BOHR MODEL OF THE HYDROGEN ATOM -

LIMITATION OF RUTHERFORD’S ATOMIC MODEL

according to Clark Maxwell, a charged particle moving under the influence of attractive force continuously loses energy in the form of E. M radiation. So electron emit radiation and loses of energy. Due to this its motion would slow down and radius become smaller and smaller and ultimately the electron would fall into the nucleus. But since atom is quite stable.

further as the revolving electron is losing energy continuously, the atom should emit E.M radiation over a continuous range. But experimentally, it was found that radiation is emit in discrete form of particular frequency.

Clearly the classical ideas are not sufficient to explain the atomic structure.

BOHR’S THEORY OF H-ATOM

Bohr, in 1913 concluded that in spite of the success of electromagnetic theory in explaining large scale phenomena, it could not be applied to the processes at the atomic scale.

Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. These are:

An electron in an atom could revolve in certain stable orbits without the emission of radiant energy. These are called the stationary states of the atom.
The electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\large \frac{h}{2\pi}$ where $\large h$ is the Planck’s constant. i.e $\fn_cm \large \left [ L=n\frac{h}{2\pi} \right ]$ or $\fn_cm \large \left [ mvr=n\frac{h}{2\pi} \right ]$
The radiation of energy occurs only when an electron jumps from one permitted orbit to another. When electron jumps from inner to higher orbit, energy absorbed and emits when electron jumps from outer to the inner orbit

$\fn_cm \large \left [ E_f-E_i=h\nu \right ]$

RADIUS OF BOHR’S STATIONARY ORBIT

We know that

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

$\fn_cm \large \therefore v=\frac{nh}{2\pi mr}..............(i)$

we know that electrostatic force between electron and nucleus serves as a centripetal force

$\fn_cm \large i.e\,\,\,\frac{mv^2}{r}=\frac{kze^2}{r^2}$

$\fn_cm \large \therefore r=\frac{kze^2}{mv^2}................(ii)$

putting the value of $\fn_cm \large v$ from equation (i) to equation (ii), we get

$\fn_cm \large r=\frac{kze^2}{m\left ( \frac{nh}{2\pi mr} \right )^2}$

$\fn_cm \large r=\frac{kze^2}{m}.\frac{4\pi^2m^2r^2}{n^2h^2}$

$\fn_cm \large r=\frac{n^2h^2}{kze^24\pi^2m}$

$\fn_cm \large r=\left ( \frac{h^2}{4\pi^2mke^2} \right ).\frac{n^2}{z}$

$\fn_cm \large \left [ r=(5.29\times10^{-11})\frac{n^2}{z} \right ]$

for H-atom

$\fn_cm \large n=1,\,\,r=0.53A^0$

$\fn_cm \large n=2,\,\,r=2.12A^0$

$\fn_cm \large n=3,\,\,r=4.177A^0$

VELOCITY OF ELECTRON

Now

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

$\fn_cm \large \therefore r=\frac{nh}{2\pi mv}..........(i)$

and also $\fn_cm \large \frac{mv^2}{r}=\frac{kze^2}{r^2}$

$\fn_cm \large \therefore r=\frac{kze^2}{mv^2}................(ii)$

from equation (i) and equation (ii)

$\fn_cm \large \frac{nh}{2\pi mv}=\frac{kze^2}{mv^2}$

$\fn_cm \large \therefore v=\frac{2\pi kze^2}{nh}$

$\fn_cm \large \left [ v=\left ( \frac{2\pi ke^2}{h} \right ).\frac{z}{n} \right ]$

