Dalton’s Atomic Theory: John Dalton, in 1803, put forward his atomic theory. The following are the important points of the theory :
(i) Matter is composed of particles called atoms.
(ii) All the atoms of an element are similar in mass and properties.
(iii) Atom is indivisible.
(iv) Atoms of different elements combine in a simple ratio to give “compound atom”.
 The discovery of phenomena of Radioactivity by Prof. Henri Becquerel showed that atom is divisible.
 The discovery of isotopes of an element by Soddy proved that the atoms of an element are different in mass and properties.
 The modern researches, such as discharge tube experiments have conclusively proved that atom is no longer an indivisible particle. It has definitely a complicated structure. Recent researches have shown that atom is composed of three, elementary, fundamental or subatomic particles, namely electron, proton and neutron.
Electron (_{–1}e^{0}): It is (discovered by J.J. Thomson), negatively charged particle. It is the component particle of cathode rays.
DISCOVERY OF ELECTRON: CATHODE RAYS
During the latter half of the nineteenth century, it was found that while normally dry gases do not conduct an electric current, they do so under very low pressure and then patches of light are seen. The passage of electricity through gases as studied by a number of physicists, particularly by Faraday, Davy, Crookes and J.J. Thomson.
When a current of high voltage (10,000 volts) is passed through a gas of air kept at a very low pressure (0.01 – 0.03 mm) blue rays are seen emerging from the case. These rays are called “Cathode Rays”.
Millikan’s Oil Drop Method: Determination of Charge on an Electron:
In 1909, Millikan measured the charge on an electron by his oil drop method. In this method a spray of oil droplets is produced by an atomizer, some of which pass through an opening into a viewing chamber, where we can observe them with a microscope. Often these droplets have an electric charge, which is picked up from the friction forming the oil droplets. A droplet may have one or more additional electrons in it, giving it a negative charge.
As the droplet falls to the bottom of the chamber, it passes between two electrically charged plates. The droplet can be suspended between them; we adjust the voltage in the plates so that the electrical attraction upward just balances the force of gravity downward. We then use the voltage needed to establish this balance to calculate the mass – to charge ratio for the droplet. Because we already know the mass of the droplet we can find the charge on it.
Millikan’s found that the charge on all droplets could be expressed as whole number multiples of e, where the value of e is 1.602´10^{19} C. By combining e/m. ratio and ‘e’ we calculate mass of the electron
\(\displaystyle {{M}_{e}}=\frac{e}{{e/m}}=\frac{{1.6022\times {{{10}}^{{19}}}}}{{1.76\times {{{10}}^{8}}}}\) = 9.104 x 10^{31} kg
This very small value shows that the electron is a subatomic particle.
Thus charge on an electron = 1.602 x 10^{–19}C
MORE ABOUT ELECTRON
 Cathode rays were discovered by William Crooke’s.
 Cathode rays originate from cathode in the discharge tube.
 Cathode rays are composed of stream of negatively charged particles.
 The magnitude of negative charge on electron was first determined by Millikan (oil drop experiment).
 The specific charge (e/m) on electron was first of all determined by J.J. Thomson using mass spectrometer.
 The value of e/m of cathode ray particle does not depend upon the material of the cathode or the nature of the gas taken in the discharge tube.
 The name electron was suggested by J.S. Stoney.
 Magnitude of charge on electron = 1.6 x 10^{–19} coulomb = 4.8 x 10^{–10} e.s.u.
 Amount of charge on one mole of electrons is one Faraday.
 Rest mass of electron = 9.1 x 10^{–28} gram= 9.1 x 10^{–31} kg= 9.1 x 10^{–33} Quintal.
 Mass of electron on atomic scale = 0.000549 a.m.u = 1/1837 of the mass of hydrogen atom.
 The mass of electron moving with a velocity ‘v’ is given as: m_{(in motion)} = \(\frac{{{{m}_{{rest}}}}}{{\sqrt{{1{{{\left( {\frac{\nu }{c}} \right)}}^{2}}}}}}\)
 Mass of a substance in motion increases with increase in velocity.
 Mass of electron moving with a velocity of light will become infinite.
 Mass of one mole of electrons is 0.55 mg.
 Electron is universal component of matter.
 Electron is the fundamental particle which takes part in chemical combination.
 The physical and chemical properties of an element depend upon the distribution of electrons in outer shells.
 Flow of electrons in a conductor means flow of electricity which can be detected by means of a Galvanometer.
Proton (_{1}H^{1}, H^{+},P) : It was discovered by Rutherford. It is a positively charged particle. The magnitude of charge on proton is the same as that on an electron. Proton, like electron, is a fundamental particle and is universal component of matter.
DISCOVERY OF PROTON: POSITIVE RAYS OR CANAL RAYS
Atoms are electrically neutral. Hence after the discovery of the negatively charged constituent (electron) of an atom, attempts were made to discover the positively charged counterpart of electrons. By using a discharge tube containing a perforated cathode. Goldstein (1886) found that some rays passed through these holes in a direction opposite to that of the cathode rays.
These are called the positive rays or canal rays. J.J. Thomson (1910) measured their charge by mass ratio from which he was able to deduce that these contain positive ions.
MORE ABOUT PROTON
 Charge on proton = 1.6 x 10^{–19} coulomb
= 4.8 x 10^{–10} e.s.u.
 Mass of proton = mass of hydrogen atom
= 1837 times the mass of electron
= 1.00747 a.m.u.
= 1.6726 x 10^{–24} gram
= 1.6726 x 10^{–27} kg.
 Volume of a proton [(4/3)Πr³] is roughly close to 1.5 x 10^{–38} cm^{3}.
 Mass of one mole of protons is about 1.007 gm.
 Proton is protium nucleus.
 Proton is ionized hydrogen atom (H^{+}).
 Removal of solitary electron from hydrogen atom gives proton.
 Hydrogen atom minus electron is proton. Neutron (_{0}n^{1}, N) : It is a chargeless particle i.e., it is neutral. It was discovered by James Chadwick. The reason of its late discovery is its charglessness.When beryllium or boron is bombarded with aparticles, highly penetrating radiations are obtained. These radiations are not deflected from their path by strong magnetic or electric field. These radiations are composed of chargeless particles, known as neutrons._{2}He^{4} + _{4}Be^{9} → _{6}C^{12} + _{0}n^{1}MORE ABOUT NEUTRON
 Mass of neutron = 1.00899 a.m.u.= 1.6749 x 10^{–27} kg
 Actual mass of neutron is slightly greater than that of proton.
 Density of neutron is of the order 10^{12} kg/cc.
 Out of the three types of fundamental particles, neutron is the most unstable. It decays as follows:$latex \displaystyle \underset{{Neutron}}{\mathop{{_{0}{{n}^{1}}}}}\,\to \text{ }\underset{{\text{Proton}}}{\mathop{{_{1}{{H}^{1}}}}}\,\text{ +}\underset{{Electron}}{\mathop{{_{{1}}{{e}^{0}}}}}\,\text{ + }\underset{{Antineutrino}}{\mathop{{_{0}{{v}^{0}}}}}\,\text{ }
Neutrino and antineutrino: These are the particles of gold which was bombarded with aparticles emitted from radium, it was observed by Rutherford that :(i) A large number of particles penetrated through the metal foil straightway.(ii) Some were deflected from their path through large angles.(iii) Very few –particles returned back after striking the metal foil.The following conclusions were drawn from these experiments.(a) A major portion of atom is empty space, because most of the aparticles pass straight through the atom.(b) The atom has a very small, rigid, positively charged body called the nucleus. aparticles are repelled from the metal foil due to this +ve ly charged part.(c) The whole mass of the atom is concentrated in the nucleus. So, it is very heavy and rigid. Due to its rigidity and +ve charge, the heavy particles are thrown back after striking the metal foil.
MORE ABOUT THE NUCLEUS
 Size of nucleus is measured in Fermi (1 fermi = 10^{–13} cm)
 Size of the nucleus is of the order of 10^{–12} to 10^{–13} cm.
 The nuclear size depends upon mass number (A). It is proportional to the cube root of mass number. Thus nuclear size (R) is :
R = R_{0} A^{1/3} where R_{0} is a constant (1.4 x 10^{–13} cm), and A is the mass number.
 Atomic size is of the order of 10^{–8} cm. Thus atom is about 10^{5} times bigger than the nucleus.
 Nucleus contains protons and neutrons. These nuclear particles are collectively known as Nucleons.
Mass Number (A) and Atomic Number (Z): The sum of protons and neutrons present in the nucleus is called mass number. It is always a whole number. The mass number is denoted by the symbol ‘A’ . Since, electrons have negligible mass, therefore, the entire mass of atom is due to nucleons (neutrons and protons).
Mass number (A) = Number of protons + Number of neutrons
The concept of atomic number was given by Moseley. He observed that when a beam of high speed electrons is bombarded on a metal, Xrays are emitted. The emitted Xrays have wavelength related to the number of protons present in the nucleus of metal atom. This number of protons present in the nucleus of the atom is called atomic number denoted by “Z”. The square root of the frequency of emitted Xrays is proportional to the atomic number of metal which is bombarded with the stream of electrons.
 Moseley’s relation is √v =a(Zb), where v = frequency of emitted Xrays, Z = atomic number of the metal, ‘a’ and ‘b’ are constants.
 Atomic number of an element is also equal to the number of electrons present in the outer shells around the nucleus since atom as a whole is electrically neutral.
Atomic number = Number of protons in the Nucleus
= Number of electrons in neutral atom
 Atomic number is fundamental property of an element.
 Two different elements can never have identical atomic numbers.
 The atom of an element ‘X’ having Mass number ‘A’ and atomic number ‘Z’ is represented by symbol, _{Z}X^{A}.
 Number of neutrons = A – Z.
Isotopes: They were discovered by F. Soddy. They are atoms of a given element which have the same atomic number but differ in their mass numbers. Thus isotopes have the same nuclear charge but differ in the number of neutrons in the nucleus. Isotopes have identical chemical properties but differ in physical properties. They possess different charge: mass ratio, i.e., e/m of isotopes are different and they can be separated on this basis. The instrument which separates the particles according to their charge: mass ratio is called Mass spectrometer.
Examples :
 _{8}O^{16}, _{8}O^{17}, _{8}O^{18},
 _{17}Cl^{35}, _{17}Cl^{37},
 _{6}C^{12}, _{6}C^{13}, _{6}C^{14}
Elements have nonintegral atomic masses due to the existence of isotopes. The atomic weight of an element is the average of weights of all the isotopes of that element.
For example, the fractional atomic weight of chlorine (35.5) is due to the existence of isotopes _{17}Cl^{35} and _{17}Cl^{37} in the ratio of 3: 1. Their average atomic weight is the atomic weight of chlorine.
Average weight \(\displaystyle =\frac{{\text{Weight of 3 atoms of}{{\text{ }}_{{\text{17}}}}\text{C}{{\text{l}}^{{\text{35}}}}\text{ +Weight of one atom of}{{\text{ }}_{{\text{17}}}}\text{C}{{\text{l}}^{{\text{37}}}}}}{{\text{Total four atoms}}}\)
= \(\frac{{3\times 35+1\times 37}}{4}=\frac{{105+37}}{4}=\frac{{142}}{4}\)
= 35.5 a.m.u
Kinetic Isotopic Effect: The effect on the rate of a reaction (kinetics of a reaction) due to the presence of an isotope of an element is called kinetic isotopic effect, for example:
During electrolysis of water (H_{2}O) containing heavy water (D_{2}O) the H—O bond in H_{2}O breaks up at a faster rate than D—O bond in D_{2}O. As a result we get heavy water. Thus, the manufacture of heavy water by the electrolysis of ordinary water is a beautiful example of kinetic isotopic effect.
Isobars : They are atoms with the same MASS NUMBER but different ATOMIC NUMBERS. Thus isobars have different number of electrons, protons and neutrons but the sum of neutrons and protons in their nucleus is the same.
Examples: _{1}H^{3} and _{2}He^{3}, _{18}Ar^{40}, _{19}K^{40} and _{20}Ca^{40}; _{52}Te^{130}, _{56}Ba^{130} and _{54}Xe^{130}.
As isobars are the atoms of different elements they possess different physical and chemical properties. In the periodic table, isobars are placed in separate groups.
Isotones (Isoneutronic species): They are the atoms possessing the same number of neutrons but different mass numbers.
Examples:
_{1}H^{3} and _{2}He^{4}; _{15}P^{31} and _{16}S^{32}; _{19}K^{39} and _{20}Ca^{40}.
Since isotones differ in the number of protons (atomic number) in their nuclei, so their physical and chemical properties are different.
Isosters: These are the species with same number of atoms with equal number of electrons and similar bonding. e.g.,
ATOMIC MODELS
We know the fundamental particles of the atom. Now let us see, how these particles are arranged in an atom to suggest a model of the atom.
Thomson’s Model:
J.J. Thomson, in 1904, proposed that there was an equal and opposite positive charge enveloping the electrons in a matrix. This model is called the plum – pudding model after a type of Victorian dessert in which bits of plums were surrounded by matrix of pudding.
Thomson’s Model of Atom
This model could not satisfactorily explain the results of scattering experiment carried out by Rutherford who worked with Thomson.
Rutherford’s Model
(1) Rutherford carried out experiment on the bombardment of thin (10^{–4} mm) Au foil with high speed positively charged particles emitted from Ra and gave the following observations based on this experiment,
(i) Most of the particles passed without any deflection.
(ii) Some of them were deflected away from their path.
(iii) Only a few (one in about 10,000) were returned back to their original direction of propagation.
(2) From the above observations he concluded that, an atom consists of
(i) Nucleus which is small in size but carries the entire mass i.e. contains all the neutrons and protons.
(ii) Extra nuclear part which contains electrons. This model was similar to the solar system.
(3) Properties of the nucleus
(i) Nucleus is a small, heavy, positively charged portion of the atom and located at the centre of the atom.
(ii) All the positive charge of atom (i.e. protons) are present in nucleus.
(iii) Nucleus contains neutrons and protons, and hence these particles collectively are also referred to as nucleons.
(iv) The size of nucleus is measured in Fermi (1 Fermi = 10^{–13} cm).
(v) The radius of nucleus is of the order of 1.5 x 10^{–13} cm to 6.5
x 10^{–13} i.e. 1.5 to 6.5 Fermi.
Generally the radius of the nucleus is given by the following relation,
\({{r}_{n}}={{r}_{o}}(=1.4\times {{10}^{{13}}}cm)\times {{A}^{{1/3}}}\)
This exhibited that nucleus is \({{10}^{{5}}}\) times small in size as compared to the total size of atom.
(vi) The Volume of the nucleus is about and that of atom is \({{10}^{{24}}}c{{m}^{3}},\) and that of atom is i.e., volume of the nucleus is \({{10}^{{15}}}\) times that of an atom.
(vii) The density of the nucleus is of the order of \({{10}^{{15}}}g\,c{{m}^{{3}}}\) .
(4) Drawbacks of Rutherford’s model
(i) It does not obey the Maxwell theory of electrodynamics, according to it “A small charged particle moving around an oppositely charged centre continuously loses its energy”. If an electron does so, it should also continuously lose its energy and should set up spiral motion ultimately failing into the nucleus.
(ii) It could not explain the line spectra of atom and discontinuous spectrum nature.
Planck’s quantum theory
When black body is heated, it emits thermal radiation’s of different wavelengths or frequency. To explain these radiations, max planck put forward a theory known as planck’s quantum theory.
(i) The radiant energy which is emitted or absorbed by the black body is not continuous but discontinuous in the form of small discrete packets of energy, each such packet of energy is called a ‘quantum‘. In case of light, the quantum of energy is called a ‘photon‘.
(ii) The energy of each quantum is directly proportional to the frequency (ν) of the radiation, i.e.
\(E\propto \nu \) or \(E=hv=\frac{{hc}}{\lambda }\)
Where, Planck’s constant = 6.62×10^{–27} erg. sec. or \(6.62\times {{10}^{{34}}}Joules\,\sec .\) .
