Bohr’s Atomic Model

To overcome the difficulties and drawbacks of Rutherford’s atomic model, a Danish physicist Neil Bohr suggested a new atomic model in 1913 known as Bohr’s atomic model.

The main postulates of Bohr’s atomic model are as follows;

  • An electron in an atom revolves around the nucleus in discrete circular orbits called stationary orbits where the electrostatic force between the electron and the nucleus provides the required centripetal force i.e.
    \[\frac{mv_n^2}{r_n}=\frac{1}{4πε_o}\frac{Ze.e}{r_n^2}\] Where, $m =$ mass of electron
    $Ze =$ charge of nucleus
    $r_n =$ radius of nth orbit
    $v_n =$ velocity of electron in nth orbit
    $ε_0 =$ permittivity of free space between nucleus and orbit
  • An electron revolves only in those orbits where its angular momentum is an integral multiple of $\frac{h}{2π}$ i.e.
    \[mv_nr_n=n\frac{h}{2π}\] Where, $n= 1, 2, 3, …$ is an integer called principal quantum number and $h$ is Planck’s constant.
    This relation shows that the angular momentum is quantized. This is known as Bohr’s quantization condition.
  • An electron jumps from lower energy level to higher energy level by absorbing certain amount of energy and it jumps from higher energy level to lower energy level by emitting certain amount of energy in the form of radiation such that
    \[E_{\text{higher}}-E_{\text{lower}}=hf\] Where, $f$ is the frequency of the radiation absorbed or emitted during the transition.
    This relation is known as Bohr’s frequency condition.

Mathematical Analysis of Bohr’s Atomic Model

Let us consider an atom having atomic number $Z$ and nuclear charge $+Ze$.

Bohr's Atomic Model (Hydrogen Atom)

If an electron of charge $e$ and mass $m$ revolves around the nucleus of an atom in $n^{\text{th}}$ orbit of radius $r_n$ with speed $v_n$, then the electrostatic force between the electron and the nucleus provides the required centripetal force to keep the electron in a stable orbit i.e.

\[\frac{mv_n^2}{r_n}=\frac{1}{4πε_o}\frac{Ze.e}{r_n^2}\] \[mv_n^2=\frac{Ze^2}{4πε_0r_n} \text{ ____(1)}\]

From Bohr’s quantization condition, \[mv_nr_n=n\frac{h}{2π}\] \[v_n=\frac{nh}{2πmr_n} \text{ ____(2)}\]

Radius of nth orbit

Putting the value of $v_n$ in equation $(1)$, \[m\left(\frac{nh}{2πmr_n}\right)^2=\frac{Ze^2}{4πεor_n}\] \[r_n=\frac{ε_on^2h^2}{πmZe^2} \text{ _(3)}\]

For an atom, all the quantities of RHS in above equation are constant except $n$. \[\therefore r_n∝n^2\]

Hence, radius of an orbit is directly proportional to the square of the principal quantum number which means that the stationary orbits are not equally spaced.

Putting $ε_0=8.86×10^{-12}$ $\text{Fm}^{-1}$, $m=9.1×10^{-31}$ $\text{kg}$, $e=1.6×10^{-19}$ $\text{C}$ and $h=6.62×10^{-34}$ $\text{Js}$ in equation $(3)$, we get, \[r_n=\frac{0.528n^2}{Z}Å \text{ ____(4)}\]

For hydrogen atom, $Z=1$,

\[r_n=0.528n^2Å\]

For first orbit, $n =1$ \[r_1=0.528Å=0.528*10^{-10}m\]

This is the radius of the first orbit of hydrogen atom and this radius is known as Bohr’s radius.

Speed of an electron in nth orbit

Putting the value of $r_n$ from equation $(3)$ in Bohr’s quantization condition, \[mv_n\left(\frac{ε_0n^2h^2}{πmZe^2}\right)=n\frac{h}{2π}\] \[v_n=\frac{Ze^2}{2ε_0nh}\text{ _(5)}\]

From this relation, \[v_n ∝ \frac{1}{n}\]

Therefore, electron moves at a higher speed in lower orbits and vice versa. The speed of an electron in first orbit is maximum and it is zero in infinite state $(n=∞)$.

