Atomic Structure Chemistry

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SUBJECT – CHEMISTRY

TOPIC – ATOMIC STRUCTURE COURSE CODE – AISM-09/C/ATS

Contents:- Atomic Structure

Electronic Magnetic Radiation...24

Atomic Spectrum...28

Planck‟s Quantum Theory...29

Bohr‟s Atomic Model...32

Quantum Numbers...50

Pauli‟s Exclusion Principle...54

Shapes Ans Sizes of Orbitals...54

Hund‟s rule of maximum multiplicity...59

Electronic configuration of elements...60

Dual Character...68

Derivation of De-Broglie Equation...69

Heisenberg‟s uncertainty principle………..76

Quantum mechanical model of model……….80

Photoelectric Effect………83

Answer to excercises………..88

FUNDAMENTAL PARTICLES

Atoms are made up-essentially, of three fundamental particles, which differ in mass and electric charge as follows:

Electron Proton Neutron

Symbol e or e- p n Approximate relative mass 1/1836 1 1 Approximate relative charge 1 +1 0 Mass in kg 9.109534 10 31 _{1.6726485 10} 27 _{1.6749543 10} 27 Mass in amu 5.4858026 10 4 _{1.007276471 1.008665012} Actual charge/C 1.6021892 10 19 _{1.6021892 10} 19 ₀

The atomic mass unit (amu) is 1/12th of the mass of an individual atom of 6C12, i.e. 1.660565 10 27 kg. The neutron and proton have approximately equal masses of 1 amu and the electron is about 1836 times lighter; its mass can sometimes be neglected as an approximation. The electron and proton have equal, but opposite, electric charges; the neutron is not charged.

The existence of electrons in atoms was first suggested, by J.J. Thomson, as a result of experimental work on the conduction of electricity through gases at low pressures, which produces cathode rays and x-rays, and a study of radioactivity by Becquerel, the Curies and Rutherford.

An atom is electrically neutral, and if it contains negatively charged electrons it must also contain some positively charged particles, and the supposition that they existed within atoms came about as a result of Rutherford's experiments in which he bombarded elements with the - rays and

-rays were given off by radioactive elements. The neutron was discovered in 1932 by James Chadwick by bombarding beryllium with 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”.

Cathode

To Vacuum Pump + (Anode)

Glass Tube Fluorescence

CATHODE RAY EXPERIMENT (DISCHARGE TUBE)

Some of the important properties of the cathode rays studied by Sir J.J. Thomson and others are given below:

 Cathode rays come out at right angles to the surface of the cathode and move in straight lines.

 Their path is independent on the position of the anode.

 They produce phosphorescence on certain salts like ZnS and fluorescence on glass.

 They blacken photographic plates.

 The rays pass through thin sheet of metals. If the metal sheet is too thick to be penetrated the rays cast a shadow.

 The rays ionize a gas through which they pass.

 They heat a substance on which they fall.

 They rotate a light wheel placed in their paths. This shows that cathode rays contain material particles having both mass and velocity.

 The mass of a particle present in cathode rays is found to be 1/1837 of H atom. This shows that the particle is of sub atomic nature.

 Cathode rays are deflected by a magnetic or an electric field showing the particle to be electrically charged, the direction of deflection shows that they are negatively charged.

 Cathode rays contain the smallest unit of negative charge.

 No cathode ray was produced when the tube was completely evacuated.

 Different gases produce same cathode rays as they have the same e/m (charge/mass) ratio. This indicates that the particles present in cathode rays are fundamental constituent of all matter.

Sir J.J. Thomson named these negatively charged sub atomic particles as electron.

“A sub atomic particle which is a fundamental constituent of all matter having a mass 1/1837th _{of a H atom and which carries the smallest unit of} negative charge is called an electron”.

Determination of Velocity and Charge/mass (e/m) ratio of Electrons: Sir J.J. Thomson (1897) extended the cathode ray experiment for the determination of velocity of electrons and their charge/mass ratio, The value of e/m for an electron = 1.76 108 C/g. For the H+ ion (proton), e/m = 96500/1.008 C/g.

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

19 e 8 e 1.6022 10 M e / m 1.76 10 = 9.104 10-31 kg

This very small value shows that the electron is a subatomic particle. Thus charge on an electron = 1.602 10–19_C.

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

containing a perforated cathode. Goldstein (1886) found that some rays passed through these holes in a direction opposite to that of the cathode rays. Anode To Vacuum Pump Cathode Positive Rays Cathode Rays POSITIVE 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. Their properties are:

 They are positively charged.

 The positive charge is either equal to or whole number multiple of the charge on an electron.

 When hydrogen gas was filled in the discharge tube the positive charge on the positive rays was equal to the negative charge on an electron, and the mass was less than the hydrogen atom.

 Unlike cathode rays the properties of positive rays are characteristics of the gas in the tube.

 The deflection of positive rays under the influence of an electric or magnetic field is smaller than that of the cathode rays for the same strength of field. This shows that the positive rays have a greater mass than that of electrons.

 The mass of the positive rays depends on the atomic weights or molecular weights of the gases in the discharge tube. The charge/mass ratio also varies because the change in positive charge on the rays. It may be either equal to or integral multiple of the charge on an electron.

 The lightest of all particles identified in positive rays from different elements was one with a mass very slightly less than that of hydrogen atom (or nearly equal to H atom). The lightest positively charged particle is called a proton (P or P+_{). Positive rays are atomic or} molecular resides from which some electrons have been removed. The removed electrons constitute the cathode rays and the positive residues form the positive or canal rays.

Positive Rays Cathode Rays

H H+ _e–

O2 O2 e

O2 O22+ 2e–

The mass of a proton is very slightly less than that of a H atom. This shows that protons are sub atomic particles. Protons are fundamental constituent of matter because positive rays are produced by all substances.

“A sub atomic particle, which is a fundamental constituent of all matter having a mass slightly less than that of H atom and which carries a positive charge equal in magnitude to the charge on an electron, is called a proton”. A proton is denoted by p or p+ _of

+1p.

Comparison of Positive (Canal) Rays and Cathode Rays:

Properties Cathode Rays Canal Rays

Sign of Charge Negative Positive

Mag. of Charge Always –1 Mostly +1, but also +2, +3…

Mass Definite value Variable, depends on ions

e/m Definite value Variable, depends on ions

DISCOVERY OF NEUTRON

After the discovery of electrons and protons. Rutherford (1920) had predicted the existence of a neutral fundamental particle. In 1932, Chadwick

bombarded the element Beryllium with particles and noticed the emission of a radiation having the following characteristics.

 The radiation was highly penetrating.

 The radiation was unaffected by magnetic and electric fields which show that it is electrically neutral.

 It was found to have approximately the same mass as the protons.

The name „neutron‟ was given to this sub atomic particle. It is denoted by n or1

on. Bombardment of beryllium by particles results in the formation of

carbon and neutrons are emitted.

Be 1

0

C n

At present there are a number of evidences which confirm that like electron, proton and neutron is also a fundamental constituent of atoms (a single exception is 1

1Hatom which does not contain any neutron)

Mass of a neutron is 1.008930 amu (1.6753 10–24_{g or 1.6753 10}–27 _kg) Neutron “A sub atomic particle, which is a fundamental constituent of matter having mass approximately equal to the hydrogen atom and which is

Illustration 1:

The neutron is attracted towards (A) Positive charged particles (B) Negative charged particles (C) Not attracted by any charge (D) None of these

Solution:

Neutron is an uncharged particle. Hence (C) is correct. ATOMIC TERM

Nuclide:

Various species of atoms in general. Nucleons:

Sub-atomic particles in the nucleus of an atom, i.e., protons and neutrons.

Atoms of an element with the same atomic number but different mass number.

Mass number (A):

Sum of the number of protons and neutrons, i.e., the total number of nucleons,

Atomic number (Z):

The number of protons in the nucleus of an atom. This, when subtracted from A, gives the number of neutrons.

Isobars:

Atoms, having the same mass numbers but different atomic numbers, e.g., 15P32 and 16S32.

Isotones:

Atoms having the same number of neutrons but different number of protons or mass number, e.g., 14 16 15

6 C, 8O, 7N.

Atoms molecules or ions having the same number of electrons, e.g., N2, CO, CN-_.

Nuclear isomers:

Atoms with the same atomic and mass numbers but different radioactive properties, e.g., uranium X (half life 1.4 min) and uranium Z (half life 6.7 hours).

Atomic mass unit:

Exactly equal to 1/12th of the mass of 6C12 atom. (a.m.u.): 1 a.m.u. = 1.66 10–24 _{g 931.5 MeV}

Illustration 2:

The ion that is isoelectronic with CO is

(A) CN- (B) O2+

(C) O2- (D) N2+

Solution:

Illustration 3.

Find the number of neutrons in a neutral atom having atomic mass 23 and number of electrons 11.

Solution:

Number of protons = number of electrons = 11

Number of neutrons = atomic mass – number of protons =23–11 = 12 Illustration 4:

How many protons, electrons and neutrons are present in 0.18 g30 15P?

Solution:

No. of neutrons in one atom = (30 – 15) = 15 0.18 g 30 15P = 0.18 30 = 0.006 mole Now, number of 30 15P atoms in 0.006 mole = 0.006 6.02 10 23 Number of electrons in 0.006 mole of 30

15P=Number of protons in 0.006 mole 30 15P = 15 0.006 6.02 10 23 _{= 5.418 10}22 _{and number of} neutrons = 5.418 1022 Exercise 1:

(i) Find the mass number of a uninegative ion having 10 neutrons and 10 electrons in one mole of the ions.

(ii) Find the total number and total mass of protons present in 34 mg of NH3.

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.

electron Positive sphere

This model could not satisfactorily explain the results of scattering experiment carried out by Rutherford who worked with Thomson.

Rutherford’s Model:

– particles emitted by radioactive substance were shown to be dipositive Helium ions (He++_{) having a mass of 4 units and 2 units of positive charge.} Rutherford allowed a narrow beam of –particles to fall on a very thin gold foil of thickness of the order of 0.0004 cm and determined the subsequent path of these particles with the help of a zinc sulphide fluorescent screen. The zinc sulphide screen gives off a visible flash of light when struck by an particle, as ZnS has the remarkable property of converting kinetic energy of particle into visible light. [For this experiment, Rutherford specifically used particles because they are relatively heavy resulting in high momentum].

Observation:

 Majority of the –particles pass straight through the gold strip with little or no deflection.

 Very few –particles are deflected backwards through angles greater than 90 .

 Some were even scattered in the opposite direction at an angle of 180 [Rutherford was very much surprised by it and remarked that “It was as incredible as if you fired a 15–inch shell at a piece of tissue paper and it came back and hit you”]. There is far less difference between air and bullet than there is between gold atoms and -particle assuming of course that density of a gold atom is evenly distributed. The distance of nucleus from where the - particle returns back through 180 is called distance of closet approach and is given by

1 2 0 2 0 q q r 1 4 mv 2 Conclusions:

 The fact that most of the - particles passed straight through the metal foil indicates the most part of the atom is empty.

 The fact that few - particles are deflected at large angles indicates the presence of a heavy positively charge body i.e., for such large deflections to occur - particles must have come closer to or collided with a massive positively charged body.

 The fact that one in 20,000 have deflected at 180° backwards indicates that volume occupied by this heavy positively charged body is very small in comparison to total volume of the atom.

Illustration 5:

An –particle is traveling towards gold nuclei returns back through 10 10 _{m from it. What is the velocity of the - particle. [Given 1} amu = 1.66 10 27 _{kg, atomic mass of He = 4 and gold = 79 and}

0

1

4

Now one He atom has charge (q1) = 2e One gold atom has charge (q2) = 79e Putting these values we get

19 19 9 2 27 10 2 2 1.6 10 79 1.6 10 9 10 v 4 1.66 10 10 v = 3.311 105 _m/s

Conclusions of -Scattering Experiment:

On the basis of the above observation, and having realized that the rebounding -particles had met something even more massive than themselves inside the gold atom, Rutherford proposed an atomic model as follows.

 All the +ve charge and nearly the total mass of an atom is present in a very small region at the centre of the atom. The atom‟s central core is called nucleus.

 The size of the nucleus is very small in comparison to the size of the atom. Diameter of the nucleus is about 10–13_{cm while the atom has a} diameter of the order of 10–8 cm. So, the size of atom is 105 times more than that of nucleus.

 Most of the space outside the nucleus is empty.

 The electrons, equal in number to the net nuclear positive charge, revolve around the nucleus with fast speed just like planets around the sun.

 The centrifugal force arising due to the fast speed of an electron balances the coulombic force of attraction of the nucleus and the

electron remains stable in its path. Thus according to him atom consists of two parts (a) nucleus and (b) extra nuclear part.

Defects in Rutherford’s Atomic Model:

 Position of electrons: The exact positions of the electrons from the nucleus are not mentioned.

 Stability of the atom: Bohr pointed out that Rutherford‟s atom should be highly unstable. According to the law of electro–dynamics, when a charged body moves under the influence of an attractive force, it loses energy continuously in the form of electromagnetic radiation. The electron should therefore, continuously emit radiation and lose energy. As a result of this a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus.

+

It was calculated that the electron should fall into the nucleus in less than 10–8 _{sec. But it is known that electrons keep moving outside the nucleus.} To solve this problem Neils Bohr proposed an improved form of Rutherford‟s atomic model.

Before going into the details of Neils Bohr model we would like to introduce you some important atomic terms.

Illustration 6:

Prove that density of the nucleus is constant. Solution:

Radius of the nucleus = 1.33 10–13 _A1/3 _{cm, where A is the mass} number

= 1.33 10–11 _A1/3 _m Density of nucleus = Mass

Volume = 27 11 1/ 3 3 A 1.66 10 kg 4 (1.33 10 A ) 3 = 1.66 1027 3 _{11 3} 4 3.14 (1.33 10 ) kg/m 3 _{= constant}

Thus density of nucleus is constant, independent of the element under consideration.

SOME IMPORTANT CHARACTERISTICS OF A WAVE

A wave is a sort of disturbance which originates from some vibrating source and travels outward as a continuous sequence of alternating crests and

troughs. Every wave has five important characteristics, namely, wavelength ( ), frequency ( ), velocity (c), wave number and amplitude (a).

a

Crest Crest

Trough Trough

Electronic Magnetic Radiation:

Ordinary light rays, X–rays, –rays, etc. are called electromagnetic radiations because similar waves can be produced by moving a charged body in a magnetic field or a magnet in an electric field. These radiations have wave characteristics and do not require any medium for their propagation.

 Wavelength ( ): The distance between two neighbouring troughs or crests is known as wavelength. It is denoted by and is expressed in cm, m, nanometers (1 nm =10–9 _{m) or Angstrom (1 Å=10}–10 _m).

 Frequency ( ): The frequency of a wave is the number of times a wave passes through a given point in a medium in one second. It is denoted by (nu) and is expressed in cycles per second (cps) or hertz (Hz) 1Hz = 1cps.

1 or = c

 Velocity: The distance travelled by the wave in one second is called its velocity. It is denoted by c and is expressed in cm sec–1_.

c = or = c

 Wave number : It is defined as number of wavelengths per cm. It

is denoted by and is expressed in cm–1_.

= 1 or =

c

 Amplitude: It is the height of the crest or depth of the trough of a wave and is denoted by a. It determines the intensity or brightness of the beam of light.

 Electromagnetic Spectrum: The arrangement of the various types of electromagnetic radiation in order of increasing or decreasing wavelengths or frequencies is known as electromagnetic spectrum. Wavelengths of Electromagnetic Radiations:

Electromagnetic radiations Wave length (Å)

Radio waves 3 1014 to 3 107

Micro waves 3 109 _{to 3 10}6

Infrared (IR) 6 106 _{to 7600}

1.1.1 Visible 7600 to 3800

Ultra violet (UV) 3800 to 150

X–rays 150 to 0.1

Gamma rays 0.1 to 0.01

Cosmic rays 0.01 to zero

Illustration 7:

Find out the longest wavelength of absorption line for hydrogen gas containing atoms in ground state.

Solution: 2 2 2 1 2 1 1 1 RZ n n

For longest wavelength E should be smallest, i.e. transition occurs from

n = 6 n = 5 n = 4 n = 3 n = 2 n = 1 i.e. 1= 109673 cm–1 ₁2 2 2 1 1 1 2 1 _{= 109673} 3 4 cm –1 = 4 ₁ 3 109673cm = 1.2157 10 –5 _{cm = 121.6 nm} Illustration 8:

Calculate the energy in kJ per mole of electronic charge accelerated by a potential of 1 volt.

Solution:

Energy in joules = charge in coulombs potential difference in volt = 1.6 10–19 _{6.02 10}23 ₁

Exercise 2:

(i) What is the ratio between energies of two radiations one with a wavelength of 6000 Å and other with 2000 Å.

(ii) Find the frequency and wave number of a radiation having wavelength of 1000 Å.

ATOMIC SPECTRUM

If the atom gains energy the electron passes from a lower energy level to a higher energy level, energy is absorbed that means a specific wave length is absorbed. Consequently, a dark line will appear in the spectrum. This dark line constitutes the absorption spectrum.

If the atom loses energy, the electron passes from higher to a lower energy level, energy is released and a spectral line of specific wavelength is emitted. This line constitutes the emission spectrum.

Types of Emission Spectra:

 Continuous spectra: When white light from any source such as sun or bulb is analyzed by passing through a prism, it splits up into seven different wide bands of colour from violet to red (like rainbow). These

colour are so continuous that each of them merges into the next. Hence the spectrum is called as continuous spectrum.

 Line spectra: When an electric discharge is passed through a gas at low pressure light is emitted. If this light is resolved by a spectroscope, It is found that some isolated coloured lines are obtained on a photographic plate separated from each other by dark spaces. This spectrum is called line spectrum. Each line in the spectrum corresponds to a particular wavelength. Each element gives its own characteristic spectrum.

PLANCK’S QUANTUM THEORY

When a black body is heated, it emits thermal radiations of different wavelengths or frequency. To explain these radiations, Max Planck put forward a theory known as Planck‟s quantum theory. The main points of quantum theory are

 Substances radiate or absorb energy discontinuously in the form of small packets or bundles of energy.

 The smallest packet of energy is called quantum. In case of light the quantum is known as photon.

 The energy of a quantum is directly proportional to the frequency of the radiation. E (or) E = h where is the frequency of radiation and h is Planck‟s constant having the value 6.626 10–27 erg – sec or 6.626 10–34 J–sec.

 A body can radiate or absorb energy in whole number multiples of a quantum h , 2h ,3h ………..nh . where „n' is the positive integer.

 Neils Bohr used this theory to explain the structure of atom.

Illustration 9.

The wave number of a radiation is 400 cm 1_{. Find out its}

(a) Wavelength (b) Frequency

(c) J per photon (d) kcal per mol of photons

(e) kJ per mol of photons Solution: (a) 1or = 1 = 1 ₁ 400 cm = 2.5 10 –3 _cm (b) = c = c = 3 10–10 cm/s 400 cm–1 = 7 1 1.2 10 s (c) Ephoton = h = hc = h c = 6.626 10–34 _{Js 3 10}8 _{cm/s 400 cm}–1

= 7.95 10 J (d) Ephoton = 7.95 10–21J

For 1 mol of photon energy = 7.95 10–21_{J 6.022 10}23 _mol–1 = (4.7875 103 _{J mol}–1_{) (1 kcal/4184 J)}

= 1.14 kcal mol–1

(e) E = (4.7875 103 _{J mol}–1_{) (1 kJ/1000J)} = 4.7875 kJ mol–1

Illustration 10:

A near ultraviolet photon of 300 nm is absorbed by a gas and then re emitted as two photons. One photon is red with wavelength 760 nm. What would be the wavelength of the second photon?

Solution:

E = hc

Wavelength of the 1st _{photon =} 1 E1 =

1

hc

And that of second photon = 2 E2 =

2

hc

ETotal = E1 + E2= emitted energy

1 2

2 1 1 1 300 760 2 1 1 1 300 760 2 1 760 300 760 300

2 = wavelength of the second photon = 496 nm

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:

 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.

 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 h 2

mvr = nh

v = velocity of the electron

n = orbit number in which electron is present r = radius of the orbit

 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.

 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.

 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, ( E) is given by E = E2 – E1 = h

where E1 and E2 are the energies of the electron in the first and second energy levels, and 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 = KZe₂ 2

r

K =

o

1

4 (where o is permittivity of free space)

K = 9 109 _Nm2 _C–2

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

The centrifugal force acting on the electron is mv2

r

Since the electrostatic force balances the centrifugal force, for the stable electron orbit. 2 mv r = 2 2 KZe r … (1) or v2 ₌ KZe2 mr … (2)

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

mvr =nh

v = nh 2 mr v2 ₌ 2 2 2 2 2 n h 4 m r … (3) Equating (2) and (3) 2 2 2 2 2 2 KZe n h mr 4 m r

Solving for r we get r = ₂n h2 2 ₂ 4π mΚΖe

Where n = 1, 2, 3 - - -

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 ro.

ro = 2 2 2 2 n h 4 me K= 2 2 34 2 2 31 19 9 1 6.626 10 4 3.14 9 10 1.6 10 9 10 =5.29 10–11 _{m=0.529 Å}

rn = 0.529 n

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 = ½ mv2

Potential energy = KZe2

r

Total energy = 1/2 mv2 _–KZe2

r … (4)

From equation (1) we know that

2 mv r = 2 2 KZe r ½ mv2 ₌ KZe2 2r

Substituting this in equation (4)

Total energy (E) =KZe2

2r – 2 KZe r = 2 KZe 2r

Substituting for r, gives us

E =2π mZ e K2 ₂ 2₂ 4 2

n h 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 10–12 2

2

Z

n erg per atom

= –21.8 10–19 2 2 Z n J per atom = –13.6 Z₂2 n eV per atom (1eV = 3.83 10–23 _kcal 1eV = 1.602 10–12 _erg 1eV = 1.602 10–19_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 = , E = 0, which corresponds to an ionized atom i.e., the electron and nucleus are infinitely separated.

H H+_{+ e}– _{(ionization).} Calculation of Velocity: We know that mvr =nh 2 ; v = nh 2 mr

By substituting for r we get

v = 2 KZe2

nh

v = 2.18 108 Z

n cm/sec.

Further application of Bohr‟s work was made, to other one electron species (Hydrogenic ion) such as He+ _{and Li}2+_{. In each case of this kind, Bohr‟s} prediction of the spectrum was correct.

Illustration 11:

The velocity of electron in the second orbit of He will be

(A) 6 2.18 10 m / s (B) 6 1.09 10 m / s (C) 6 4.36 10 m / s (D) None of these Solution: 0 n v Z v

n . Hence (A) is correct.

Illustration 12:

Calculate the velocity of an electron in Bohr‟s first orbit of hydrogen atom

(Given r = 0.53 10–10 _m) Solution:

According to Bohr‟s theory mvr = nh nh or v 2 2 mr r = 0.53 10–10_m n = 1, h = 6.62 10–34kg m2 s–1, m = 9.1 10–31 kg v = 34 2 1 31 10 1 (6.62 10 kgm s ) 22 2 (9.1 10 kg) (0.53 10 m) 7 = 2.18 106 _m/s Illustration 13:

The velocity of e in first Bohr‟s orbit is 2.17 106 _{m/s. Calculate the} velocity in 3rd _{orbit of He}2+ _ion.