Photoelectric effect
(1) When radiations with certain minimum frequency \(({{\nu }_{0}})\) strike the surface of a metal, the electrons are ejected from the surface of the metal. This phenomenon is called photoelectric effect and the electrons emitted are called photoelectrons. The current constituted by photoelectrons is known as photoelectric current.
(2) The electrons are ejected only if the radiation striking the surface of the metal has at least a minimum frequency called Threshold frequency. The minimum potential at which the plate photoelectric current becomes zero is called stopping potential.
(3) The velocity or kinetic energy of the electron ejected depend upon the frequency of the incident radiation and is independent of its intensity.
(4) The number of photoelectrons ejected is proportional to the intensity of incident radiation.
(5) Einstein’s photoelectric effect equation
According to Einstein,
Maximum kinetic energy of the ejected electron = absorbed energy – threshold energy
\(\frac{1}{2}mv_{{\max }}^{2}=h\nu h{{\nu }_{0}}=hc\left[ {\frac{1}{\lambda }\frac{1}{{{{\lambda }_{0}}}}} \right]\)
Where \(({{\nu }_{0}})\) and \({{\lambda }_{0}}\) are threshold frequency and threshold wavelength.
Bohr’s atomic model
Bohr retained the essential features of the Rutherford model of the atom. However, in order to account for the stability of the atom he introduced the concept of the stationary orbits. The Bohr postulates are,
(1) An atom consists of positively charged nucleus responsible for almost the entire mass of the atom (This assumption is retention of Rutherford model).
(2) The electrons revolve around the nucleus in certain permitted circular orbits of definite radii.
(3) The permitted orbits are those for which the angular momentum of an electron is an integral multiple of \(h/2\pi \) where \(h\) is the Planck’s constant. If is the mass and is the velocity of the electron in a permitted orbit of radius \(r,\) then
\(L=mvr=\frac{{nh}}{{2\Pi }}\) ; \(n=1\) ,2, 3…
Where \(L\)
is the orbital angular momentum and is the number of orbit. The integer is called the principal quantum number. This equation is known as the Bohr quantization postulate.
(4) When electrons move in permitted discrete orbits they do not radiate or lose energy. Such orbits are called stationary or nonradiating orbits. In this manner, Bohr overcame Rutherford’s difficulty to account for the stability of the atom. Greater the distance of energy level from the nucleus, the more is the energy associated with it. The different energy levels were numbered as 1,2,3,4 .. and called as K, L, M, N…etc.
(5) Ordinarily an electron continues to move in a particular stationary state or orbit. Such a state of atom is called ground state. When energy is given to the electron it jumps to any higher energy level and is said to be in the excited state. When the electron jumps from higher to lower energy state, the energy is radiated.
Advantages of Bohr’s theory
(i) Bohr’s theory satisfactorily explains the spectra of species having one electron, viz. hydrogen atom, \(H{{e}^{+}},L{{i}^{{2+}}}\) etc.
(ii) Calculation of radius of Bohr’s orbit : According to Bohr, radius of n^{th} orbit in which electron moves is
\({{r}_{n}}=\left[ {\frac{{{{h}^{2}}}}{{4{{\pi }^{2}}m{{e}^{2}}k}}} \right].\frac{{{{n}^{2}}}}{Z}\)
Where, \(n=\) Orbit number, m=Mass number, e = Charge on the electron, Z = Atomic number of element, k = Coulombic constant
After putting the values of m,e,k,h, we get.
r_{n} =( n^{2}/Z) x 0.529 Å
(iii) Calculation of velocity of electron
\({{V}_{n}}=\frac{{2\pi {{e}^{2}}ZK}}{{nh}},\,{{V}_{n}}={{\left[ {\frac{{Z{{e}^{2}}}}{{mr}}} \right]}^{{1/2}}}\)
\({{V}_{n}}=\frac{{2.188\times {{{10}}^{8}}Z}}{n}cm.{{\sec }^{{1}}}\)
(iv) Calculation of energy of electron in Bohr’s orbit
Total energy of electron = K.E. + P.E. of electron
\(=\frac{{kZ{{e}^{2}}}}{{2r}}\frac{{kZ{{e}^{2}}}}{r}=\frac{{kZ{{e}^{2}}}}{{2r}}\)
Substituting of r, gives us \(E=\frac{{2{{\pi }^{2}}\,m{{Z}^{2}}{{e}^{4}}{{k}^{2}}}}{{{{n}^{2}}{{h}^{2}}}}\)
Where, n=1, 2, 3……….
Putting the value of m, e, k, h, we get
\(E=21.8\times {{10}^{{12}}}\times \frac{{{{Z}^{2}}}}{{{{n}^{2}}}}erg\,per\,atom\)
\(=21.8\times {{10}^{{19}}}\times \frac{{{{Z}^{2}}}}{{{{n}^{2}}}}J\,per\,atom\,(1J=\text{1}{{\text{0}}^{\text{7}}}erg)\)
\(E=13.6\times \frac{{{{Z}^{2}}}}{{{{n}^{2}}}}eV\,per\,atom\text{(1eV}=\text{1}\text{.6}\times \text{1}{{\text{0}}^{{19}}}J)\)
\(=13.6\times \frac{{{{Z}^{2}}}}{{{{n}^{2}}}}k.cal/mole\) (1 cal = 4.18J)
or \(\frac{{1312}}{{{{n}^{2}}}}{{Z}^{2}}kJmo{{l}^{{1}}}\)
When an electron jumps from an outer orbit (higher energy) \({{n}_{2}}\) to an inner orbit (lower energy) \({{n}_{1}}\) then the energy emitted in form of radiation is given by
\(\Delta E={{E}_{{{{n}_{2}}}}}{{E}_{{{{n}_{1}}}}}=\frac{{2{{\pi }^{2}}{{k}^{2}}m{{e}^{4}}{{Z}^{2}}}}{{{{h}^{2}}}}\left( {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right)\,\)
\(\Rightarrow \ \Delta E=13.6{{Z}^{2}}\left( {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right)eV/atom\)
As we know that \(E=h\bar{\nu },\) \(c=\nu \lambda \) \(\bar{\nu }=\frac{1}{\lambda }\) \(=\frac{{\Delta E}}{{hc}},\)
\(=\frac{{2{{\pi }^{2}}{{k}^{2}}m{{e}^{4}}{{Z}^{2}}}}{{c{{h}^{3}}}}\left( {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right)\)
This can be represented as \(\frac{1}{\lambda }=\bar{\nu }=R{{Z}^{2}}\left( {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right)\)
Where \(R=\frac{{2{{\pi }^{2}}{{k}^{2}}m{{e}^{4}}}}{{c{{h}^{3}}}}\) R is known as Rydberg constant. Its value to be used is \(109678c{{m}^{{1}}}.\)
The negative sign in the above equations shows that the electron and nucleus form a bound system, i.e., the electron is attracted towards the nucleus. Thus, if electron is to be taken away from the nucleus, energy has to be supplied. The energy of the electron in n =1 orbit is called the ground state energy; that in the n=2 orbit is called the first excited state energy, etc. When n=∞, E=0 which corresponds to ionized atom i.e., the electron and nucleus are infinitely separated.
(6) Spectral evidence for quantisation (Explanation for hydrogen spectrum on the basis of bohr atomic model)
(i) The light absorbed or emitted as a result of an electron changing orbits produces characteristic absorption or emission spectra which can be recorded on the photographic plates as a series of lines, the optical spectrum of hydrogen consists of several series of lines called Lyman, Balmar, Paschen, Brackett, Pfund and Humphrey. These spectral series were named by the name of scientist who discovered them.
(ii) To evaluate wavelength of various Hlines Ritz introduced the following expression,
\(\bar{\nu }=\frac{1}{\lambda }=\frac{\nu }{c}=R\left[ {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right]\)
This remarkable agreement between the theoretical and experimental value was great achievment of the Bohr model.
(iii) Although Hatom consists of only one electron yet it’s spectra consist of many spectral lines.
(iv) Comparative study of important spectral series of Hydrogen is shown in following table.
(v) If an electron from n^{th }excited state comes to various energy states, the maximum spectral lines obtained will be \(\frac{{n(n1)}}{2}.\)
(vi) Thus, at least for the hydrogen atom, the Bohr theory accurately describes the origin of atomic spectral lines.
