Bohr's atomic model: Class 12 Physics Notes, Definition, Working Principle, Formula & Real-Life Applications

The following are the postulates given by the Bohr Model of the Hydrogen Atom:

1. Bohr said that electrons don’t just move anywhere—they follow certain fixed paths around the nucleus. These paths don’t change unless the electron gains or loses energy.
2. Each path (or orbit) has a fixed amount of energy. If an electron stays in one path, its energy stays the same. That’s why atoms don’t collapse or give off energy all the time.
   
3. When electrons jump from one orbit to another, they either take in energy or give it out. If they go to a higher level, they absorb energy. If they drop to a lower one, they release energy as light.
   
4. Not all orbits are allowed. Only certain ones work, and that’s based on a rule about the electron’s motion—it has to fit in a way that depends on a number called Planck’s constant.

In 1913, Prof. Niel Bohr removed the difficulties of Rutherford's atomic model by the application of Planck's quantum theory. For this he proposed the following postulates
(1) An electron moves only in certain circular orbits, called stationary orbits. In stationary orbits electron does not emit radiation, contrary to the predictions of classical electromagnetic theory.

(2) According to Bohr, there is a definite energy associated with each stable orbit and an atom radiaties energy only when it makes a transition from one of these orbits to another. If the energy of electron in the higher orbit be $E_2$ and that in the lower orbit be $E_1$ , then the frequency $v$ of the radiated waves is given by $\begin{array}{c}hv={\mathrm{E}}_2-{\mathrm{E}}_1 or v=\frac{{\mathrm{E}}_2-{\mathrm{E}}_1}{\mathrm{}\mathrm{h}}\#\left(i\right)\end{array}$

(3) Bohr found that the magnitude of the electron's angular momentum is quantized, and this magnitude for the electron must be integral multiple of $\frac{h}{2\pi }$ . The magnitude of the angular momentum is $L=mvr$ for a particle with mass $m$ moving with speed $v$ in a circle of radius $r$ . So, according to Bohr's postulate, $(\mathrm{n}=\mathrm{1,2},3\dots$ )

Each value of $n$ corresponds to a permitted value of the orbit radius, which we will denote by $r_n$ The value of $n$ for each orbit is called principal quantum number for the orbit. Thus, $\begin{array}{c}m{v}_n{r}_n=mvr=\frac{nh}{2\pi }\#\left(ii\right)\end{array}$

According to Newton's second law a radially inward centripetal force of magnitude $F=\frac{m{v}^2}{r_n}$ is needed by the electron which is being provided by the electrical attraction between the positive proton and the negative electron.
Thus, $\frac{{\mathrm{m}\mathrm{v}}_n^2}{{\mathrm{r}}_{\mathrm{n}}}=\frac{1}{4\pi {\epsilon }_0}\frac{{\mathrm{e}}^2}{{\mathrm{r}}_n^2}$
Solving Eqs. (ii) and (iii), we get

$\begin{array}{c}{r}_n=\frac{{\epsilon }_0{n}^2{h}^2}{\pi m{e}^2}\#\left(iv\right)\\ and v_n=\frac{e^2}{2{\epsilon }_{0}nh}\#\left(v\right)\end{array}$

The smallest orbit radius corresponds to $n=1$ . We'll denote this minimum radius, called the Bohr radius as ao. Thus, ${\mathrm{a}}_0=\frac{{\epsilon }_0{\mathrm{}\mathrm{h}}^2}{\pi {\mathrm{m}\mathrm{e}}^2}$

Substituting values of ${\epsilon }_0,\mathrm{}\mathrm{h},\mathrm{p},\mathrm{m}$ and e , we get $\begin{array}{c}{\mathrm{a}}_0=0.529\times {10}^{-10}m=0.529Å\#\left(vi\right)\end{array}$

Eq. (iv), in terms of ao can be written as, $\begin{array}{c}{r}_n=n^2{a}_0 or r_n\propto n^2\#\left(vii\right)\end{array}$

Similarly, substituting values of $e,{\epsilon }_0$ and $h$ with $n=1$ in Eq. (v), we get $\begin{array}{c}{v}_1=2.19\times {10}^{6}m/s\#\left(viii\right)\end{array}$

This is the greatest possible speed of the electron in the hydrogen atom. Which is approximately equal to $c/137$ where $c$ is the speed of light in vacuum.
Eq. (v), in terms of $v_1$ can be written as, ${\mathrm{v}}_{\mathrm{n}}=\frac{{\mathrm{v}}_1}{\mathrm{n}}\mathrm{}\mathrm{o}\mathrm{r}\mathrm{}{\mathrm{v}}_{\mathrm{n}}\propto \frac{1}{\mathrm{n}}$