Solution: vn = 0 Z v n Hence v = 2.19 106 2 3 = 1.45 10 6 _m/s 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. 1 _{= R} 2 2 1 2 1 1 n n

Where, = wave number = wave length

R = Rydberg constant (109678 cm–1₎ n1 and n2 have integral values as follows

Series n1 n2 Main spectral lines

Lyman Balmer Paschen Brackett 1 2 3 4 2, 3, 4, etc 3, 4, 5 etc 4, 5, 6 etc 5, 6, 7 etc Ultra – violet Visible Infra – red Infra – red

Note:

All lines in the visible region are of Balmer series but reverse is not true. i.e., all Balmer lines will not fall in visible region]

The pattern of lines in atomic spectrum is characteristic of hydrogen. Illustration 14:

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

Solution: H 2 2 1 2 1 1 1 R n n 1 2 2 1 1 1 109678 cm 2 4 16 cm 3 109678 = 4863 0 A

Merits of Bohr’s Theory:

 The experimental value of radii and energies in hydrogen atom are in good agreement with that calculated on the basis of Bohr‟s theory.

 Bohr‟s concept of stationary state of electron explains the emission and absorption spectra of hydrogen like atoms.

 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

 It does not explain the spectra of atoms having more than one electron.

 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.

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

 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 15:

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: 2 0 n r n r Z and 2 0 n 2 E Z E n . Hence (B) is correct. Illustration 16:

An electron in a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum required for its escape from the atom. What is the wavelength of the emitted electron? (me = 9.11 10–31 _kg)

(A) 4.7 Å (B) 4.7 nm

(C) 9.4 Å (D) 9.40 nm

Since 13.6 eV is need for ionization

Total energy absorbed = 1.5 13.6eV 20.4eV

6.8 eV is converted into K.E.

34 31 19 h 6.625 10 2m KE 2 9.11 10 6.8 1.6 10 = 10 4.7 10 m 4.7Å

Hence (A) is correct. Illustration 17:

The observation that electrons can be diffracted is an evidence that electrons

(A) have particle properties (B) have wave properties (C) are emitted by atoms (D) are absorbed by ions Solution:

Illustration 18:

A series of lines in the spectrum of atomic hydrogen lies at wavelengths 656.46, 482.7, 434.17, 410.29 nm. What is the wavelength of next line in this series?

Solution:

The given series of lines are in the visible region and thus appears to be Balmer series

Therefore n1 = 2 and n2=? for next line If = 410.29 10–7 _{cm and n}

1 = 2 n2 may be calculated for the last line

1_{= R} 2 2 1 2 1 1 n n 7 1 410.29 10 = 109673 2 2 2 1 1 2 n n2 = 6

Thus next line will be obtained during the jump of electron from 7th _to 2nd _{shell i.e,} 1_{= R} 2 2 1 1 2 7 = 109673 1 1 4 49 = 397.2 10–7 _{cm = 397.2 nm} Illustration 19:

Find the energy released in (ergs) when 2.0 gm atom of hydrogen undergo transition giving spectral line of lowest energy in visible region of its atomic spectrum.

Solution: E = 2.178 10–18 (Z2) _{2 2} 1 2 1 1 J / atom n n

For visible photon, n1 = 2

For lowest energy transition n2 = 3 E = 2.178 10–18 ₁2 2 2 1 1 2 3 J/atom For 2.0g atom E = (2 6.023 1023_{) 2.178 10}–18 5 36 joules = 3.63 105J = 3.63 1012 ergs Illustration 20:

Which hydrogen like ionic species has wavelength difference between the first line of Balmer and first line of Lyman series equal to 59.3 10–9 _{m? Neglect the reduced mass effect.}

Wave number of first Balmer line of an species with atomic number Z is given by 2 2 2 1 1 RZ 2 3 2 5RZ

36 ; Similarly wave number of v of first Lyman line is given by

= RZ2 2 2 1 1 1 2 =3₄RZ2; 1 and 1 – = 36₂ 4 ₂ 5RZ 3RZ = 2 1 36 4 5 3 RZ = 2 88 15RZ Z2 ₌ 9 7 88 59.3 10 15 1.097 10 = 9 or Z = 3 ionic species is Li2+ Exercise 3:

(i) Calculate of the radiations when the electron jumps from III to II orbit of hydrogen atom. The electronic energy in II and III Bohr orbit of hydrogen atoms are –5.42 10–12 _{and –2.41 10}–12 _erg respectively.

(ii) In a hydrogen atom, the energy of an electron in first Bohr‟s orbit is – 13.12 105 _{J mol}–1_{. What is the energy required for its excitation to} Bohr‟s second orbit.

(iii) What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n =? What is the colour corresponding to this wavelength? (Rydberg constant = 109,677 cm 1₎

QUANTUM NUMBERS

An atom contains large number of shells and subshells. These are distinguished from one another on the basis of their size, shape and orientation (direction) in space. The parameters are expressed in terms of different numbers called quantum numbers.

Quantum numbers may be defined as a set of four numbers with the help of which we can get complete information about all the electrons in an atom. It tells us the address of the electron i.e., location, energy, the type of orbital occupied and orientation of that orbital.

 Principal quantum number (n): It tells the main shell in which the electron resides, the approximate distance of the electron from the nucleus and energy of that particular electron. It also tells the maximum number of electrons that a shell can accommodate is 2n2_, where n is the principal quantum number.

Maximum number of electrons 2 8 18 32

 Azimuthal or angular momentum quantum number  : This

represents the number of subshells present in the main shell. These subsidiary orbits within a shell will be denoted as 0, 1, 2, 3, 4,… or s, p, d, f… This tells the shape of the subshells. The orbital angular momentum of the electron is given as   1 h

2 or   1 for a

particular value of „n‟ h

where 2

 . For a given value of n values of

possible  vary from 0 to n – 1.

 The magnetic quantum number (m): An electron due to its angular

motion around the nucleus generates an electric field. This electric field is expected to produce a magnetic field. Under the influence of external magnetic field, the electrons of a subshell can orient themselves in certain preferred regions of space around the nucleus called orbitals. The magnetic quantum number determines the number of preferred orientations of the electron present in a subshell. The values allowed depends on the value of l, the angular momentum quantum number, m can assume all integral values between – to +

including zero. Thus m can be –1, 0, +1 for  = 1. Total values of m

associated with a particular value of  is given by 2+ 1.

 The spin quantum number (s): Just like earth which not only revolves around the sun but also spins about its own axis, an electron in an atom not only revolves around the nucleus but also spins about

its own axis. Since an electron can spin either in clockwise direction or in anticlockwise direction, therefore, for any particular value of magnetic quantum number, spin quantum number can have two values, i.e., +1/2 and –1/2 or these are represented by two arrows pointing in the opposite directions, i.e., and . When an electron goes to a vacant orbital, it can have a clockwise or anti clockwise spin i.e., +1/2 or –1/2. This quantum number helps to explain the magnetic properties of the substances.

Illustration 21:

If the principal quantum number n has a value of 3, what are permitted values of the quantum numbers „‟ and „m‟?

Solution:

If principal quantum number (n) has the value of 3, subsidary quantum number () will also have three values i.e. 0, 1 and 2 and magnetic quantum number (m) will have n2_{, i.e. 3}2 _{i.e. 9 values in all} and they are designated as under:

n = 3  = 0 m = 0 (s)

 = 1 m = – 1 (p) m = 0

m = – 1 m = 0 m = +1 m = +2 Illustration 22:

In which orbital the electron will reside if it has the following values of quantum numbers.

(i) n = 3,  1, m 0

(ii) n = 4,  0, m 0

Solution:

(i)  1for p subshell and m = 0 for pz orbital. Hence electron will residue in 3pz orbital.

(ii)  0for s subshell. Hence electron will residue in 4s orbital.

Illustration 23:

Write all the quantum numbers for the following orbitals.

(i) 3d_z2 (ii) 3s (iii) 4px

(i) n 3, 2, m 0, s 1/ 2

(ii) n 3, 0, m 0, s 1/ 2

(iii) n 4, 1, m 1 or 1, s 1/ 2

Pauli’s Exclusion Principle:

According to this principle, an orbital can contain a maximum number of two electrons and these two electrons must be of opposite spin.

Two electrons in an orbital can be represented by or

SHAPES AND SIZE OF ORBITALS

An orbital is the region of space around the nucleus within which the probability of finding an electron of given energy is maximum (90–95%). The shape of this region (electron cloud) gives the shape of the orbital. It is basically determined by the azimuthal quantum number, while the

orientation of orbital depends on the magnetic quantum number (m). Let us now see the shapes of orbitals in the various subshells.

s–orbitals: These orbitals are spherical and

symmetrical about the nucleus. The probability of finding the electron is maximum near the nucleus and keep on decreasing as the distance from the nucleus increases. There is vacant space between two successive s–orbitals known as radial node. But there is no radial node for 1s orbital since it is starting from the nucleus.

nucleus Z radial node 2s 1s x y

The size of the orbital depends upon the value of principal quantum number (n). Greater the value of n, larger is the size of the orbital. Therefore, 2s–orbital is larger than 1s orbital but both of them are non-directional and spherically symmetrical in shape.

p–orbitals (=1):

The probability of finding the p–electron is maximum in two lobes on the opposite sides of the nucleus. This gives rise to a dumb–bell shape for the p–orbital. For p–orbital  = 1. Hence, m = –1, 0, +1. Thus, p–orbital have

three different orientations. These are designated as px, py & pz depending upon whether the density of electron is maximum along the x y and z axis respectively. As they are not spherically symmetrical, they have directional character. The two lobes of p–orbitals are separated by a nodal plane, where the probability of finding electron is zero.

pz

py

px

Z X

The three p-orbitals belonging to a particular energy shell have equal energies and are called degenerate orbitals.

d–orbitals (= 2):

For d–orbitals, l = 2. Hence m = –2,–1, 0, +1, +2. Thus there are 5d

orbitals. They have relatively complex geometry. Out of the five orbitals, the three (dxy, dyz,dzx) project in between the axis and the other two d_z2 and

2 _y2 d

x lie along the axis.

Y X Z 2 2 x y d 2 z d Dough–nut shape or

Baby soother shape Clover leaf shape

Z Y dxy X X Y Z dyz dxz

Illustration 24:

An electron has a spin quantum number 1

2and a magnetic quantum

number 1. It cannot occupy a/an

(A) d orbital (B) f orbital

(C) p orbital (D) s orbital

Solution:

For s orbital m = 0. Hence (D) is correct. Illustration 25:

Which d-orbital has electron density in all the planes? Solution:

2

z

d orbital has lobe along Z-axis and a belt of electrons in xy plane.

Hence it has electron density in all xy, xz and yz plane. RULES FOR FILLING OF ELECTRONS IN VARIOUS ORBITALS

The atom is built up by filling electrons in various orbitals according to the following rules.

Aufbau Principle: This principle states that the electrons are added one by

one to the various orbitals in order of their increasing energy starting with the orbital of lowest energy. The increasing order of energy of various orbital is 1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,5f,6d,7p………

How to remember such a big sequence? To make it simple we are giving you the method to write the increasing order of the orbitals. Starting from the top, the direction of the arrow gives the order of filling of orbitals.

1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 4f

Alternatively, the order of increasing energies of the various orbitals can be calculated on the basis of (n +) rule.

The energy of an orbital depends upon the sum of values of the principal quantum number (n) and the azimuthal quantum number (). This is called

(n +) rule. According to this rule,

“In neutral isolated atom, the lower the value of (n +) for an orbital, lower

is its energy. However, if the two different types of orbitals have the same value of (n +), the orbitals with lower value of n has lower energy‟‟.

Illustration of (n +) Rule: Type of orbitals Value of n Values of  Values of (n+) Relative energy 1s 1 0 1 + 0 = 1 Lowest energy

2s 2 0 2 + 0 = 2 Higher energy than 1s orbital

2p 2 1 2 + 1 = 3 2p orbital (n=2) have lower energy

than 3s orbital (n=3)

3s 3 0 3 + 1 = 4

Hund’s Rule of Maximum Multiplicity:

This rule deals with the filling of electrons in the equal energy (degenerate) orbitals of the same sub shell (p, d and f). According to this rule,

“Electron pairing in p,d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied”.

This is due to the fact that electrons being identical in charge, repel each other when present in the same orbital. This repulsion can, however, be minimized if two electrons move as far apart as possible by occupying different degenerate orbitals. All the electrons in a degenerate set of orbitals will have same spin.

Electronic Configuration of Elements:

Electronic configuration is the distribution of electrons into different shells, subshells and orbitals of an atom.

Keeping in view the above mentioned rules, electronic configuration of any orbital can be simply represented by the notation]

n

Number of electrons in the subshell

Symbol of subshell or orbitals (s,p,d,f) Principal quantum number

Alternatively:

Orbital can be represented by a box and an electron with its direction of spin by arrow. To write the electronic configuration, just we need to know (i) the atomic number (ii) the order in which orbitals are to be filled (iii) maximum number of electrons in a shell, sub–shell or orbital.

 The number of electrons to be accomodated in a subshell is 2 numbers of degenerate orbitals.

Subshell Maximum number of electrons

s 2

p 6

d 10

f 14

 The maximum number of electrons in each shell (K, L, M, N…) is given by 2n2 _{where n is the principal quantum number.}

 The maximum number of orbitals in a shell is given by n2 where n is the principal quantum number.

Importance of Knowing the Electronic Configuration:

The chemical properties of an element are dependent on the relative arrangement of its electrons.

Illustration 26:

Solution:

1s2 2s2 2p3

Illustration 27:

Write the electronic configuration of following:

(i) 2 2 S z 16 (ii) Fe z 26 Solution: (i) 2 2 6 2 6 1s 2s 2p 3s 3p (ii) 2 2 6 2 6 6 1s 2s 2p 3s 3p 3d Exceptional Configurations:

Stability of Half Filled and Completely Filled Orbitals: Cu has 29 electrons. Its expected electronic configuration is

orbital will make the distribution of electron symmetrical and hence will impart more stability.

Thus the electronic configuration of Cu is 1s2 2s22p63s23p64s13d10

Fully filled and half filled orbitals are more stable. Illustration 28:

Which of the following metals has the highest value of exchange energy required for exchange stabilization?

(A) Mn (B) Cr

(C) Cu (D) Zn

Solution:

Cu has full filled d-subshell. Hence (C) is correct. Illustration 29:

The total spin resulting from a d7 _{configuration is}

(A) 1 (B) 2

(C) 5

Solution:

S = n/2, where n = number of unpaired electrons = 3. Hence (D) is correct.

Illustration 30:

The electronic configuration of an element is 1s2_2s2_2p6_3s2_3p6_3d5_4s1_. This represents its

(A) excited state (B) ground state

(C) cationic form (D) anionic form

Solution:

It is ground state electronic configuration of Cr. Hence (B) is correct. Illustration 31:

Write the electronic configuration of the following ions:

(A) H+ _{(B) Na}+

(C) O2 _{(D) F}

(A) 1s

(B) 1s2_2s2_2p6 (C) 1s2_2s2_2p6 (D) 1s2_2s2_2p6 Illustration 32:

We know the Hund‟s rule. Explain how to arrange three electrons in p orbitals.

Solution:

The important point to be remembered is, all the singly occupied orbitals should have parallel spins, i.e in the same direction either-clockwise or antieither-clockwise.

2px 2py 2pz

½ ½ ½ = 1½

The maximum multiplicity means that the total spin of unpaired electrons is maximum.

Illustration 33:

We know that fully filled and half filled orbitals are more stable. Can you write the electronic configuration of Cr(Z = 24)?.

Solution:

Cr (Z = 24)

1s2_{, 2s}2_,2p6_,3s2_,3p6_,4s1_,3d5_.

Since half filled orbital is more stable one 4s electron is shifted to 3d orbital.

Illustration 34:

The configuration of potassium (19) is 1s2_2s2_2p6_3s2_4s1 _{and not} 1s2_2s2_2p6_3s2_3p6_3d1_{. Why?}

Solution:

According to Aufbau Principle‟s electron first fill in a sub-level of lower energy level (lower n+). The energy of 4s(4 + 0 = 4) is lower than that of 3d (3 + 2 = 5), so the electronic configuration of K is 1s2_2s2_2p6_3s2_3p6_4s1_.

Illustration 35:

How many maximum number of electrons can be present in 3rd _{shell of} an atom?

Solution:

Total number of electrons in a shell = 2n2

= 2 32 = 18 Illustration 36:

How many maximum number of electrons a d-orbital can have? Solution:

Any orbital can have maximum of 2 electrons. Illustration 37:

Calculate the total number of d-electrons in molybdenum (atomic number = 42).

Electronic configuration of Mo is

2 6 2 6 2 10 6 2 4

1s 2p 3s 3p 4s 3d 4p 5s 4d

Hence, total number of d-electrons = 10 + 4 = 14 Exercise 4:

(i) Write the electronic configuration of the following: (a) Cr3+

(b) Mn2+ (c) Cu

(ii) Write the orbital rotation for the following quantum numbers. (a) n = 4, = 2

(b) n = 2, = 1

(c) n = 3, = 0

DUAL CHARACTER (PARTICLE AND WAVE CHARACTER OF MATTER AND RADIATION)

In case of light some phenomenon like diffraction and interference can be explained on the basis of its wave character. However, the certain other phenomenon such as black body radiation and photoelectric effect can be

explained only on the basis of its particle nature. Thus, light is said to have a dual character. Such studies on light were made by Einstein in 1905.