(7) Failure of Bohr model
(i) Bohr theory was very successful in predicting and accounting the energies of line spectra of hydrogen i.e. one electron system. It could not explain the line spectra of atoms containing more than one electron.
(ii) This theory could not explain the presence of multiple spectral lines.
(iii) This theory could not explain the splitting of spectral lines in magnetic field (Zeeman effect) and in electric field (Stark effect). The intensity of these spectral lines was also not explained by the Bohr atomic model.
(iv) This theory was unable to explain of dual nature of matter as explained on the basis of De broglies concept.
(v) This theory could not explain uncertainty principle.
(vi) No conclusion was given for the concept of quantisation of energy.
Examples
Question The quantum theory assumes that energy changes are not continuous. Why do not we notice this effect in our every day activities?
Solution In every day activities, we deal with macroscopic particles such as our bodies, or cars which gains and loss total amount of energy much larger than a quantum. The gain and loss of the relatively miniscule quantum of energy is unnoticed.
Question Calculate the energy per mole of photon of electromagnetic radiations of wavelength 4000Å.
Solution E / Photon = hc/λ
⇒ \(E/{{N}_{{Photon}}}=Nh\,c/\lambda \)
⇒\(\frac{{6.023\times {{{10}}^{{23}}}\times 6.626\times {{{10}}^{{27}}}\times 3.0\times {{{10}}^{{10}}}}}{{4000\times {{{10}}^{{8}}}}}\)
= 2.99 × 10^{12} erg
Question Suppose 10^{–17} J of light energy is needed by the interior of the human eye to see an object. How many photons of green light (λ = 550 nm) are needed to generate this minimum amount of energy?
Solution The energy needed to see object = 10^{–17} J
Photon energy used to see object = hc/λ
\(\frac{{6.626\times {{{10}}^{{34}}}\times 3.0\times {{{10}}^{8}}}}{{550\times {{{10}}^{{9}}}}}\)
= 3.61 × 10^{–19} J/photon
Number of photons=\(\frac{{{{{10}}^{{17}}}}}{{3.61\times {{{10}}^{{19}}}}}=27.7\)
Minimum number of photons required to see object = 28
Question For any given series in the spectrum of Hatom, there is wavelength limit beyond which the spectrum becomes continuous (not discrete). Explain.
Solution The difference in energy is maximum between the first and second orbits. As the value of principal quantum number ‘n’ increases, the difference between two successive orbits decreases. After a certain value of n, the energy levels becomes so closely spaced that they seem to be continuous.
Question Calculate the ratio of radius of 4^{th} Bohr’s to 2^{nd} Bohr’s orbit for He^{+} ion.
Solution \({{r}_{{nH{{e}^{+}}}}}={{r}_{{{{n}_{H}}}}}/Z\)
\({{r}_{{4H{{e}^{+}}}}}=\frac{{{{r}_{{4H}}}}}{2}\) and \({{r}_{{2H{{e}^{+}}}}}=\frac{{{{r}_{{2H}}}}}{2}\)
\({{\left( {\frac{{{{r}_{4}}}}{{{{r}_{2}}}}} \right)}_{{H{{e}^{+}}}}}={{\left( {\frac{{{{r}_{4}}}}{{{{r}_{2}}}}} \right)}_{H}}\)
\(\left( {\frac{{{{r}_{{1H}}}\times {{4}^{2}}}}{{{{2}^{2}}\times {{r}_{{1H}}}}}} \right)=\frac{4}{1}\) \(({{r}_{{{{n}_{H}}}}}={{r}_{{1H}}}\times {{n}^{2}})\)
Question The ionization energy of Hatom is 13.6 eV. What will be ionization energy of He^{+} and Li^{2+} ions?
Solution IE_{1} for He^{+} = 1E_{1} for H × Z^{2}
= 13.6 × 4 = 54.4 eV
Also, IE_{1} for Li^{2+} = 1E_{1} for H × Z^{2}
= 13.6 × 9 = 122.4 eV
Question The ionization energy of He^{+} is 19.6 × 10^{–18} J per atom. Calculate the energy of the second stationary state in Li^{2+} ion.
Solution The energy of the first orbit in one electron species is related to that in hydrogen atom by an expression
E = Z^{2} E_{H}
For He^{+} ion, z = 2. Hence, \({{E}_{H}}=\frac{E}{{{{Z}^{2}}}}=\frac{{19.6\times {{{10}}^{{18}}}}}{4}J\)
Now for Li^{2+} ion, where Z = 3, we will have
\(E=(9){{E}_{H}}=9\left( {\frac{{19.6\times {{{10}}^{{18}}}}}{4}J} \right)=44.1\times {{10}^{{18}}}J\)
Now for the second orbit, the energy is
\(E\acute{\ }=\frac{E}{{{{n}^{2}}}}=\frac{{44.1\times {{{10}}^{{18}}}J}}{{{{2}^{2}}}}=11.025\times {{10}^{{18}}}J\)
THE HYDROGEN SPECTRUM: When energy is supplied to a sample of hydrogen gas, the atoms present in the sample, absorb energy and are excited from the ground state to different higher energy states, depending upon the amount of energy absorbed. When these excited electrons return to the normal state, energy is emitted in the form of radiations of different wave lengths which appear in the form of different lines in the emission spectrum. The wave length (or frequency) of the lines depends upon the amount of energy emitted. Greater the amount of energy liberated, shorter will be the wave length of radiation emitted.
The hydrogen spectrum contains a large number of closely spaced lines, hence, it is a line spectrum. Although, hydrogen atom contains only one electron in its atom, yet it gives a large number of lines in its spectrum. Its explanation is that a sample of hydrogen gas contains a very large number of atoms.
The electrons in these atoms are excited to different energy states on absorption of energy, which give very large number of lines in the spectrum on deexcitation. Its lines in the spectrum are grouped into five series each of which is named after the name of its discoverer. The various series of lines of the hydrogen spectrum are shown in the following figure.
Series of Lines of Hydrogen Spectrum:
(a) Lyman Series: Lines of this series are obtained when electrons fall from 2, 3, 4 …. etc. orbit to first orbit (ground state)
For Lyman series,
n_{1} = 1, n_{2} = 2, 3, 4, 5, 6, 7,…….
The wave length of the lines in Lyman series can be calculated by using equation (13).
\(\frac{1}{\lambda }=R\left[ {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right]\)
For the first line of Lyman series,
n_{2} = 2, n_{1} = 1
\(\frac{1}{\lambda }=109678\left[ {\frac{1}{{{{1}^{2}}}}\frac{1}{{{{2}^{2}}}}} \right]\)
\(=109678\times \frac{3}{4}c{{m}^{{1}}}\)
λ =1215.7Å
The wave length of the second line of Lyman series is less than that of the first line. Thus, the wave length of the other lines decreases continuously. The wave length of the last (limiting) line of the Lyman series is obtained when \({{n}_{2}}=\infty \)
\(\frac{1}{\lambda }=109678\left[ {\frac{1}{{{{1}^{2}}}}\frac{1}{{{{\infty }^{2}}}}} \right]c{{m}^{{1}}}\)
λ = 911.7 Å
The lines of the Lyman series appear in the ultraviolet region of the spectrum.
The intensities of spectral lines decreases with increase in the value of n. e.g., the intensity of first Lyman line (2→1) is greater than second line (3→1)
(b) Balmer Series: Lines of this series are obtained when electrons fall from outer orbits to second orbit. Thus, for the various lines of this series, n_{1} = 2, n_{2} = 3, 4, 5, 6, 7, 8……
The wave length of the first line in the Balmer series can be calculated by putting
n_{1} = 2 and n_{2} = 3
\(\frac{1}{\lambda }=109678\left[ {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right]\)
\(\frac{1}{\lambda }=109678\left[ {\frac{1}{{{{2}^{2}}}}\frac{1}{{{{3}^{3}}}}} \right]\)
=\(109678\times \frac{5}{{36}}c{{m}^{{1}}}\)
\(\lambda =\frac{{36}}{{109678\times 5}}=\frac{{36}}{{548390}}\)
λ =6564.6 Å
The wave length of the last (limiting) line of Balmer series is:
\(\frac{1}{\lambda }=109678\left[ {\frac{1}{{{{2}^{2}}}}\frac{1}{{{{\infty }^{2}}}}} \right]c{{m}^{{1}}}\)
= \(\frac{{109678}}{4}c{{m}^{{1}}}\)
\(\lambda =\frac{4}{{109678}}=3.647\times {{10}^{{5}}}cm\)
λ =3647 Å
The lines of Balmer series fall in the visible region of the spectrum.
