Bohr Model of the Hydrogen Atom: Explanation, Formula, Limitation

The first atomic model was proposed by J.J Thomson. As per this model, the positively charged particles are distributed throughout the atom, and negatively charged particles are embedded in it like seeds of a watermelon. Later Rutherford revised the atomic model; with the help of the alpha-scattering experiment, which proved that all positively charged particles are present at the centre of the nucleus. He also proposed that the electron revolves around the nucleus in paths of varying radii. However, his model couldn’t explain the stability of atoms and discrete wavelengths in the hydrogen spectrum. Then Bohr gives a model for hydrogen-like atoms and explains the stability and discrete wavelengths in the hydrogen spectrum. This article discusses the Bohr Model of the Hydrogen Atom in detail. Read on to learn more.

Bohr Model of the Hydrogen Atom

As the Rutherford model does not explain about structure of the atom completely, it gives only a part of the story implying that the classical ideas. Niels Bohr made some modifications to the Rutherford model by adding the ideas of the newly developing quantum hypothesis. To explain the position, motion of the electron, and stability of the atom, he gave three postulates.

Bohr’s Postulates

(i) The electrons in an atom revolve around the nucleus in circular orbits without the emission of radiant energy. In these orbits of special radii, the electron does not radiate energy as expected from Maxwell’s laws. These orbits are called the stationary states of the atom.
(ii) The electron revolves around the nucleus only in that stationary orbits for which the angular momentum is some integral multiple of \(\frac{h}{2π}\), where \(h\) is the Planck’s constant \((h = 6.6 × 10^{-34}\,\rm{J}\,\rm{s})\). Thus the angular momentum \((L)\) of the orbiting electron is quantized. That is \(L = \frac{nh}{2π}\). This assumption is known as Bohr’s quantization rule.
(iii) The energy of the atoms has a defined value in a given stationary orbit. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, it emits a photon of radiation having energy equal to the energy difference between the initial and final states. The wavelength of the emitted photon is then given by,
\(\lambda = \frac{{hc}}{{{E_i} – {E_f}}}\)
where \(E_i\) and \(E_f\) are the energies of the initial and final states and \(E_i > E_f.\)

Energy Electron in Hydrogen Atom

We will use the above postulate to find the allowed energies of the atom for different allowed orbits of the electron. This theory is only developed for the hydrogen atom and ions having just one electron. These ions are often called hydrogen-like ions. Let us assume that nucleus having the positive charge \(Ze^+\) (\(Z\) is the number of protons) and electron moves in a circular orbit of radius \(r\) with velocity \(v\). The Coulomb attraction force acting on the electron, and it is given by,
\(F = \frac {Ze^2}{4π ε_0 r^2}\)
The acceleration of electron is towards the centre and has a magnitude of \(\frac{v^2}{r}\). If \(m\) is the mass of the electron, from Newton’s law,
\(\frac {mv^2}{r} = \frac {Ze^2}{4πε_0 r^2}\)
\(r = \frac{Ze^2}{4πε_0 mv^2}\) …….(i)
Also, from Bohr’s quantization rule, the angular moment is,
\(mvr = n \frac {h}{2π}\) …….(ii)
where \(n\) is a positive integer.
Eliminating \(r\) from (i) and (ii), we get
\(v = \frac{Ze^2}{2ε_0 hn}\) ……….(iii)
Substituting this in equation (ii),
\(r = \frac {ε_0 h^2 n^2}{πmZe^2}\) ………(iv)
From equation (iii), the kinetic energy of the electron in the \(n^{th}\) orbit is given by,
\(K = \frac {1}{2} mv^2 = \frac {mZ^2 e^4}{8ε_0^2 h^2 n^2}\)
The potential energy of the atom is, \(V = – \frac {Ze^2}{4πε_0 r} = – \frac {mZ^2 e^4}{4ε_0^2 h^2 n^2}\)
We have taken the potential energy of the widely separated nucleus and electron to be zero. The total energy of the atom is,
\(E = K + V\)
\(E = – \frac {mZ^2 e^4}{8ε_0^2 h^2 n^2} = – \frac {RhcZ^2}{n^2}\)
where \(R = \frac {me^4}{8ε_0^2 h^3 c}\) is the Rydberg constant, and its value is \(1.097 \times 10^{7}\,\rm{m}^{−1}\).
For hydrogen and hydrogen-like atoms, the Bohr model of hydrogen gives the energy \((E)\) of an electron present in the \(n\) energy level (orbit) of hydrogen as:
\(E(n) = – \frac {13.6\,\rm{eV}}{n^2}\),
where ‘\(n\)’ is the principal quantum.