Energy levels: Kinetic and potential energies $K_n$ and $U_n$ in $n$ th orbit are given by ${\mathrm{K}}_{\mathrm{n}}=\frac{1}{2}{\mathrm{m}\mathrm{v}}_{\mathrm{n}}{}^2=\frac{{\mathrm{m}\mathrm{e}}^4}{8{\epsilon }_0^2{\mathrm{n}}^2{\mathrm{}\mathrm{h}}^2}\mathrm{a}\mathrm{n}\mathrm{d}{U}_{\mathrm{n}}=-\frac{1}{4\pi {\epsilon }_0}\frac{{\mathrm{e}}^2}{{\mathrm{r}}_{\mathrm{n}}}=-\frac{{\mathrm{m}\mathrm{e}}^4}{4{\epsilon }_0^2{\mathrm{n}}^2{h}^2}$

(assuming infinity as a zero potential energy level)
The total energy $E_n$ is the sum of the kinetic and potential energies.
so, ${\mathrm{E}}_{\mathrm{n}}={\mathrm{K}}_{\mathrm{n}}+{\mathrm{U}}_{\mathrm{n}}=-\frac{{\mathrm{m}\mathrm{e}}^4}{8{\epsilon }_0^2{\mathrm{n}}^2{\mathrm{}\mathrm{h}}^2}$

Substituting values of $m,e,{\epsilon }_0$ and $h$ with $n=1$ , we get the least energy of the atom in first orbit, which is -13.6 eV . Hence,

$\begin{array}{c}{E}_1=-13.6eV\#\left(x\right)\\ E_n=\frac{E_1}{n^2}=-\frac{13.6}{n^2}eV\#\left(xi\right)\end{array}$

and
Substituting $n=\mathrm{2,3},4,\dots$ , etc., we get energies of atom in different orbits.

$E_2=-3.40\mathrm{e}\mathrm{V},E_3=-1.51\mathrm{e}\mathrm{V},\dots .{\mathrm{E}}_{\mathrm{\infty }}=0$

The Bohr model of hydrogen can be extended to hydrogen like atoms, i.e., one electron atoms, the nuclear charge is +ze, where $z$ is the atomic number, equal to the number of protons in the nucleus. The effect in the previous analysis is to replace $e^2$ every where by $z^2$ . Thus, the equations for, $r_n,v_n$ and $E_n$ are altered as under:

$\begin{array}{c}{r}_n=\frac{{\epsilon }_0{n}^2{h}^2}{nmz{e}^2}= ao \frac{n^2}{z} or r_n\propto \frac{n^2}{z}\#\left(i\right)\end{array}$

where $a_0=0.529\text{Å}\mathrm{}$ (radius of first orbit of $H$ )

$\begin{array}{c}{v}_n=\frac{z{e}^2}{2{\epsilon }_{0}nh}=\frac{z}{n}{v}_1 or v_n\propto \frac{z}{n}\#\left(ii\right)\end{array}$

where ${\mathrm{v}}_1=2.19\times {10}^6\mathrm{}\mathrm{m}/\mathrm{s}\mathrm{}$ (speed of electron in first orbit of $H$ )

$\begin{array}{c}{E}_n=-\frac{m{z}^2{e}^4}{8{\epsilon }_0^2{n}^2{h}^2}=\frac{z^2}{n^2}{E}_1 or E_n\propto \frac{z^2}{n^2}\#\left(iii\right)\end{array}$

where $E_1=-13.60\mathrm{e}\mathrm{V}\mathrm{}$ (energy of atom in first orbit of H )

(1) Ground state : Lowest energy state of any atom or ion is called ground state of the atom.

Ground state energy of H atom $=-13.6\mathrm{e}\mathrm{V}$
Ground state energy of ${\mathrm{H}\mathrm{e}}^{+}$ Ion $=-54.4\mathrm{e}\mathrm{V}$
Ground state energy of ${\mathrm{L}\mathrm{i}}^{++}$ Ion $=-122.4\mathrm{e}\mathrm{V}$

(2) Excited State : State of atom other than the ground state are called its excited states.

 

$n=2$	first excited state
$n=3$	second excited state
$n=4$	third excited state
$n=n_0+1$	excited state

 

(3) Ionisation energy (I.E.) : Minimum energy required to move an electron from ground state to  is called ionisation energy of the atom or ion
Ionisation energy of H atom $=13.6\mathrm{e}\mathrm{V}$
Ionisation energy of ${\mathrm{H}\mathrm{e}}^{+}$ Ion $=54.4\mathrm{e}\mathrm{V}$
Ionisation energy of ${\mathrm{L}\mathrm{i}}^{++}$ Ion $=122.4\mathrm{e}\mathrm{V}$

(4) Ionisation potential (I.P.) : Potential difference through which a free electron must be accelerated from rest such that its kinetic energy becomes equal to ionisation energy of the atom is called ionisation potential of the atom.
I. P of H atom $=13.6\mathrm{}\mathrm{V}$
I.P. of ${\mathrm{H}\mathrm{e}}^{+}$ Ion $=54.4\mathrm{}\mathrm{V}$

(5) Excitation energy : Energy required to move an electron from ground state of the atom to any other exited state of the atom is called excitation energy of that state.
Energy in ground state of H atom $=-13.6\mathrm{e}\mathrm{V}$
Energy in first excited state of H -atom $=-3.4\mathrm{e}\mathrm{V}$
$I^{\mathrm{s}\mathrm{t}}$ excitation energy $=10.2\mathrm{e}\mathrm{V}$ .

(6) Excitation Potential : Potential difference through which an electron must be accelerated from rest so that its kinetic energy becomes equal to excitation energy of any state is called excitation potential of that state.
$I^{\mathrm{s}\mathrm{t}}$ excitation energy $=10.2\mathrm{e}\mathrm{V}$ .
$I^{\mathrm{s}\mathrm{t}}$ excitation potential $=10.2\mathrm{}\mathrm{V}$ .

(7) Binding energy or Seperation energy : Energy required to move an electron from any state to $\mathrm{n}=\mathrm{\infty }$ is called binding energy of that state. or energy released during formation of an H -like atom/ion from $\mathrm{n}=\mathrm{\infty }$ to some particular n is called binding energy of that state.
Binding energy of ground state of H -atom $=13.6\mathrm{e}\mathrm{V}$

Under normal conditions the single electron in hydrogen atom stays in ground state ( $n=1$ ). It is excited to some higher energy state when it acquires some energy from external source. But it hardly stays there for more than ${10}^{-8}$ second. A photon corresponding to a particular spectrum line is emitted when an atom makes a transition from a state in an excited level to a state in a lower excited level or the ground level.

Let $n_i$ be the initial and $n_f$ the final energy state, then depending on the final energy state following series are observed in the emission spectrum of hydrogen atom.

A photograph of spectral lines of the Lyman, Balmer, Paschen series of atomic hydrogen.

1, 2, 3.... represents the I, II & III line of Lyman, Balmer, Paschen series.

The hydrogen spectrum (some selected lines)

Name of series	Number of Line	Quantum Number
		(Lower State	(Upper State)	Wavelength ( nm )	Energy
Lymen	I	1	2	121.6	10.2 eV
	II	1	3	102.6	12.09 eV
	III	1	4	97	12.78 eV
	series limit	1	∞ (series limit)	91.2	13.6 eV
Balmer	1	2	3	656.3	1.89 eV
	II	2	4	486.1	2.55 eV
	III	2	5	434.1	2.86 eV
	series limit	2	∞(series limit)	364.6	3.41 eV
Paschen	1	3	4	1875.1	0.66 eV
	II	3	5	1281.8	0.97 eV
	III	3	6	1093.8	1.13 eV
	series limit	3	∞ (series limit)	822	1.51 eV

Series limit : Line of any group having maximum energy of photon and minimum wavelength of that group is called series limit.

For the Lyman series ${\mathrm{n}}_{\mathrm{f}}=1$ , for Balmer series ${\mathrm{n}}_{\mathrm{f}}=2$ and so on.

According to Bohr when an atom makes a transition from higher energy level to a lower level it emits a photon with energy equal to the energy difference between the initial and final levels. If $E_i$ is the initial energy of the atom before such a transition, ${\mathrm{E}}_{\mathrm{f}}$ is its final energy after the transition, and the photon's energy is $h\nu =\frac{hc}{\lambda }$ , then conservation of energy gives,

$\begin{array}{c}h\nu =\frac{hc}{\lambda }=E_i-E_f \left(energyofemittedphoton\right) \#\left(i\right)\end{array}$

By 1913, the spectrum of hydrogen had been studied intensively. The visible line with longest wavelength, or lowest frequency is called ${\mathrm{H}}_{\alpha }$ , the next line is called ${\mathrm{H}}_{\beta }$ and so on.
In 1885, Johann Balmer, a Swiss teacher found a formula that gives the wave lengths of these lines. This is now called the Balmer series. The Balmer's formula is, $\begin{array}{c}\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{{\mathrm{n}}^2}\right)\#\left(ii\right)\end{array}$

Here, $n=\mathrm{3,4},5\dots$ , etc.
$\mathrm{R}=$ Rydberg constant $=1.097\times {10}^7{\mathrm{}\mathrm{m}}^{-1}$ and $\lambda$  is the wavelength of light/photon emitted during transition,
For n = 3, we obtain the wavelength of ${\mathrm{H}}_{\alpha }$ line.
Similarly, for n = 4, we obtain the wavelength of ${\mathrm{H}}_{\alpha }$ line. For n = ∞, the smallest wavelength (=3646Å) of this series is obtained. Using the relation,E = hc/ $\lambda$ we can find the photon energies corresponding to the wavelength of the Balmer series.

This formula suggests that,

The wavelengths corresponding to other spectral series (Lyman, Paschen, (etc.) can be represented by formula similar to Balmer's formula.

The Lyman series is in the ultraviolet, and the Paschen. Brackett and Pfund series are in the infrared region.