In the Balmer series of hydrogen spectrum, the first line (n_{2} = 3 →n_{1} = 2) is \({{L}_{\beta }}\) line. The line from \({{n}_{2}}=\infty \,\,to\,\,{{n}_{1}}=2\) is called limiting line.
(c) Paschen Series: This series is produced when the electrons fall from outer orbits to third orbit i.e. for the lines of this series,
n_{1} = 3, n_{2} = 4, 5, 6, 7……
For the first line of this series,
n_{1} = 3, and n_{2} = 4
then, \(\frac{1}{\lambda }=R\left[ {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right]\)
\(\frac{1}{\lambda }=109678\left[ {\frac{1}{{{{3}^{2}}}}\frac{1}{{{{4}^{4}}}}} \right]c{{m}^{{1}}}\)
\(\frac{1}{\lambda }=109678\times \left[ {\frac{1}{9}\frac{1}{{16}}} \right]c{{m}^{{1}}}\)
\(\lambda =\frac{{144}}{{7\times 109678}}=\frac{{144}}{{767746}}cm\)
λ =18756.2 Å
Lines of this series lie in the infrared region of the spectrum.
(d) Brackett series: When the electronic transition takes place from outer orbits to the fourth orbit, lines of this series are obtained, i.e., for this series, n_{1} = 4, n_{2} = 5, 6, 7, 8, 9………
The wave length of the first line of the series can be calculated by putting n_{1} = 4 and n_{2} = 5 in the equation
\(\frac{1}{\lambda }=R\left[ {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right]\)
\(\frac{1}{\lambda }=109678\left[ {\frac{1}{{{{4}^{2}}}}\frac{1}{{{{5}^{2}}}}} \right]c{{m}^{{1}}}\)
λ =40523 Å
This series also falls in the infrared region of the spectrum.
Pfund Series: The lines in this series of hydrogen spectrum are obtained when electrons are deexcited from higher orbits to fifth orbit, i.e., for this series, n_{1} = 5, n_{2} = 6, 7, 8, 9, 10……….
The wave length of the first line of the series can be calculated by putting, n_{1} = 5 and n_{2} = 6.
\(\frac{1}{\lambda }=R\left[ {\frac{1}{{n_{1}^{2}}}\frac{1}{{n_{2}^{2}}}} \right]\)
λ = 74600 Å
This series also falls in the infrared region of the spectrum.
In the Rydberg formula when, the line produced is called the limiting line of that series.
The number of spectral lines produced when an electron jumps from nth level to ground level.
\(\displaystyle \sum{{(n1)=\frac{{n(n1)}}{2}}}\)
In the line spectrum of H–like systems, certain frequencies are common. These frequencies correspond to a transition between two energy levels which are divisible by their respective atomic numbers.
WAVEPARTICLE DUALITY, DUAL NATURE OF ELECTRON:
It was suggested by Loius de Broglie in 1924 that a particle in motion also behaves like a wave. The wave length associated with the moving particle is given by the following equation,
λ =h/mv ……….(i)
where, h = Planck’s constant, m = mass of the moving particle, v = velocity of the particle.
This above equation (i) is known as de Broglie equation. De Broglie equation was derived on the basis of Einstein’s equation, E = mc^{2} and Planck’s equation E = hv. From both of these relation,
E = hv = mc^{2}
or \(\lambda =\frac{h}{{mv}},\) or \(\lambda =\frac{h}{p}\)
where p is the momentum of the moving particle.
The wave length associated with a particle in motion is inversely proportional to its momentum.
As the mass of the moving particle increases, the momentum also increases; the wave length of the matter wave (associated with the matter in motion) decreases.
For the particles of finite size, i.e., having appreciable mass, the momentum is very high, is very small and it can be said that macroscopic bodies in motion do not possess matter waves.
When different particles move with the same velocity, the wave length of the matter wave is inversely proportional to the mass of the particle.
HEISENBERG’S UNCERTAINTY PRINCIPLE
This principle was proposed by Werner Heisenberg in 1927. According to it, “It is impossible to determine simultaneously the exact position and momentum of a moving particle like electron, proton, neutron, etc.”
If there is certainty of position of the particle, the momentum becomes uncertain and vice versa. Mathematically, the uncertainty principle is represented as:
\(\Delta x\times \Delta p\ge \frac{h}{{4\pi }}\)
where \(\Delta x\) = uncertainty in position of the particle, \(\Delta p\) = uncertainty in the momentum of the particle
Now \(\Delta p=m\,\Delta v\) , so equation (ii) takes the form
\(\Delta x\times \Delta v\ge \frac{h}{{4\pi m}}\)
Equation (iii) can be used to calculate the uncertainty in the position or velocity of the particle.
The uncertainty principle is not in agreement with the Bohr theory as the latter gives a proper position to the electron with respect to the nucleus while according to the former, the position of electron is not certain. Hence, the idea of definite orbits for an electron is meaningless as suggested by Bohr. The waveparticle dual nature of electron and the uncertainty principle gave rise to a new concept called as “probability concept” for an electron.
Thus, we can predict the probability of locating an electron of particular energy in a given region of space around the nucleus at a given time. It gives rise to the concept of atomic orbital.
Question The λ of H_{a} line of Balmer series is 6500Å. What is the λ of \({{H}_{\beta }}\) line of Balmer series?
Solution For H_{a} line of Balmer series n_{1} = 2, n_{2}=3
For line of Balmer series n_{1} = 2, n_{2} = 4
\(\frac{1}{{{{\lambda }_{\alpha }}}}={{R}_{H}}\left[ {\frac{1}{{{{2}^{2}}}}\frac{1}{{{{3}^{2}}}}} \right]\) ……(i)
\(\frac{1}{{{{\lambda }_{\beta }}}}={{R}_{H}}\left[ {\frac{1}{{{{2}^{2}}}}\frac{1}{{{{4}^{2}}}}} \right]\) ……(ii)
By Eqs. (i) and (ii)
\({{\lambda }_{\beta }}={{\lambda }_{\alpha }}\times (80/108)=6500\times (80/108)\)
= 4814.8 Å
Question Show that circumference of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving about that orbit.
Solutions According to be Broglie:
\(\lambda =\frac{h}{{mu}}\) ….(i)
Also, \(mur=\frac{{nh}}{{2\pi }}\) ….(ii)
By Eqs. (i) and (ii) \(\lambda =\frac{{h.2\pi r}}{{n.h}}\)
or \(\mathbf{2\pi r = n\lambda }\)
Question Why does a moving cricket ball not show wave like nature?
Solution λ = h/mu and thus λ decreases if m and u increases. Thus, high mass object like a cricket ball in flight would have very small wavelength and thus, do not show any appreciable wave nature.
Question Calculate the speed and de Broglie wavelength of an electron that has been accelerated by a potential difference of 500 V.
Solution Since, \(\frac{1}{2}m{{u}^{2}}=eV\)
\(u={{\left( {\frac{{2eV}}{m}} \right)}^{{1/2}}}={{\left[ {\frac{{2\times 1.602\times {{{10}}^{{19}}}\times 500}}{{9.108\times {{{10}}^{{31}}}}}} \right]}^{{1/2}}}\)
= 1.326×10^{7} m sec^{–1}
Also, \(\lambda =\frac{h}{{mu}}\)
= \(\frac{{6.626\times {{{10}}^{{34}}}}}{{9.108\times {{{10}}^{{31}}}\times 1.326\times {{{10}}^{7}}}}\)
= 5.5 × 10^{–11} m
Question The de Broglie wave length of an alpha particle is 6.28 × 10^{–15}m. Calculate the velocity of the alpha particle (m_{p} = 1.672 × 10^{–27}kg, m_{n} = 1.675 × 10^{–27} kg).