TOTAL ENERGY OF ELECTRON

Kinetic energy of electron is given by

$\fn_cm \large K.E=\frac{1}{2}mv^2$

$\fn_cm \large \because \frac{mv^2}{r}=\frac{kze^2}{r^2}$

$\fn_cm \large \therefore K.E=\frac{1}{2}.\frac{kze^2}{r}$

Potential energy of electron is given by

$\fn_cm \large P.E=\frac{k(ze)(-e)}{r}$

$\fn_cm \large P.E=-\frac{kze^2}{r}$

$\fn_cm \large \therefore$ Total energy of electron in the orbit is

$\fn_cm \large E=K.E+P.E$

$\fn_cm \large =\frac{1}{2}\frac{kze^2}{r}-\frac{kze^2}{r}$

$\fn_cm \large \left [ E=-\frac{kze^2}{2r} \right ]$

$\fn_cm \large E=-\frac{kze^2}{2\times(5.29\times10^{-11})\frac{n^2}{z}}$

$\fn_cm \large E=-\frac{kz^2e^2}{(2\times 5.29\times10^{-11})n^2}$

$\fn_cm \large \left [ E=(-21.76\times 10^{-19}).\frac{z^2}{n^2}\,Joule \right ]$

-ve sign indicate electron is bound to the nucleus and not free to leave.

NOTE:

The above equations involves the assumption that the electronic orbits are circular, though orbits under inverse square force are , in general elliptical. However, it was shown by the German physicist Arnold Sommerfeld that, when the restriction of circular orbit is relaxed, these equations continue to hold even for circular orbits.

EXCITATION ENERGY/EXCITATION POTENTIAL

At room temperature most of H-atoms are in ground state. The minimum energy required to excite an atom in the ground state to one of the higher stationary states is called excitation energy.

Let the electron jumps from (n=1 i.e ground state) to (n=2)

$\fn_cm \large excitation \;energy\;\;=E_2-E_1$

$\fn_cm \large =\;-3.4eV-(-13.6eV)$

$\fn_cm \large =\;10.2eV$

The minimum accelerating potential which provides an electron energy sufficient to jump from the inner most orbit (ground state) to one of the outer orbits is called excitation potential.

i.e electron jumps from n=1 to n=2

Then $\fn_cm \large excitation\;potential=\frac{12.2eV}{e}\;\;\;(V=\frac{U}{q})$

$\fn_cm \large =12.2volt$

IONISATION ENERGY/ IONISATION POTENTIAL

If the energy supplied is so large that it can remove an electron from the outermost orbit of an atom, the process is called ionisation, and the energy required is called ionisation energy.

the minimum accelerating potential which would provide an electron energy sufficiently just to remove it from the atom is called ionisation potential.

for example, Total energy of electron in ground state of H-atom is $\fn_cm \large -13.6eV$

$\fn_cm \large \therefore ionisation\;energy=0-(-13.6eV)=13.6eV$

$\fn_cm \large and\;ionisation\;potential=13.6Volt$

NOTE:-

An electron can have any total energy above E= 0 eV. In such situation the electron is free. Thus there is a continuum of energy states above E=0 eV as shown in above figure.

FRANCK- HERTZ EXPERIMENT

This is the first experimental verification of the existence of discrete energy level in atom performed on 1914 by James Franck and Gustav Hertz. they studied the spectrum of mercury vapour when electrons having different (increasing) kinetic energy passed through the vapour. The electrons collide with the mercury atoms and can transfer energy to the mercury atoms. This can only happen when the energy of the electron is higher than the energy difference between an energy level of Hg occupied by an electron and a higher unoccupied level.

If an electron of having an energy 4.9 eV (for Hg) or more passes through mercury, an electron in mercury atom can absorb energy from the bombarding electron and get excited to the higher level. the colliding electron’s K.E would reduce by this amount. The excited electron would subsequently fall back to the ground state by emission of radiation (λ=253 nm)

For this experiment verification of Bohr’s basic ideas of discrete energy levels in atoms and the process of photon emission. Frank and Hertz were awarded the Nobel prize in 1925.

IMPORTANT LINKS OF WAVE OPTICS
Introduction	Alpha Particle Scattering and Rutherford’s Nuclear Model of Atom
Atomic Spectra	Bohr Model of the Hydrogen Atom
The Line Spectra of the Hydrogen Atom	De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation