# Chapter Notes: Atomic Structure Physics Class 12

Notes for Atomic Structure chapter of class 12 physics. Dronstudy provides free comprehensive chapterwise class 12 physics notes with proper images & diagram.

Structure Of An Atom

All matter is made up of tiny particles known as atoms. There are only about 105 different kinds of atoms, and they combine with each other in different ways to form groups called molecules. All matter has been found to be composed of atoms or molecules, and some knowledge of how atoms are made will give us valuable information about the behaviour of matter.

Thomson's Atomic Model

On the basis of his experiments J.J. Thomson proposed a model of internal atomic structure according to which atom consisted of positively charged substance (+ve electric field) distributed uniformly over the entire body of the atom, with negative electrons embedded in this continuous positive charge like seeds in a watermelon. It was a good effort to reveal mystery of an atom but it was not the true picture of an atom

Rutherford's Atomic Model

The correct description of the distribution of positive and negative charges within an atom was made in 1911 by a New Zealander when working at Manchester University in England. This was Ernest Rutherford, who was later called as Lord Rutherford for his many scientific achievements. He entered into physics during that crucial period of its development when the phenomenon of natural radioactivity had just been discovered, and he was first to realize that radioactivity represents a spontaneous disintegration of heavy unstable atoms.
Rutherford realized that important information about the inner structure of atoms could be obtained by the study of collisions between on rushing a particles and the atoms of various materials forming the target.

The basic idea of the experimental arrangement used by Rutherford in his studies was explained as follows:
A speck of $\alpha$-emitting radioactive material; a lead shield with a hole that allowed a narrow beam of theÂ $\alpha$Â - particles to pass through; a thin metal foil to deflect or scatter them; and a pivoted fluorescent screen with a magnifier through which the tiny flashes of light were observed
whenever anÂ $\alpha$Â -particle struck the screen. The whole apparatus was evacuated, so that the particles would not collide with air molecules.

Observations of Rutherford

Most of theÂ $\alpha$-particles penetrated the foil with very little deflection. An appreciable fraction of them were deflected through large angles - a few were turned back almost as though they had been reflected from the foil. This was a deflection of nearly 180Â° and a completely impossible phenomenon according to the Thomson's model
Such large deflections required strong forces to be acting, such as those between very smaller charged particles very close together. This would be possible, Rutherford reasoned, if all the positive charge, along with most of the atomic mass, were concentrated in a very small central region which Rutherford called the atomic nucleus.

Because there would be a coulomb force of attraction between the positive nucleus and the negative electrons, the two would be down together and the atom would vanish unless some provisions were made to prevent it. It was suggested that the electrons might be orbiting rapidly around the nucleus, so that the electrostatic attraction would merely provide the necessary centripetal force.

Drawbacks of the Rutherford model

(i)Â Â  Rutherford's atomic model was unable to make any predictions about the light that an atom would emit
(ii)Â  More serious than this was its conflict with the accepted laws of electromagnetic theory. An electron revolving rapidly around a nucleus must have a continual centripetal acceleration, and this acceleration would cause a continuous loss of energy by radiation. Bohr calculated that this emission of radiation would cause the electrons in an atom to lose all their energy and fall into the nucleus within a hundred - millionth of a second. Since matter composed of atoms exists permanently, as far as we know, there was obviously something wrong here.

Bohrâ€™s Theory of The Hydrogen Like Atoms

Bohrâ€™s theory applies to hydrogen atom and species like He+, Li++ etc. Here a single electron revolves around a stationary nucleus of positive charge Ze where Z = 1 for hydrogen atom, Z = 2 for He+ etc.

Bohr in defiance of the well - established laws of classical mechanics and electrodynamics, proposed the following postulates:
1. Of all the infinite number of mechanically possible orbits for an electron revolving around a nucleus, only a few are permitted. These are the orbits in which the angular momentum of the electron is an integral multiple of ${h \over {2\pi }}$, where h is the planckâ€™s constant
2. While circling around these permitted orbits, the electrons do not emit any electromagnetic radiation, even though conventional electrodynamics holds that they should
3.Electrons may jump from one orbit to another, in such case the difference in energy between the two states of motion is radiated as a photon whose frequency is determined by the quantum rule

$\Delta E = hv$

Bohr's Orbits

For an electron orbiting in a hydrogen like atom, the necessary centripetal force is the electrostatic attraction between the negative electron and the massive, positively-charged nucleus
Thus,Â Â  $k{{Z{e^2}} \over {{r^2}}} = {{m{v^2}} \over r}$Â orÂ $r = {{kZ{e^2}} \over {m{v^2}}}$Â (i)
According to Bohr's theory
$mvr = {{nh} \over {2\pi }}$ Â  Â  Â Â  where, n = 1, 2, 3
$r = {{nh} \over {2\pi mv}}$ Â  Â  Â  Â  Â  Â (ii)
From (i) and (ii)
$v = {{2\pi kZ{e^2}} \over {nh}}$ Â Â and Â  Â  Â $r = {{{n^2}{h^2}} \over {4{\pi ^2}kmZ{e^2}}}$