Solution Mass of alpha particle = mass of two protons + mass of two neutrons
= 2 × 1.672 × 10^{–27} + 2 × 1.675 ×10^{–27} kg
= 3.344 × 10^{–27} + 3.35 × 10^{–27}
= 6.694 × 10^{–27} kg
de Broglie wave length is:
\(\lambda =\frac{h}{{mv}}\)
or \(v=\frac{h}{{\lambda m}}\)
= \(\frac{{6.626\times {{{10}}^{{34}}}}}{{6.28\times {{{10}}^{{15}}}\times 6.694\times {{{10}}^{{27}}}}}\) meter sec^{–1}
= 1.576 × 10^{7} m sec^{–1}.
Question What is the de Broglie wave length associated with a proton moving with 25% of the velocity of light?
Solution Mass of proton, m = 1.6 × 10^{–24} gram
Velocity of proton = 3 × 10^{10} (cm/sec) x 0.25 = 0.75 × 10^{10} cm sec^{–1}
The de Broglie wave length associated is given by
\(\lambda =\frac{h}{{mv}}=\frac{{6.626}}{{1.6\times {{{10}}^{{24}}}gm}}\times \frac{{{{{10}}^{{27}}}erg\,\sec }}{{0.75\times {{{10}}^{{10}}}cm\,{{{\sec }}^{{1}}}}}\)
= 5.52 × 10^{–13} cm.
Question What conclusion may be drawn from the following results of (a) and (b)?
(a) If a 1×10^{–3} – kg body is traveling along the xaxis at 1 m/s within 0.01 m/s. Calculate the theoretical uncertainty in its position.
(b) If an electron is traveling at 100 m/s within 1 m/s, calculate the theoretical uncertainty in its position.
[h = 6.63 × 10^{–34} J. s, mass of electron = 9.109 × 10^{–31} kg]
Solution (a) The velocity has an uncertainty of 0.02 m/s (from 0.99 to 1.01 m/s)
⇒ \(\Delta x=\frac{h}{{4\pi m\Delta v}}=\frac{{6.63\times {{{10}}^{{34}}}(J.s)}}{{(4\times 3.14)(1\times {{{10}}^{{3}}}kg)(0.02m/s)}}\)
≅ 3×10^{–30} meter.
(b) \(\Delta x=\frac{{6.63\times {{{10}}^{{34}}}(J.s)}}{{(4\times 3.14)(9.109\times {{{10}}^{{31}}}kg)(2m/s)}}\)
Question Find out the number of waves made by a Bohr electron in one complete revolution in its 3^{rd} Bohr orbit of Hatom.
Solution \({{r}_{n}}=\frac{{{{n}^{2}}{{h}^{2}}}}{{4{{\pi }^{2}}m{{e}^{2}}}};\,\,\,{{u}_{n}}=\frac{{2\pi {{e}^{2}}}}{{nh}}\)
∴No. of waves in one round =
= \(\frac{{2\pi {{r}_{3}}\times {{u}_{3}}\times m}}{h}\)
= \(\frac{{2\pi \times {{n}^{2}}{{h}^{2}}}}{{4{{\pi }^{2}}m{{e}^{2}}}}\times \frac{{2\pi {{e}^{2}}}}{{nh}}\times \frac{m}{h}\)
⇒ n = 3
QUANTUM MECHANICAL MODEL OF ATOM
In 1920, a new model of atom was developed by Ervin Schrodinger. In the atomic model proposed by Schrodinger, idea of quantization and conclusions of de Broglie principle and Heisenberg uncertainty principle were incorporated. In this model the behaviour of the electron in an atom is described by the mathematical equation known as Schrodinger Wave Equation, given below:
\(\frac{{{{\partial }^{2}}\psi }}{{\partial {{x}^{2}}}}+\frac{{{{\partial }^{2}}\psi }}{{\partial {{y}^{2}}}}+\frac{{{{\partial }^{2}}\psi }}{{\partial {{z}^{2}}}}+\frac{{8{{\pi }^{2}}m(EU)\psi }}{{{{h}^{2}}}}=0\)
Here, in this equation, x,y and z are the three space coordinates, m = mass of electron, h = Planck’s constant, E = total energy, U = Potential energy, = wave function of electron wave. The permitted solutions of Schrodinger wave equation are known as wave functions which correspond to a definite energy state of an electron known as orbital. Thus, the discrete Bohr orbits are replaced by orbitals, i.e., “three dimensional region of definite shape about the nucleus where the electron density is maximum or where the probability of finding an electron is maximum or where the electron passes its maximum time.”
The Schrodinger wave equation may simply be interpreted by stating that a particle or body of mass m, energy E and velocity v possesses wave like properties associated with it, with amplitude given by the wave function (Psi).
 Ψ(Psi) gives the three dimensional amplitude of electron wave.
 Ψ ^{2} dV is the probability of finding an electron in a volume dV about the nucleus of an atom.
 The particular wave of Ψ is called eigen function and the value of energy corresponding to this is called eigen value.
 The eigen function of an electron is called atomic orbital.
 The wave equation is applicable to atoms as well as molecules.
 The solution of wave equation gives regions in space where Ψ is +ve as well as —ve . But Ψ^{2} (probability of finding an electron) is always positive.
Probability Distribution : In wave mechanics, an electron in motion is described by a wave function,ѱ. Ѱ has no physical significance and refers to the amplitude of the electron wave. However, ѱ^{2} is a significant term and give the probability of finding an electron or intensity of electron. An atomic orbital is a three dimensional region of definite shape about the nucleus where there is more intensity of electrons. An atomic orbital is considered as a diffused electron cloud having more electron density close to the nucleus. The probability of finding an electron in a given volume about the nucleus is understood best in the form of radial probability distribution curves. The probability distribution curves for some orbitals are given below. The distance of maximum radial probability is the radius of an atom. The point at which radial probability becomes zero is known as Nodal point. In general there are (n –1) nodal points for sorbitals; (n –2) for porbitals; (n –3) for dorbitals and (n –4) for forbitals (n = principal quantum number).
 The radius of maximum probability of 1s electron is 0.53 Å (Bohr radius).
 The number of regions of maximum probability for 1s, 2p, 3d, and 4f orbitals are one each.
 For 2s, 3p, 4d and 5f –atomic orbitals there are two regions of maximum probability.
 The small humps in the distribution curves show that the electron has a tendency to penetrate closer to the nucleus.
 In between the regions of maximum electron density, there is a region of zero electron density known as nodal point. Greater the number of nodal points, higher is the energy of an orbital.
QUANTUM NUMBERS : In order to define the ‘state’ of an electron in an atom, a set of four numbers is required known as Quantum numbers. The term ‘state’ includes, the energy, position with respect to the nucleus, orientation in space and the interaction of the electron with other electrons.
(i) Principal Quantum Number (n) : This quantum number was introduced by Bohr. It gives the average distance of the electron from the nucleus. It also indicates the average volume of the electron cloud. It determines the main energy shell in which the electron is revolving round the nucleus. ‘n’ will have positive integral values only .
The main energy level (shell) corresponding to different values of n are:
Principal Quantum Number (n)  Main Energy Level 
n =1  K shell 
n =2  L shell 
n =3  M shell 
n =4  N shell 
Energy of electron in ‘n’th shell of hydrogen atom and like ions is \({{E}_{n}}=\frac{{13.6{{Z}^{2}}}}{{{{n}^{2}}}}\)
eV/ atom where ‘Z’ is the atomic number.
As the distance of the electron from the nucleus increases, energy of electron also increases.
Energy of electron increases with increasing values of “n”.
Energy of electron at infinite distance from the nucleus is zero.
Total number of electrons in nth shell is 2n^{2}.
The angular momentum of an electron in an orbit depends upon its principal quantum number and is given by \(mvr=\frac{{nh}}{{2\pi }}\) where ‘n’ is principal quantum number.
(ii) Azimuthal, Angular, Secondary, Subsidiary or Serial Quantum Number (l) : It was given by Sommerfeld. It explains the fine spectrum of hydrogen atom. It gives the angular momentum of electron in elliptical orbit while in motion round the nucleus. It also gives the shape of the subshell in which the electron is located. It (l) may have any +ve integral value ranging from 0 to ( n –1).
