2.4 Bohr’s Model for Hydrogen Atom

Neils Bohr (1913) was the first to explain quantitatively the general features of the structure of hydrogen atom and its spectrum. He used Planck’s concept of quantisation of energy. Bohr’s model for hydrogen atom is based on the following postulates:

(i) The electron in the hydrogen atom can move around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus.

(ii) The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state when required amount of energy is absorbed by the electron or energy is emitted when electron moves from higher stationary state to lower stationary state (equation 2.16). The energy change does not take place in a continuous manner.

iii) The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by \(\Delta E\), is given by:
\(
f=\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h} \dots ….(2.10)
\)
Where \(E_{1}\) and \(E_{2}\) are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr’s frequency rule.
iv) The angular momentum of an electron is quantised. In a given stationary state it can be expressed as in equation (2.11)
\(
m_{e} \mathrm{v} r=n \cdot \frac{h}{2 \pi} \quad n=1,2,3 \ldots \ldots … (2.11)
\)

Where \(m_{e}\) is the mass of electron, \(v\) is the velocity and \(r\) is the radius of the orbit in which electron is moving. Thus an electron can move only in those orbits for which its angular momentum is integral multiple of \(h / 2 \pi\). That means angular momentum is quantised. 

According to Bohr’s theory for hydrogen atom:

(a) The stationary states for electron are numbered \(n=1,2,3 \ldots \ldots \ldots\) These integral numbers are known as Principal quantum numbers.

(b) The radii of the stationary states are expressed as:
\(r_{\mathrm{n}}=n^{2} a_{0} \dots ….(2.12)\)
where \(a_{0}=52.9 \mathrm{pm}\). Thus the radius of the first stationary state, called the Bohr orbit, is \(52.9 \mathrm{pm}\). Normally the electron in the hydrogen atom is found in this orbit (that is \(n=1\) ). As \(n\) increases the value of \(r\) will increase. In other words the electron will be present away from the nucleus.

(c) The most important property associated with the electron, is the energy of its stationary state. It is given by the expression.
\(
E_{n}=-\mathrm{R}_{H}\left(\frac{1}{n^{2}}\right) \quad n=1,2,3 \ldots \dots ….(2.13)
\)
where \(R_{H}\) is called Rydberg constant and its value is \(2.18 \times 10^{-18} \mathrm{~J}\). The energy of the lowest state, also called as the ground state, is \(E_{1}=-2.18 \times 10^{-18}\left(\frac{1}{1^{2}}\right)=-2.18 \times 10^{-18} \mathrm{~J}\). The energy of the stationary state for \(n=2\), will be : \(E_{2}=-2.18 \times 10^{-18} \mathrm{~J}\left(\frac{1}{2^{2}}\right)=-0.545 \times 10^{-18} \mathrm{~J}\).

Fig. 2.11 depicts the energies of different stationary states or energy levels of hydrogen atom. This representation is called an energy level diagram. When the electron is free from the influence of nucleus, the energy is taken as zero. The electron in this situation is associated with the stationary state of Principal Quantum number \(=n=\infty\) and is called as ionized hydrogen atom. When the electron is attracted by the nucleus and is present in orbit n, the energy is emitted and its energy is lowered. That is the reason for the presence of negative sign in equation (2.13) and depicts its stability relative to the reference state of zero energy and \(n=\infty\).

(d) Bohr’s theory can also be applied to the ions containing only one electron, similar to that present in hydrogen atom. For example, \(\mathrm{He}^{+} \mathrm{Li}^{2+}, \mathrm{Be}^{3+}\) and so on. The energies of the stationary states associated with these kinds of ions (also known as hydrogen-like species) are given by the expression.
\(E_{\mathrm{n}}=-2.18 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right) \mathrm{J} \dots ….(2.14)\)
and radii by the expression
\(
r_{n}=\frac{52.9\left(n^{2}\right)}{Z} \mathrm{pm} \dots ….(2.15)
\)
where \(Z\) is the atomic number and has values 2,3 for the helium and lithium atoms respectively. From the above equations, it is evident that the value of energy becomes more negative and that of radius becomes smaller with increase of \(Z\). This means that electron will be tightly bound to the nucleus.

(e) It is also possible to calculate the velocities of electrons moving in these orbits. Although the precise equation is not given here, qualitatively the magnitude of velocity of electron increases with the increase of positive charge on the nucleus and decreases with the increase of principal quantum number.

Angular Momentum

Just as linear momentum is the product of mass \((m)\) and linear velocity (v), angular momentum is the product of moment of inertia \((I)\) and angular velocity \((\omega)\). For an electron of mass \(m_{e}\), moving in a circular path of radius \(r\) around the nucleus,
angular momentum \(=I \times \omega\)
Since \(I=m_{\mathrm{e}} \mathrm{r}^{2}\), and \(\omega=\mathrm{v} / r\) where \(\mathrm{v}\) is the linear velocity,
\(\therefore\) angular momentum \(=m_{e} r^{2} \times \mathrm{v} / r=m_{\mathrm{e}} \mathrm{v} r\)

Explanation of Line Spectrum of Hydrogen

Line spectrum observed in case of hydrogen atom, can be explained quantitatively using Bohr’s model. The energy gap between the two orbits is given by equation (2.16)

\(\Delta E=E_{\mathrm{f}}-E_{1} \dots ……(2.16)\)
Combining equations (2.13) and (2.16)

\(\Delta E=\left(-\frac{\mathrm{R}_{\mathrm{H}}}{n_{\mathrm{f}}^{2}}\right)-\left(-\frac{\mathrm{R}_{\mathrm{H}}}{n_{i}^{2}}\right)\) (where \(n_{i}\) and \(n_{\mathrm{f}}\) stand for initial orbit and final orbits)
\(
\Delta E=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{\mathrm{f}}^{2}}\right)=2.18 \times 10^{-18} \mathrm{~J}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{\mathrm{f}}^{2}}\right)
\dots ….(2.17)\)
The frequency \((f)\) associated with the absorption and emission of the photon can be evaluated by using equation
\(
f=\frac{\Delta E}{h}=\frac{\mathrm{R}_{\mathrm{H}}}{h}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{\mathrm{f}}^{2}}\right)
\)
\(
=\frac{2.18 \times 10^{-18} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{\mathrm{f}}^{2}}\right) \dots …..(2.18)
\)

\(
=3.29 \times 10^{15}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{\mathrm{f}}^{2}}\right) \mathrm{Hz} \dots …..(2.19)
\)
and in terms of wavenumbers \((\bar{\nu})\)
\(
\begin{aligned}
\bar{\nu} &=\frac{v}{\mathrm{c}}=\frac{\mathrm{R}_{\mathrm{H}}}{h \mathrm{c}}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{\mathrm{f}}^{2}}\right) \dots ….(2.20) \\
&=\frac{3.29 \times 10^{15} \mathrm{~s}^{-1}}{3 \times 10^{8} \mathrm{~ms}^{-\mathrm{s}}}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{\mathrm{f}}^{2}}\right) \\
&=1.09677 \times 10^{7}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{\mathrm{f}}^{2}}\right) \mathrm{m}^{-1} \dots …..(2.21)
\end{aligned}
\)

In case of absorption spectrum, \(n_{\mathrm{f}}>n_{i}\) and the term in the parenthesis is positive and energy is absorbed. On the other hand in case of emission spectrum \(n_{i}>n_{\mathrm{f}}, \Delta E\) is negative and energy is released.

The expression (2.17) is similar to that used by Rydberg (2.9) derived empirically using the experimental data available at that time. Further, each spectral line, whether in absorption or emission spectrum, can be associated to the particular transition in hydrogen atom.  The brightness or intensity of spectral lines depends upon the number of photons of same wavelength or frequency absorbed or emitted.

Limitations of Bohr’s Model

Bohr’s model of the hydrogen atom was no doubt an improvement over Rutherford’s nuclear model, as it could account for the stability and line spectra of hydrogen atom and hydrogen like ions (for example, \(\mathrm{He}^{+}, \mathrm{Li}^{2+}, \mathrm{Be}^{3+}\), and so on). However, Bohr’s model was too simple to account for the following points.

i) It fails to account for the finer details (doublet, that is two closely spaced lines) of the hydrogen atom spectrum observed by using sophisticated spectroscopic techniques. This model is also unable to explain the spectrum of atoms other than hydrogen, for example, helium atom which possesses only two electrons. Further, Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).

ii) It could not explain the ability of atoms to form molecules by chemical bonds.

Problem 2.10
What are the frequency and wavelength of a photon emitted during a transition from \(n=5\) state to the \(n=2\) state in the hydrogen atom?

Solution: Since \(n_{i}=5\) and \(n_{\mathrm{f}}=2\), this transition gives rise to a spectral line in the visible region of the Balmer series. From equation (2.17)
\(
\begin{aligned}
\Delta E &=2.18 \times 10^{-18} \mathrm{~J}\left[\frac{1}{5^{2}}-\frac{1}{2^{2}}\right] \\
&=-4.58 \times 10^{-19} \mathrm{~J}
\end{aligned}
\)
It is an emission energy
The frequency of the photon (taking energy in terms of magnitude) is given by
\(
\begin{aligned}
&f=\frac{\Delta E}{h} \\
&=\frac{4.58 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}} \\
&=6.91 \times 10^{14} \mathrm{~Hz} \\
&\lambda=\frac{\mathrm{c}}{f}=\frac{3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}}{6.91 \times 10^{14} \mathrm{~Hz}}=434 \mathrm{~nm}
\end{aligned}
\)

Problem 2.11
Calculate the energy associated with the first orbit of \(\mathrm{He}^{+}\). What is the radius of this orbit?

Solution:
\(E_{\mathrm{n}}=-\frac{\left(2.18 \times 10^{-18} \mathrm{~J}\right) Z^{2}}{n^{2}}\) atom \(^{-1}\)
For \(\mathrm{He}^{+}, n=1, \mathrm{Z}=2\)
\(E_{1}=-\frac{\left(2.18 \times 10^{-18} \mathrm{~J}\right)\left(2^{2}\right)}{1^{2}}=-8.72 \times 10^{-18} \mathrm{~J}\)
The radius of the orbit is given by the equation 
\(
\mathrm{r}_{n}=\frac{(0.0529 \mathrm{~nm}) n^{2}}{Z}
\)
Since \(n=1\), and \(Z=2\)
\(
r_{n}=\frac{(0.0529 \mathrm{~nm}) 1^{2}}{2}=0.02645 \mathrm{~nm}
\)