Energy of the Atom

The electron revolving around the nucleus has kinetic energy

$k = {1 \over 2}m{v^2} = {m \over 2}{\left( {{{2\pi kZ{e^2}} \over {nh}}} \right)^2} = {{2{\pi ^2}{k^2}m{Z^2}{e^4}} \over {{n^2}{h^2}}}$

The electrostatic potential energy is

$U = {{k{e^2}} \over r} =- k{e^2}\left( {{{4{\pi ^2}kmZ{e^2}} \over {{n^2}{h^2}}}} \right) = {{ - 4{\pi ^2}{k^2}m{Z^2}{e^4}} \over {{n^2}{h^2}}}$

Notice that the magnitude ofÂ  potential energy is double the magnitude ofÂ  kinetic energy.
Total energy of the atom is
E = K + U=${1 \over {{n^2}}}\left( {{{2{\pi ^2}{k^2}m{Z^2}{e^4}} \over {{h^2}}}} \right)$
Putting the values of$k = {\rm{ }}9 \times {10^9}$Â Nm2/C2
$e = 1.6 \times {10^{ - 19}}$ $C$ and $h = 6.63 \times {10^{ - 34}}$ $Js$,Â  we get
$E = {{{Z^2}} \over {{n^2}}}(2.18{10^{ - 18}}){\rm{ }}J =-13.6{{{Z^2}} \over {{n^2}}}eV$

$(2.18{10^{ - 18}}){\rm{ }}J{\rm{ }} =- 13.6{{{Z^2}} \over {{n^2}}}eV$
An energy level diagram for the hydrogen atom (Z = 1) is shown in the figure. The vertical axis represents energy. The (arbitrary) zero of energy is taken as the energy of a stationary electron, infinitely far from the positive nucleus. The lowest energy level

(n = 1) is known as the ground state. The energy level corresponding to n = 2 is called theÂ  first excited state and so on .Â  In this diagram zero energy level corresponds to
n =$\infty$ which is the ionized state of the atom

From the third rule of Bohrâ€™s, a photon is emitted when an electron makes a transition from a higher sate toÂ  a lower state. The difference between the two energy states is the energy of the emitted photon. Thus
$\Delta E = {E_2} - {E_1} = hv$
Using the energy equation
DE = 2.18$\times$10-18 Z2
${\rm{ v = }}{{\Delta E} \over h}{\rm{ =3}}{\rm{.29}} \times {\rm{1015 Z}}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$Hz
Dividing the above equation by c = 3$\times$${10^8}$Â m/s, we get
${1 \over \lambda } ={{3.29 \times {{10}^{15}}} \over {3 \times {{10}^8}}}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]\,\,$m-1
orÂ ${1 \over \lambda } = {R_\infty }\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$
whereÂ ${R_\infty }$= Rydberg constantÂ $= 1.097 \times {10^7}$${m^{ - 1}}$

The various transitions for the hydrogen atomÂ  are shown in the following figure. All transitions starting from n = 2 onwards and ending at n = 1 belong to the Lyman Series. Likewise, all transitions starting from n = 3 onwards and ending at n = 2 belong to the Balmer Series. The other spectral seriesâ€™ names are mentioned in the figure

Successes and Limitations of the Bohr Model

Bohr showed that Planck's quantum idea was a necessary part of the atom and its inner mechanism; he introduced the idea of quantized energy levels and explained the emission or absorption of radiation as being due to the transition of an electron from one level to another. As a model for even multielectron atoms, the Bohr picture is still useful. It leads to a good, simple, rational ordering of the electrons in larger atoms and qualitatively helps to predict a great deal about chemical behaviour and spectral details.

Bohr's theory is unable to explain the following facts:
1.The spectral lines of hydrogen atom is not a single line but a collection of several lines very close together.
2.The structure of multielectron atoms is not explained.
3.No explanation for using the principle of quantisation of angular momentum.
4.No explanation for Zeeman effect
If a substance which gives a line emission spectrum is placed in a magnetic field, the lines of the spectrum get split up into a number of closely spaced lines

This phenomenon is known as Zeeman effect.
The above discussion may be summarized as follows:
The atom consists of a heavy positively charged nucleus and negatively charged electrons moving around it. The electron is an elementary particle having a mass ${m_e} \approx 9.1 \times {10^{ - 31}}$kg and a charge -e, e being an elementary charge approximately equal toÂ $1.60 \times {10^{ - 19}}C.$

The nuclear charge is equal to +Ze, where Z is the atomic number. The atom contains Z electrons, their total charge being -Ze. Consequently, the atom is an electrically neutral system. The size of the nucleus varies depending on Z from 10-13 cm to 10-12 cm. The size of the atom is a quantity of the order ofÂ  10-8 cm.

