}}{{\rm{e}}^{\rm{2}}}}}…..\left( {{\rm{ii}}} \right)\)

Since the values of \({\rm{h,m,}}\) and \({\rm{e}}\) were obtained experimentally and are constant, we can replace them in equation \(\left( {{\rm{ii}}} \right).\)

\({\rm{r = }}\frac{{{{\rm{r}}_{\rm{0}}}{{\rm{n}}^{\rm{2}}}}}{{\rm{z}}}\) or \({\rm{r = 0}}.{\rm{529}} \times {\rm{1}}{{\rm{0}}^{{\rm{ – 8}}}}\frac{{{{\rm{n}}^{\rm{2}}}}}{{\rm{z}}}{\rm{cm}}\)

where, \({{\rm{r}}_{\rm{0}}}{\rm{ = }}\frac{{{{\rm{h}}^{\rm{2}}}}}{{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}{\rm{mK}}{{\rm{e}}^2}}}{\rm{ = 0}}{\rm{.529 \times 1}}{{\rm{0}}^{{\rm{ – 8}}}}{\rm{\;cm}}\)

Where \({\rm{n}}\) denotes the primary quantum number and thus the orbit’s number. When \({\rm{n = 1}}\) and for the hydrogen atom, \({\rm{Z = 1,}}\) the equation \(\left( {{\rm{iii}}} \right)\) becomes

\({\rm{r = 0}}{\rm{.529 \times 1}}{{\rm{0}}^{{\rm{ – 8}}}}{\rm{\;cm = }}{{\rm{\alpha }}_{\rm{0}}}\)

Bohr took this last quantity, \({{\rm{\alpha }}_{\rm{0}}}\) (the first Bohr radius), to be the radius of the hydrogen atom in the ground state. This figure is fairly consistent with other data on atom size. For \({\rm{n = 2,3,4,}}\) and so on, the value of the second and third orbits of hydrogen-containing the electron in the excited state may be calculated.

Energy of Electron in Each Orbit

The energy of a revolving electron in a hydrogen atom is equal to the sum of its kinetic and potential energy.

\({\rm{E = K \times E + P}}{\rm{.E}}{\rm{. = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}{\rm{ – }}\frac{{{\rm{Kz}}{{\rm{e}}^{\rm{2}}}}}{{\rm{r}}}…….\left( {{\rm{iv}}} \right)\)

Also, we know:

\(\frac{{{\rm{KZ}}{{\rm{e}}^{\rm{2}}}}}{{\rm{r}}}{\rm{ = m}}{{\rm{v}}^{\rm{2}}}\)

Or  we can write:

\({\rm{E = }}\frac{{\rm{1}}}{{\rm{2}}}\frac{{{\rm{KZ}}{{\rm{e}}^{\rm{2}}}}}{{\rm{r}}}{\rm{ – }}\frac{{{\rm{KZ}}{{\rm{e}}^{\rm{2}}}}}{{\rm{r}}}\)

\({\rm{E = – }}\frac{{\rm{1}}}{{\rm{2}}}\frac{{{\rm{KZ}}{{\rm{e}}^{\rm{2}}}}}{{\rm{r}}}………\left( {\rm{v}} \right)\)

Substituting the value of \({\rm{r}}\) from equation \(\left( {\rm{ii}} \right)\) in \(\left( {\rm{v}} \right)\)

\({\rm{E = – }}\frac{{{\rm{2}}{{\rm{\Pi }}^{\rm{2}}}{\rm{m}}{{\rm{K}}^{\rm{2}}}{\rm{Z}}{{\rm{e}}^{\rm{4}}}}}{{{{\rm{n}}^{\rm{2}}}{{\rm{h}}^{\rm{2}}}}}…….\left( {{\rm{vi}}} \right)\)

Substituting the values of \({\rm{m,e,}}\) and \({\rm{h}}\) in \(\left( {{\rm{vi}}} \right){\rm{,}}\)

\({\rm{E = }}\frac{{{\rm{ – 2}}{\rm{.179 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{{\rm{Z}}^{\rm{2}}}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{\;J per}}\,{\rm{atom}}\)

We can calculate the energy for each orbit by using a correct integer for \({\rm{n}}\) (quantum or orbit number).

Bohr’s theory may also be applied to ions with only one electron, which is analogous to hydrogen atoms. For instance, \({\rm{H}}{{\rm{e}}^{\rm{ + }}}{\rm{,L}}{{\rm{i}}^{\rm{ + }}},{\rm{B}}{{\rm{e}}^{\rm{ + }}},\) and so forth. The formulae give the energy and radii of the stationary states associated with these types of ions (also known as hydrogen-like species).

\({\rm{E = }}\frac{{{\rm{ – 2}}{\rm{.179 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{ \times }}{{\rm{Z}}^{\rm{2}}}\,{\rm{J}}\,{\rm{per}}\,{\rm{atom}}\)

\({\rm{r = 0}}{\rm{.529 \times 1}}{{\rm{0}}^{{\rm{ – 8}}}}\frac{{{{\rm{n}}^{\rm{2}}}}}{{\rm{Z}}}\,{\rm{cm}}\)

Where \({\rm{Z}}\) is the atomic number and \({\rm{n = 1,2,3}}…..\)

Significance of Negative Value of Energy

The energy of an electron is arbitrarily set to zero at infinity. This is known as a zero-energy situation. When an electron moves and makes contact with a nucleus, it does work and expends energy. As a result, the energy of the electron decreases and falls below zero, giving rise to a negative value.