Absorption and Emission

According to Bohr’s model, an electron will absorb energy in the form of photons to reach an excited or a higher energy level. After escaping to the higher energy level, also known as the excited state, the excited electron will be less stable, and therefore, would rapidly emit a photon to come back to a lower, more stable energy level. The energy of the emitted photon is adequate to the difference in energy between the two energy levels for a selected transition. The energy emitted or absorbed is calculated using the equation,
\(hv = ∆E = (\frac {1}{n^2} – \frac {1}{m^2})13.6\,\rm{eV}\)
where \(h\) is the Plank constant and \(v\) is the frequency of the photons.

Radii of Different Orbits

From equation (iv), the radius of the smallest circle allowed to the electron is \((n = 1)\).
\(r = \frac {ε_0 h^2 n^2}{πmZe^2}\)
For a hydrogen atom, \(Z = 1\), and plugging the values of all constants we will get, \(r_1 = 53\) picometers \((1\,\rm{pm} = 10^{-12}\,\rm{m})\) or \(0.053\,\rm{nm}\). This length is also called Bohr’s radius. It is denoted by the symbol \(a_0\). For a hydrogen atom, \(2^{nd}\) allowed radius is \(4a_0\), \(3^{rd}\) allowed radius is \(9a_0\), and so on. The general equation of the radius of the \(n^{th}\) orbit of the hydrogen atom is
\(r_n = n^2 a_0\)
The atomic radius of the hydrogen-like ion with \(Z\) proton in the nucleus is,
\(r_n = \frac {n^2 a_0}{Z}\)

Limitations of Bohr’s Theory

1. It does not explain the spectra of atoms having more than one electron.
2. It does not talk about the fine structure of the spectral lines of hydrogen atoms.
3. It will not give any information about the relative intensities of an atom’s spectral lines.
4. It does not explain to account for the splitting of spectral lines in a magnetic (Zeeman effect) or electric field (Stark effect).
5. It does not follow Heisenberg’s uncertainty principle.

Summary

J.J Thomson’s model and Rutherford’s model came before Bohr’s model, but they were unable to explain the proper structure of the atom. Rutherford’s model was not able to explain the stability of atoms. Bohr proposed an atomic model and explained the stability of an atom. Bohr proposed that electrons move around the nucleus in specific circular orbits. The radius of those specific orbits is given by,
\(r = \frac {Ze^2}{4πε_0 mv^2}\)
According to Bohr’s calculation, for the hydrogen atom, the energy for an electron in the specific orbit is given by the expression:
\(E(n) = -\frac {13.6\,\rm{eV}}{n^2}\)
The hydrogen spectrum can be explained in terms of electrons absorbing and emitting photons to change energy levels, where the photon energy is:
\(hv = ∆E = (\frac {1}{n^2} – \frac {1}{m^2}) × 13.6\,\rm{eV}\)
Bohr’s Model of the Hydrogen Atom doesn’t apply to atoms having more than one electron.

Bohr Model for Hydrogen: Sample Problems

Q.1. Calculate the wavelength of radiation emitted when an electron in a hydrogen atom makes a transition from the state \(n = 3\) to the state \(n = 2\):
Ans: The wavelength of the emitted photon is given by,
\(\frac{1}{λ} = \frac {E_i – E_f}{hc} = RZ^2 (\frac {1}{n^2} – \frac {1}{m^2})\)
where ‘\(R\)’ is the Rydberg constant and \(Z\) is the atomic number of the atom.
\(\frac{1}{λ} = R × 1^2 (\frac{1}{2^2} – \frac {1}{3^2})\)
\(λ = \frac {36}{5R} = \frac {36}{5 × 1.09737 × 10^7\,\rm{m}^{-1}} = 656\,\rm{nm}\)

Q.2. If \(a_0\) is the Bohr radius, the radius of the \(n = 2\) electron’s orbit in triply ionized beryllium is:
Ans: Given,
Orbit no \(n = 2\)
We know the atomic number of beryllium is \(Z = 4\),
To calculate the radius of the hydrogen equivalent system, we use the following formula:
\(r_n = \frac {n^2 a_0}{Z}\)
where \(Z\) is the atomic number of elements, \(n\) is the number of orbits and \(r\) is the total radius.
\(r_2 = \frac {2^2 a_0}{4} = a_0\)
The radius of the second orbit of the triply ionized beryllium is: \(a_0\).

FAQs on Bohr Model of Hydrogen Atom