Total energy of an electron in nth orbit

The kinetic energy of electron due to its motion is given by, \[E_K=\frac{1}{2}mv^2\]

Putting the value of $v_n$ from equation $(5)$, \[E_K=\frac{1}{2}m\left(\frac{Ze^2}{2ε_0nh}\right)^2\] \[E_K=\frac{me^4Z^2}{8ε_0^2n^2h^2}\]

This gives the kinetic energy of the electron and this relation shows that \[E_K ∝ \frac{1}{n^2}\]

Therefore, kinetic energy decreases in higher orbits.

The potential energy of electron due to its position in the orbit under the action of electrostatic force of attraction is given by, \[E_P=-\frac{1}{4πε_0}\frac{Ze.e}{r_n}\]

Putting the value of $r_n$ from equation $(3)$, \[E_P=-\frac{1}{4πε_0}\frac{Ze.e}{\frac{ε_0n^2h^2}{πmZe^2}}\] \[E_P=-\frac{me^4Z^2}{4ε_0^2n^2h^2}\]

This gives the potential energy of the electron. This relation shows that \[E_P ∝ -\frac{1}{n^2}\]

The negative sign means that the potential energy increases in higher orbits.

Now, the total energy of the electron is the sum of the kinetic energy and the potential energy of the electron. \[E_n=E_K+E_P\] \[E_n=\frac{me^4Z^2}{8ε_0^2n^2h^2}-\frac{me^4Z^2}{4ε_0^2n^2h^2}\] \[E_n=-\frac{me^4Z^2}{8ε_0^2n^2h^2}\]

The negative sign shows that the electron is bound to the nucleus i.e. energy is required to separate an electron away from the nucleus. This relation shows that \[E_n∝-\frac{1}{n^2}\]

Therefore, the energy of an electron increases with the increase in principal quantum number $(n)$. For $n=∞$, $E_{\infty}=0$ which is the maximum value of energy.

Putting the value of $ε_0$, $m$ and $h$, we get, \[E_n=-\frac{13.6(Z^2)}{n^2}eV\]

For hydrogen atom, $Z=1$, \[E_n=-\frac{13.6}{n^2}eV\] \[\text{For n=1, } E_n=-13.6 eV\]

Hence, the energy of electron in the ground state of hydrogen atom is $-13.6$ $e\text{V}$.

Ionization Energy

An atom of hydrogen contains a proton in its nucleus and an electron revolving around it. There is electrostatic force between the nucleus and the electron. When they are at infinite distance from each other, they do not interact with each other and in this case, the energy of the electron and nucleus is taken as zero. When they are brought closer to each other, the electrostatic attraction between them starts developing due to which a certain amount of energy is released. Therefore, the energy of the system falls below zero i.e. becomes negative.

This means that the energy of electron in hydrogen atom is less than an isolated electron. To remove the electron from the hydrogen atom, we have to give $+13.6$ $e\text{V}$ of energy. This energy required to remove an electron from the valence shell of a gaseous atom is known as ionization energy.
Hence, the ionization energy of hydrogen atom is $+13.6$ $e\text{V}$.

Frequency and Wavelength of Emitted Radiation

According to Bohr’s atomic model, when an electron jumps from higher energy state $(n_2)$ to lower energy state $(n_1)$ then the radiation of frequency $f$ is emitted such that,

\[hf=E_{n_2}-E_{n_1}\] \[hf=\frac{me^4Z^2}{8ε_0^2h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\] \[f=\frac{me^4Z^2}{8ε_0^2h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\] \[\frac{c}{λ}=\frac{me^4Z^2}{8ε_0^2h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\] \[\frac{1}{λ}=\frac{me^4Z^2}{8ε_0^2h^3c}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\] \[\frac{1}{λ}=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{ ____(6)}\] \[\text{Where, }R=\frac{me^4Z^2}{8ε_0^2h^3c}\]

This $R$ is known as Rydberg constant named after Swedish physicist Johannes Rydberg. It is a physical constant related to atomic spectra and its value is found to be $1.01×10^7$ $\text{m}^{-1}$.
For hydrogen atom, \[\frac{1}{λ}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\]


© 2022 AnkPlanet | All Rights Reserved