The energy of the atom is quantized. This means that it can assume only discrete (i.e. separated by finite gaps) values: E1, E2, E3,â€¦, which are called the energy levels of the atom
(E1 < E2 < E3 < â€¦). Atoms with different Z's have different sets of energy levels

In a normal (unexcited) state, the atom is on the lowest possible energy level. In such a state, the atom may stay for an infinitely long time. By imparting an energy to the atom, it is possible to transfer it to an excited state with an energy higher than the energy of the ground state. A transition of the atom to a higher energy level may occur as a result of absorption of a photon or as a result of a collision with another atom or a particle, say, an electron

Excited states of the atom are unstable. The atom can stay in an excited state for about 10-8 s. After that the atom spontaneously (by its own) goes over to a lower energy level, emitting in this process a photon with an energy
${E_{ik}} = h{v_{ik}}$ (i > k),

where i is the number of the energy level in the initial state and k is the number of the level to which a spontaneous transition of the atom occurred. For example, an atom which is in an excited state with the energy E3 can return to the ground state either directly, by emitting a photon of frequency
n31 = (E3 - E1)/h, or through an intermediate state with the energy E2, as a result of which two photons with frequencies n32 = (E3 - E2)/h and n21 = (E2 - E1)/h are emitted.

Important Formulae

Â Â Â  ${r_n} = 0.53{{{n^2}} \over Z}{A^ \circ }$Â  Â where Z = atomic numberÂ

2. Velocity of the electron in the nth orbit

Â  Â  Â ${v_n} = {Z \over n}\left( {{c \over {137}}} \right)$ Â where c = 3 x 108 m/sÂ

3.Â  Energy of the electron in the nth orbit

${E_n} =- 13.6{{{Z^2}} \over {{n^2}}}$Â (eV)
${E_n} =- (2.18{10^{ - 18}}){{{Z^2}} \over {{n^2}}}$Â (J)
$E = K + U$
$K =- E =- {U \over 2}$
$U = 2E = -2K$

4. Wavelength of photon emitted for a transition from n2 to n1

${1 \over \lambda } = {R_\infty }{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$
where {R_\infty }Â = 1.096 x 107 m-1 (for a stationary nucleus)
If nucleus is not considered to be stationary
$R = {{{R_\infty }} \over {1 + {m \over M}}}$
where m is the mass of electron and M is the mass of nucleus

Â 5. Wavelength (Ã…) of a photon of energy E (eV) is given by

$\lambda= {{12400} \over {E\,(eV)}}{A^0}$

Â 6.Momentum of a photon of energy E

$\rho= {E \over c}$

Example 1

A single electron orbits around a stationary nucleus of charge +Ze, where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from second Bohr orbit to the third Bohr orbit.
(a) Find the value of Z
(b)Â  Find the energy required to excite the electron from n = 3 to n = 4
(c)Â  Find the wavelength of radiation required to remove electron from first Bohrâ€™s Orbit to infinity.
(d)Â  Find the kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.

Solution

(a) Given DE23 = 47.2 eV
Â  Â  Â  Since $\Delta E = 13.6{Z^2}{{12400} \over {\Delta E}} = {{12400} \over {340}} =36.47{A^0}$eV
Â  Â  Â Â $47.2{\rm{ }} = {\rm{ }}13.6{Z^2}$ Â $\Delta E = 13.6{Z^2}{{12400} \over{\Delta E}} = {{12400} \over {340}} = 36.47{A^0}$Â $\Rightarrow Z = {\rm{ }}5$

(b) To find DE34;Â Â  n1 = 3;Â Â Â Â  n2 = 4
Â  Â  Â $\Delta E = 13.6{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$
Â  Â  Â $\Delta E = {13.65^2}\left( {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right)$=16.53eV

(c) Â Ionization energy is the energy required to excite the electron from n = 1 to n =$\infty$
Â  Â  Â  Thus,$\Delta E = 13.6 \times {5^2}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$= 340 eVÂ
Â  Â  Â  The respective wavelength is
Â  Â  Â  $\lambda =\infty= {{12400} \over {\Delta E}} = {{12400} \over {340}}$=36.47${A^0}$z

(d)Â K = -E = +340 eV
Â  Â  Â U = 2E = -680 eV
Â  $L = {h \over {2\pi }} = {{6.63 \times {{10}^{ - 34}}} \over {2\pi }} = 1.056 \times {10^{ - 34}}J - s$=${{6.63 \times {{10}^{ - 34}}} \over {2\pi }}$=${{6.63 \times {{10}^{ - 34}}} \over {2\pi }}$=$1.056 \times {10^{ - 34}}$ J-s

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