Bohr's Atomic Model

Bohr developed a model for hydrogen atom and hydrogen like one–electron species (hydrogenic species). He applied quantum theory in considering the energy of an electron bound to the nucleus.

Important Postulates:

$\bullet$An atom consists of a dense nucleus situated at the centre with the electron revolving around it in circular orbits without emitting any energy. The force of attraction between the nucleus and an electron is equal to the centrifugal force of the moving electron.

$\bullet$Of the finite number of circular orbits around the nucleus, an electron can revolve only in those orbits whose angular momentum (mvr) is an integral multiple of factor

$\frac{h}{2\pi }$

mvr = $\frac{nh}{2\pi }$

where, m = mass of the electron

v = velocity of the electron

n = orbit number in which electron is present

r = radius of the orbit

$\bullet$As long as an electron is revolving in an orbit it neither loses nor gains energy. Hence these orbits are called stationary states. Each stationary state is associated with a definite amount of energy and it is also known as energy levels. The greater the distance of the energy level from the nucleus, the more is the energy associated with it. The different energy levels are numbered as 1,2,3,4, ( from nucleus onwards) or K,L,M,N etc.

$\bullet$Ordinarily an electron continues to move in a particular stationary state without losing energy. Such a stable state of the atom is called as ground state or normal state.

$\bullet$If energy is supplied to an electron, it may jump (excite) instantaneously from lower energy (say 1) to higher energy level (say 2, 3, 4, etc.) by absorbing one or more quanta of energy. This new state of electron is called as excited state. The quantum of energy absorbed is equal to the difference in energies of the two concerned levels.

Since the excited state is less stable, atom will lose it's energy and come back to the ground state.

Energy absorbed or released in an electron jump, (DE) is given by

DE = E₂ – E₁ = h$\nu$

where E₁ and E₂ are the energies of the electron in the first and second energy levels, and $\nu$ is the frequency of radiation absorbed or emitted.

Note: If the energy supplied to hydrogen atom is less than 13.6 eV, it will accept or absorb only those quanta which can take it to a certain higher energy level i.e., all those photons having energy less than or more than a particular energy level will not be absorbed by hydrogen atom. But if energy supplied to hydrogen atom is more than 13.6 eV then all photons are absorbed and excess energy appears as kinetic energy of emitted photo electron.

Radius and Energy Levels of Hydrogen Atom:

Consider an electron of mass 'm' and charge 'e' revolving around a nucleus of charge Ze (where, Z = atomic number and e is the charge of the proton) with a tangential velocity v. r is the radius of the orbit in which electron is revolving.

By Coulomb's Law, the electrostatic force of attraction between the moving electron

and nucleus is

Coulombic force = $\frac{\text{KZ}{\text{e}}^{\text{2}}}{{\text{r}}^{\text{2}}}$

K = $\frac{1}{4\pi {\in }_o}$ (where ${\in }_0$ is permittivity of free space)

K = 9 x10⁹ Nm² C^–2

In C.G.S. units, value of K = 1 dyne cm² (esu)^–2

The centrifugal force acting on the electron is $\frac{\text{m}{\text{v}}^{\text{2}}}{\text{r}}$

Since the electrostatic force balances the centrifugal force, for the stable electron orbit.

$\frac{\text{m}{\text{v}}^{\text{2}}}{\text{r}}$=$\frac{\text{KZ}{\text{e}}^{\text{2}}}{{\text{r}}^{\text{2}}}$ … (1)

(or) v² = $\frac{\text{KZ}{\text{e}}^{\text{2}}}{\text{mr}}$ … (2)

According to Bohr's postulate of angular momentum quantization, we have

mvr =$\frac{\text{nh}}{\text{2}\pi }$

v = $\frac{\text{nh}}{\text{2 }\pi \text{mr}}$

v² =$\frac{n^2{h}^2}{4{\pi }^2{m}^2{r}^2}$ … (3)

Equating (2) and (3)

$\frac{KZ{e}^2}{mr}=\frac{n^2{h}^2}{4{\pi }^2{m}^2{r}^2}$

Solving for r we get r = $\frac{n^2{h}^2}{4{\pi }^{2}mKZ{e}^2}$

Where n = 1, 2, 3 - - - - - $\infty$

Hence only certain orbits whose radii are given by the above equation are available for the electron. The greater the value of n, i.e., farther the energy level from the nucleus the greater is the radius.

The radius of the smallest orbit (n=1) for hydrogen atom (Z=1) is r_o.

r_o = $\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}K}$ = $\frac{1^2\times {\left(6.626\times {10}^{-34}\right)}^2}{4\times {\left(3.14\right)}^2\times 9\times {10}^{-31}\times {\left(1.6\times {10}^{-19}\right)}^2\times 9\times {10}^9}$ = 5.29 x10^–11 m = 0.529 Å

Radius of n^th orbit for an atom with atomic number Z is simply written as

r_n = 0.529 x $\frac{n^2}{Z}$Å

Calculation of Energy of an Electron:

The total energy, E of the electron is the sum of kinetic energy and potential energy.

Kinetic energy of the electron = $\frac{m{v}^2}{2}$

Potential energy = $\frac{-\text{K}Z{\text{e}}^{\text{2}}}{\text{r}}$

Total energy = $\frac{m{v}^2}{2}$ –$\frac{\text{K}Z{\text{e}}^{\text{2}}}{\text{r}}$ … (4)

From equation (1) we know that

$\frac{m{v}^2}{r}$= $\frac{KZ{e}^2}{r^2}$

$\therefore$ $\frac{m{v}^2}{2}$= $\frac{KZ{e}^2}{2r}$

Substituting this in equation (4)

Total energy (E) =$\frac{KZ{e}^2}{2r}$– $\frac{KZ{e}^2}{r}$ = $-\frac{KZ{e}^2}{2r}$

Substituting for r, gives us

E =$\frac{2{\pi }^{2}m{Z}^2{e}^4{K}^2}{n^2{h}^2}$ where n = 1, 2, 3……….

This expression shows that only certain energies are allowed to the electron. Since this energy expression consist of so many fundamental constant, we are giving you the following simplified expressions.

E = –21.8 x10^–12 x $\frac{Z^2}{n^2}$ erg per atom

= –21.8 x10^–19 x $\frac{Z^2}{n^2}$ J per atom

= –13.6 x$\frac{Z^2}{n^2}$ eV per atom

(1eV = 3.83 x10^–23 kcal

1eV = 1.602 x10^–12 erg

1eV = 1.602 x10^–19J)

E = –313.6 x$\frac{Z^2}{n^2}$kcal / mole (1 cal = 4.18 J)

The energies are negative since the energy of the electron in the atom is less than the energy of a free electron (i.e., the electron is at infinite distance from the nucleus) which is taken as zero. The lowest energy level of the atom corresponds to n=1, and as the quantum number increases, E becomes less negative.

When n = $\infty$, E = 0, which corresponds to an ionized atom i.e., the electron and nucleus are infinitely separated.

H $\to$ H⁺+ e^– (ionization).

Calculation of Velocity:

We know that

mvr =$\frac{nh}{2\pi }$; v = $\frac{nh}{2\pi mr}$

By substituting for r we get

v = $\frac{2\pi KZ{e}^2}{nh}$

Where except n and Z all are constants

v = 2.18 x10⁸ $\frac{Z}{n}$cm/sec.

Further application of Bohr's work was made, to other one electron species (Hydrogenic ion) such as He⁺ and Li²⁺. In each case of this kind, Bohr's prediction of the spectrum was correct.

Illustration 1. The velocity of electron in the second orbit of $H{e}^{+}$ will be

(A) $2.18\times 10^{6}m/s$ (B) $1.09\times 10^{6}m/s$

(C) $4.36\times 10^{6}m/s$ (D) None of these

Solution: $v_n=\frac{v_{0}Z}{n}$. Hence (A) is correct.

HYDROGEN ATOM

If an electric discharge is passed through hydrogen gas taken in a discharge tube under low pressure, and the emitted radiation is analyzed with the help of spectrograph, it is found to consist of a series of sharp lines in the UV, visible and IR regions. This series of lines is known as line or atomic spectrum of hydrogen. The lines in the visible region can be directly seen on the photographic film.

Each line of the spectrum corresponds to a light of definite wavelength. The entire spectrum consists of six series of lines, each series, known after their discoverer as the Balmer, Paschen, Lyman, Brackett, Pfund and Humphrey series. The wavelength of all these series can be expressed by a single formula.

$\frac{1}{\lambda }=\bar{\nu }$= R $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Where, $\bar{\nu }$= wave number

$\lambda$ = wave length

R = Rydberg constant (109678 cm^–1)

n₁ and n₂ have integral values as follows

The pattern of lines in atomic spectrum is characteristic of hydrogen.

Illustration 2. Find the wavelength of a spectral line produced when an electron in the H-atoms jumps from 4^th level to 2^nd level.

Solution: $\frac{1}{\lambda }=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

$\frac{1}{\lambda }=109678\left(\frac{1}{2^2}-\frac{1}{4^2}\right)c{m}^{-1}$ Þ $\lambda =\frac{16}{3\times 109678}cm$

= 4863 $\stackrel{0}{A}$

Merits of Bohr's Theory:

$\bullet$The experimental value of radii and energies in hydrogen atom are in good agreement with that calculated on the basis of Bohr's theory.

$\bullet$Bohr's concept of stationary state of electron explains the emission and absorption spectra of hydrogen like atoms.

$\bullet$The experimental values of the spectral lines of the hydrogen spectrum are in close agreement with that calculated by Bohr's theory.

Limitations of Bohr's Theory

$\bullet$It does not explain the spectra of atoms having more than one electron.

$\bullet$Bohr's atomic model failed to account for the effect of magnetic field (Zeeman effect) or electric field (Stark effect) on the spectra of atoms or ions. It was observed that when the source of a spectrum is placed in a strong magnetic or electric field, each spectral line further splits into a number of lines. This observation could not be explained on the basis of Bohr's model.

$\bullet$De Broglie suggested that electrons like light have dual character. It has particle and wave character. Bohr treated the electron only as particle.

$\bullet$Another objection to Bohr's theory came from Heisenberg's Uncertainty Principle. According to this principle "It is impossible to determine simultaneously the exact position and momentum of a small moving particle like an electron". The postulate of Bohr, that electrons revolve in well defined orbits around the nucleus with well defined velocities is thus not tenable.

Illustration 3. The radii of two of the first four Bohr orbits of the hydrogen atom are in the ratio

1 : 4. The energy difference between them may be

(A) 0.85 eV (B) 2.55 eV

(C) 3.40 eV (D) 8.20 eV

Solution: $r_n=\frac{r_0{n}^2}{Z}$and$E_n=\frac{E_0{Z}^2}{n^2}$. Hence (B) is correct.