Bohr’s Explanation of Hydrogen Spectrum

The solitary electron in a hydrogen atom lives in the first orbit \(\left( {{\rm{n = 1}}} \right)\) and has the lowest energy state at ambient temperature (ground state). Depending on how much energy is absorbed by hydrogen gas in the discharge tube, the electron moves to higher energy levels, such as \(2, 3, 4, 5, 6, 7,\) and so on. The electron jumps to a lower energy level to recuperate from these high energy levels. As a result, the electron emits a photon, which contains additional energy.

The Lyman series is created when an electron returns to the ground state, \({{{\rm{n}}_1} = 1,}\) from higher energy levels \(\left( {{{\rm{n}}_2} = 2,3,4,5,} \right)\) etc.). The Balmer, Paschen, Brackett, and Pfund series are created when the electron returns to the second, third, fourth, and fifth energy levels, as seen in Fig.

Shortcomings of the Bohr’s Model of Atom 

(1) The Bohr theory’s greatest achievement was its ability to anticipate lines in the spectrum of hydrogen atoms. Every other atom containing more than one electron, on the other hand, was a spectacular failure.

(2) We no longer believe in Bohr’s assumption of well-defined electron orbits. In reality, any mechanical description of the atom is rejected in light of recent advances such as the dual nature of matter and the uncertainty principle. 

(3) The capacity of atoms to form molecules through chemical bonding was not accounted for by Bohr’s model of electronic structure. No one has been able to explain atomic spectra without numerical quantisation, and no one has attempted to characterise atoms using classical physics; therefore, we only accept Bohr’s thoughts on quantisation today.

(4) The influence of magnetic fields (Zeeman effect) and electric fields (Stark effect) on the spectra of atoms was not explained by Bohr’s theory.

Summary

Rutherford’s nuclear model basically claimed that each atom had a nucleus and that negative electrons existed outside of it. It didn’t say how or where the electrons were organised. It also couldn’t explain why electrons didn’t fall into the nucleus despite electrical attraction.  Niels Bohr suggested a new atom model in \(1913\) that described some of these phenomena as well as the hydrogen emission spectrum. Bohr’s hypothesis was founded on certain postulates and was based on Planck’s quantum theory. Bohr was able to determine the energy of electrons for hydrogen and hydrogen-like ions. He was also able to calculate the radius of different orbits

Frequently Asked Questions (FAQs) 

Q.1. Discuss the postulates given by Bohr to explain the Hydrogen atom.
Ans: The postulates of Bohr’s theory are as follows:
(i) Electrons can only travel around the nucleus in certain circular orbits that are allowed.
(ii) An electron does not emit (or lose) energy when revolving in these precise orbits.
(iii) Only quantum or photon jumps allow an electron to shift from one energy level to another.
(iv) An electron orbiting around the nucleus has an angular momentum (mvr) that is an integral multiple of Planck’s constant divided by \({\rm{2\pi }}.\)

Q.2. What is Bohr’s explanation of the hydrogen spectrum?
Ans: When energy is supplied to hydrogen gas in the discharge tube, the electron goes to higher energy levels. When an electron returns to the ground state, \({{\rm{n}}_1} = 1,\) from higher energy levels, the Lyman series is formed. Similarly, the Balmer, Paschen, Brackett, and Pfund series are created when the electron returns to the second, third, fourth, and fifth energy levels.

Q.3. Give the formulas to calculate the energy and radius of Bohr’s orbit.
Ans: The formulae are given as:
\({\rm{r = 0}}{\rm{.529 \times 1}}{{\rm{0}}^{{\rm{ – 8}}\,}}{\rm{cm}}\)
\({\rm{E = }}\frac{{{\rm{ – 2}}{\rm{.179 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}}}{{{{\rm{n}}^{\rm{2}}}}}\,{\rm{J}}\,{\rm{per}}\,{\rm{atom}}\)

Q.4. What are the shortcomings of Bohr’s model of the atom?
Ans: Following are the shortcomings of Bohr’s model:
(i) The Bohr theory’s greatest achievement was its ability to anticipate lines in the spectrum of hydrogen atoms. Every other atom containing more than one electron, on the other hand, was a spectacular failure.
(ii) The capacity of atoms to form molecules through chemical bonding was not accounted for by Bohr’s model of electronic structure.
(ii) The influence of magnetic fields (Zeeman effect) and electric fields (Stark effect) on the spectra of atoms was not explained by Bohr’s theory.

Q.5. In which region of the electromagnetic spectrum does the Lyman series of lines fall?
Ans: Ultraviolet region.

Study Law Of Conservation Of Mass Here  

We hope this article on Bohr’s Model for Hydrogen and Hydrogen Like Atoms has helped you. If you have any queries, drop a comment below, and we will get back to you at the earliest.