The total values of l are equal to the Principle quantum number of ‘n’.
Principal Quantum number (n) and azimuthal Quantum number (l) can never have identical numerical values.
The orbital angular momentum of an electron depends upon the azimuthal quantum number (l) and is given by: \(\sqrt{{l(l+1)}}.\frac{h}{{2\pi }}\) = \(\sqrt{{l(l+1)}}.h\) \(\left( {where\,\,h\,\,cross\,+h=\frac{h}{{2\pi }}} \right)\)
The total number of subshells (l) in a shell (n) is equal to shell number.
The various sub shells corresponding to different values of ‘l’ are as follows:
Azimuthal Q. Number (l)  Subshell  Shape  Max. number of electrons 
l = 2  s–  Symmetrically spherical shape  2 
l = 1  p–  Dumbbell  6 
l = 2  d–  Double dumbbell  10 
l = 3  f–  Complicated shape  14 
l = 4  g–  Highly complicated shape  18 
The various shells are comprised of the following subshells:
(a)  n =1  K –shell  Designation 
l = 0  ssubshell  1s  
(b)  n = 2,  L shell  
l = 0
l = 1 
ssubshell
psubshell 

(c)  n = 3
l = 0 = 1 =2 
Mshell
ssubshell psubshell dsubshell 

(d)  n= 4,
l = 0 =1 =2 =3 
Nshell
ssubshell psubshell dsubshell fsubshell 
µ Increasing order of energy of subshells is :
s < p < d < f.
µ Decreasing order of screening effect :
s > p > d > f
(iii) Magnetic or Orientation Quantum Number (m): It explains Zeeman effect. It gives the atomic orbital in which the electron is present. It specifies the orientations of atomic orbitals in a magnetic field. The values of ‘m’ vary from – l, through 0 to +l. Thus, the total values of m are (2l + 1).
A subshell is made up of atomic orbitals which are described as follows:
Subshell (l)  Values of m (magnetic quantum number)  Atomic orbitals  Designations 
l = 0  m = 0  1  s 
l = 1  m =–1,0,+1  3  p_{x},p_{z},p_{y} 
l = 2  m = – 2,–1, 0, +1, +2  5  d_{xy},d_{yz},d_{zx},d_{x}^{2} –y^{2},d_{z}^{2} 
l = 3  m =0, ±1, ±2, ±3  7  Complicated 
The atomic orbitals p_{x}, p_{y} and p_{z} are dumbbell shaped and posses equal energies but differ in their orientations in space. They are called degenerate orbitals.
The plane where the electron density is almost zero is called nodal plane.
Number of nodal plane for np orbital = 1.
Pictorial representation of nodal plane:
For p_{y} atomic orbital xz plane is the nodal plane :
Orbital Designation of nodal plane
2p_{x} or 3p_{x} etc. yz
2p_{y} or 3p_{y} etc. xz
2p_{z} or 3p_{z} etc. xy
The atomic orbitals dxy, dxz, dyz, \({{d}_{{{{x}^{2}}{{y}^{2}}}}}\) and dz^{2} are also degenerate (possess equal energies).
The probability of finding the electron in the xy plane in the atomic orbital, \({{d}_{{{{x}^{2}}{{y}^{2}}}}}\) is not zero.
The atomic orbital having dough nut or a belly band or baby soother like shape is dz^{2}. It is dumbbell shaped with a collar of high electron density in the xy plane.
Nodal planes for d_{xy} d_{yz} and d_{xz} orbitals are 2 each.
Orbital Nodal plane
d_{xy} xz and yz planes
d_{xz} xy and yz planes
d_{yz} xy and xz planes
forbitals are seven in number designated as,
\({{f}_{{{{x}^{3}}}}},{{f}_{{{{y}^{3}}}}},{{f}_{{{{z}^{3}}}}},{{f}_{{{{x}^{2}}{{y}^{2}}}}},{{f}_{{{{x}^{2}}{{z}^{2}}}}},{{f}_{{{{z}^{2}}{{x}^{2}}}}},{{f}_{{xyz}}}\)
(iv) Spin quantum Number (s) : While in motion around the nucleus, the electron spins about its own axis. The spin may be clockwise or anticlockwise. The spinning electron would add to the angular momentum of the electron and therefore changes the energy associated with the electron. Assuming the spin to be quantized, there are two possible values of s, i.e. s = +1/2 or 1/2 depending upon whether the electron spins clockwise or anticlockwise. As a convention, the clockwise and anticlockwise spins are represented by an arrow ↑ and ↓ respectively. Two electrons having the same direction of spin are said to have parallel spins while the two having different direction of spins are said to have antiparallel spins.
For each value of ‘m’ there are two values of spin quantum number, s = +1/2 or 1/2 i.e in any atomic orbital only two electrons can be accommodated having antiparallel (↑↓) spin.
The solution of Schrodinger wave equation gives the principal (n), azimuthal (l) and magnetic quantum numbers (m) but not the spin quantum number (s). It was introduced on account of the spin of revolving electron.
Significance of Quantum Numbers. The four quantum numbers are of physical significance. They give the address of an electron i.e they are capable of indicating the probable position (shell, subshell, atomic orbital) and energy of an electron in the atom. For example, if for an electron.
n = 3, l = 1, m = –1, and , then it indicates that the electron is:
— present in the third shell (Mshell)
— present in the 3p subshell (since for p, l = 1)
— present in the 3p_{x} or 3p_{y} atomic orbital
— spinning in clockwise direction.
Pauli’s Exclusion Principle: This principle states, “No two electrons in an atom can have an identical set of all the four quantum numbers”. If three quantum numbers are the same, the fourth will definitely be different. This principle shows that an atomic orbital cannot have more than two electrons and if there are two electrons in any atomic orbital, they will have anti parallel spins. This principle is very helpful in determining the maximum number of electrons in a shell or a subshell. For example:
For First Energy Level (KShell):
n = 1, l = 0, (1 ssubshell), m = 0 (1 s –atomic orbital) s(Two electrons having opposite spins) s = ½ or 1/2. These electrons are designated as 1s^{2}
For Second Energy Level (LShell):
Principal Azimuthal Q. Magnetic Q. spin Q. Designation
 number(n) number (I) number (m) number(s)
n = 2 l = 0 (2s) m = 0 (2s) ½ or 1/2 2s^{2}
1 (2p) m =–1(2p_{x}) ½ or 1/2
m = 0 (2p_{z}) ½ or 1/2 2p^{6}
m = 1 (2p_{y} ½ or 1/2
Total electrons in second shell are eight.
µ Similarly, it can be shown that dsubshell (l = 2) can accommodate 10 and fsubshell can have a maximum of fourteen (14) electrons.
Shielding or Screening Effect: According to the screening rule, “the electrons in the completely filled inner shells screen the outer electrons against the attraction by the nucleus”, i.e. the outer electrons are not attracted by the nucleus so effectively as they would have been attracted had the inner shell electrons not been present. This is known as Shielding or Screening Effect. Due to this effect, the ns orbitals are filled with electrons earlier than the (n – 1) dorbitals. In a similar way the 5s, 5p and 6sorbitals are occupied by electrons before the 4forbitals.
µ In a given shell, the decreasing order of screening effect is : s > p > d > f .
ELECTRONIC CONFIGURATION OF ELEMENTS:
The distribution of electrons in various shells and subshells is called electronic configuration of elements. This arrangement of electrons in the atom decides the properties of an element. The following rules are used for writing the electronic configuration:
 Aufbau’s Principle: Aufbau is not the name of any Scientist. It is a German word which means ‘building up’ or ‘construction’. According to this principle, “subshells are filled with electrons in the increasing order of their energies”, i.e. Subshell of lower energy will be filled first with electrons.
µ Subshell having lower value of (n + l) will be of lower energy, where n is the principal and l, the azimuthal quantum number for the subshell.
µ When the values of (n + l) for two or more subshells available for electrons, are the same, then that having lower value of ‘n’ will be of lower energy and hence will be occupied by the electrons first.
 Hund’s Rule of Maximum Multiplicity: According to this rule, “pairing of electrons in a subshell starts after all the available atomic orbitals of that subshell are singly filled (halffilled) with electrons having parallel spins” or pairing of electrons in a subshell is impossible in the presence of vacant atomic orbitals in that subshell”.
µ In psub shell, the fourth electron starts pairing, and the sixth electron starts pairing in dsubshell.
µ In fsubshell, pairing starts with eight electron.
µ This rule gives the number of unpaired electrons in an atom, ion or molecule.
 Exactly halffilled subshells have lesser energy and thus assume more stability than any other arrangement. Thus, p^{3} is more stable arrangement than p^{2}, p^{4} or p^{5}.
 When the electronic configuration ns^{2}np^{6} is attained in the outermost shell of an atom, the next incoming electron enters the (n + 1) ssubshell. The nd and nf –sub shells will be vacant.
Electronic configuration of the Elements
Element  Symbol  At. No.  Electronic Configuration 
Hydrogen  H  1  1s^{2} 
Helium  He  2  1s^{2} 
Lithium  Li  3  1s^{2}, 2s^{1} 
Beryllium  Be  4  1s^{2}, 2s^{2} 
Boron  B  5  1s^{2}, 2s^{2} 2p^{1} 
Carbon  C  6  1s^{2}, 2s^{2} 2p^{2} 
Nitrogen  N  7  1s^{2}, 2s^{2} 2p^{3} 
Oxygen  O  8  1s^{2}, 2s^{2} 2p^{4} 
Fluorine  F  9  1s^{2}, 2s^{2} 2p^{5} 
Neon  Ne  10  1s^{2}, 2s^{2} 2p^{6} 
Sodium  Na  11  1s^{2}, 2s^{2} p^{6}, 3s^{1} 
Magnesium  Mg  12  1s^{2}, 2s^{2} p^{6}, 3s^{2} 
Aluminium  Al  13  1s^{2}, 2s^{2} p^{6}, 3s^{2 }3p^{1} 
Silicon  Si  14  1s^{2}, 2s^{2} p^{6}, 3s^{2 }3p^{2} 
Phosphorus  P  15  1s^{2}, 2s^{2} p^{6}, 3s^{2 }3p^{3} 
Sulphur  S  16  1s^{2}, 2s^{2} p^{6}, 3s^{2 }3p^{4} 
Chlorine  Cl  17  1s^{2}, 2s^{2} p^{6}, 3s^{2 }3p^{5} 
Argon  Ar  18  1s^{2}, 2s^{2} p^{6}, 3s^{2 }3p^{6} 
Potassium  K  19  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6}, 4s^{1} 
Calcium  Ca  20  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6}, 4s^{2} 
Scandium  Sc  21  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{1}, 4s^{2} 
Titanium  Ti  22  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{2}, 4s^{2} 
Vanadium  V  23  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{3}, 4s^{2} 
Chromium  Cr  24  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{5}, 4s^{1} 
Manganese  Mn  25  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{5}, 4s^{2} 
Iron  Fe  26  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{6}, 4s^{2} 
Cobalt  Co  27  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{7}, 4s^{2} 
Nickel  Ni  28  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{8}, 4s^{2} 
Copper  Cu  29  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{10}, 4s^{1} 
Zinc  Zn  30  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{10}, 4s^{2} 
Gallium  Ga  31  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{10}, 4s^{2 }p^{1} 
Germanium  Ge  32  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{10}, 4s^{2 }p^{2} 
Arsenic  As  33  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{10}, 4s^{2 }p^{3} 
Selenium  Se  34  1s^{2}, 2s^{2} p^{6}, 3s^{2 }p^{6} d^{10}, 4s^{2 }p^{4} 
Question (a) A compound of vanadium has a magnetic moment of 1.73 BM. Find the electronic configuration of vanadium ion in the compound
(b) Name the orbitals corresponding to given set of quantum numbers
(a) \(n=3,\ \ l=2,\ m=\pm \ 2\) (b) \(n=4,\ \ l=0,\ m=0\)
(c) \(n=2,\ \ l=1,\ m\pm \ 1\) (d) \(n=2,\ \ l=1,\ m=\ 2\)
Solution (a) Magnetic moment = \(\sqrt{{n\,(n+2)}}\ \) where n = number of unpaired electrons
√[n(n+2)]=1.73⇒ n² + 2n = 1.73², n = 1
Therefore vanadium atom must have one unpaired electron and thus its electronic configuration \(_{{23}}{{V}^{{4+}}}\ :\ 1{{s}^{2}}2{{s}^{2}}\ 2{{p}^{6}}\ 3{{s}^{2}}\ 3{{p}^{6}}\ 3{{d}^{1}}\)
(b) (a) 3dx^{2} – y^{2} or 3dxy (b) 4s (c) 2px or 2py (d) no such orbital
Question Why electron cannot exist inside the nucleus according to Heisenberg’s uncertainty principle?
Solution Diameter of the atomic nucleus is of the order of \({{10}^{{15}}}\ m\) ,The maximum uncertainty in the position of electron is \({{10}^{{15}}}\ m\) Mass of electron \(=9.1\times {{10}^{{31}}}\ kg\)
\(\Delta x.\ \Delta p=\frac{h}{{4\pi }}\)
\(\Delta v=5.80\times {{10}^{{10}}}m{{s}^{{1}}}\)
This value is much higher than the velocity of light and hence not possible.
Question How many 7s electron are there in an atom with Z = 104?
Solution The electronic configuration of the said element is:
\(\displaystyle 1{{s}^{2}},\ 2{{s}^{2}},\ 2{{p}^{6}},\ 3{{s}^{2}},\ 3{{p}^{6}},\ 3{{d}^{{10}}},\ 4{{s}^{2}},\ 4{{p}^{6}},\)
\(\displaystyle 4{{d}^{{10}}},\ 4{{f}^{{14}}},\ 5{{s}^{2}},\ 5{{p}^{6}},\ 5{{d}^{{10}}},\ 5{{f}^{{14}}},\ 6{{s}^{2}},\ 6{{p}^{6}},\ 6{{d}^{2}},\ 7{{s}^{2}}\)
Hence, there are two 7s electrons.
MISCELLANEOUS EXERCISES
Exercise 1: Calculate the ratio of specific charge (e/m) of a proton and that of an aparticle.
Exercise 2: What is the fraction of volume occupied by the nucleus with respect to the total volume of an atom?
Exercise 3: Calculate the minimum and maximum values of wavelength in Balmer series of a H atom.
Exercise 4: The wave number of the first line of Balmer series of hydrogen is 15,200 cm^{–}^{1}. What is the wave number of the first line of Balmer series for the Li^{2+} ion?
Exercise 5: What possible can be the ratio of the de Broglie wavelengths for two electrons having the same initial energy and accelerated through 50 volts and 200 volts?
Exercise 6: What is likely to be the principle quantum number for a circular orbit of diameter 20 nm of the hydrogen atom if we assume Bohr orbit to be the same as that represented by the principle quantum number?
Exercise 7: Calculate the uncertainty in the velocity of a particle weighing 25.0 g if the uncertainty in position is 10^{–}^{5} m. (given Planck constant h = 6.6 x 10^{–}^{34} Js).
Exercise 8: Calculate the de Broglie wavelength of a tennis ball of mass 60.0 g moving with a velocity of 10 metres per second (Planck constant, h = 6.63 x 10^{–}^{34} Js).
Exercise 9: Suppose 10^{–17}J of energy is needed by the interior of human eye to see an object. How many photons of green light (l=550 nm) are needed to generate this minimum amount of energy?
Exercise 10: Ionization potential of hydrogen atom 13.6 eV. If hydrogen atom in ground state excited by monochromatic light of energy 12.1 eV, then what will be the total spectral lines emitted according to Bohr’s theory?
ANSWERS TO MISCELLANEOUS EXERCISES
Exercise 1: 1 : 2
Exercise 2: 10^{–}^{15}
Exercise 3: 3647 nm and 6564 nm
Exercise 4: 1,36,800 cm^{–}^{1}
Exercise 5: 2 : 1
Exercise 6: 14
Exercise 7: 2.1 x 10^{–}^{28}
Exercise 8: 10^{–}^{33} m
Exercise 9: 28
Exercise 10: 3