Louis de Broglie, in 1924 extended the idea of photons to material particles such as electron and he proposed that matter also has a dual character-as wave and as particle.

Derivation of de-Broglie Equation:

The wavelength of the wave associated with any material particle was calculated by analogy with photon.

In case of photon, if it is assumed to have wave character, its energy is given by

E = h …(i)

(according to the Planck‟s quantum theory)

where is the frequency of the wave and „h‟ is Planck‟s constant If the photon is supposed to have particle character, its energy is given by

(according to Einstein‟s equation)

where „m‟ is the mass of photon, „c‟ is the velocity of light. By equating (i) and (ii)

h = mc2 But = c/

h c = mc2

(or) = h /mc

The above equation is applicable to material particle if the mass and velocity of photon is replaced by the mass and velocity of material particle. Thus for any material particle like electron.

= h/mv or = h

p where mv = p is the momentum of the particle.

Relation Between Kinetic Energy and Wavelength: K.E (E)=1/2 mv2

v = 2E m h mv h 2E.m

Derivation of Angular Momentum from de Broglie Equation:

According to Bohr‟s model, the electron revolves around the nucleus in circular orbits. According to de Broglie concept, the electron is not only a particle but has a wave character also.

If the wave is completely in phase, the circumference of the orbit must be equal to an integral multiple of wave length ( )

Therefore 2 r = n

where „n‟ is an integer and „r‟ is the radius of the orbit But = h/mv

2 r = nh /mv or mvr = n h/2

which is Bohr‟s postulate of angular momentum, where „n‟ is the principal quantum number.

“Thus, the number of waves an electron makes in a particular Bohr orbit in one complete revolution is equal to the principal quantum number of the orbit”. Alternatively: Number of waves „n‟ = 2 r = 2 r h mv = 2 mvr h

Where v and r are the velocity of electron and radius of that particular Bohr orbit in which number of waves are to be calculated, respectively.

The electron is revolving around the nucleus in a circular orbit. How many revolutions it can make in one second?

Let the velocity of electron be v m/sec. The distance it has to travel for one revolution 2 r, (i.e., the circumference of the circle).

Thus, the number of revolutions per second is = v

2 r

Common unit of energy is electron volt which is amount of energy given when an electron is accelerated by a potential of exactly 1 volt. This energy equals the product of voltage and charge. Since in SI units

coulombs volts = joules, 1 eV numerically equals the electronic charge except that joule replaces coulombs.

Illustration 38:

An e-_{, a proton and an alpha particle have K.E of 16 E, 4 E and E} respectively. What‟s the qualitative order of their Broglie wavelengths? (A) e > P > a (B) e > p = a

(C) P < e < a (D) a < e = P Solution:

= h

2mK.E . Hence (B) is correct.

Illustration 39:

Calculate the de Broglie wavelength of a ball of mass 0.1 kg moving with a speed of 60 ms–1_. Solution: h mv= 34 6.6 10 0.1 60 = 1.1 10–34 _m.

This is apparent that this wavelength is too small for ordinary observation. Although the de Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles.

Since, we come across macroscopic objects in our everyday life, de Broglie relationship has no significance in everyday life.

[Distinction between the wave- particle nature of a photon and the particle- wave nature of a sub atomic particle]:

Photon Sub Atomic Particle

1. Energy = h _{Energy =} 1 2mv 2 2. Wavelength = c Wavelength = h mv Note:

We should never interchange any of the above Illustration 40:

Number of waves made by electron in nth _{orbit = n} Hence the number of waves in 4rd _{orbit = 4}

Illustration 41:

What is the de-Broglie wavelength of electron having K.E. of 5 eV? Solution: K.E. = 1 2 mv 2 2K.E. v m h Now, mv = h 2K.E. m = 34 19 31 6.62 10 2 5 1.6 10 9.1 10 = 5.486 10 10 _m Illustration 42:

Through what potential difference an electron must be accelerated to have a de-Broglie wavelength of 1Ao .

Solution: = 10 10 m We know K.E. = 1 2 mv eV 2 v = 2eV m Now h h mv 2eVm Hence h₂2 v 2 em 2 34 10 2 19 31 6.62 10 2 10 h 1.6 10 9.1 10 = 1.5 10 6 _volt

HEISENBERG’S UNCERTAINTY PRINCIPLE

All moving objects that we see around us e.g., a car, a ball thrown in the air etc., move along definite paths. Hence their position and velocity can be measured accurately at any instant of time. Is it possible for subatomic particle also?

As a consequence of dual nature of matter, Heisenberg, in 1927 gave a principle about the uncertainties in simultaneous measurement of position and momentum (mass velocity) of small particles.

“It is impossible to measure simultaneously the position and momentum of a small microscopic moving particle with absolute accuracy or certainty” i.e., if an attempt is made to measure any one of these two quantities with higher accuracy, the other becomes less accurate.

The product of the uncertainty in position ( x) and the uncertainty in the momentum ( p = m. v where m is the mass of the particle and v is the uncertainty in velocity) is equal to or greater than h/4 where h is the Planck‟s constant.

Thus, the mathematical expression for the Heisenberg‟s uncertainty principle is simply written as

x. p h/4

Explanation of Heisenberg’s uncertainty principle

Suppose we attempt to measure both the position and momentum of an electron, to pinpoint the position of the electron we have to use light so that the photon of light strikes the electron and the reflected photon is seen in the microscope. As a result of the hitting, the position as well as the velocity of the electron are disturbed. The accuracy with which the position of the particle can be measured depends upon the wavelength of the light used. The uncertainty in position is . The shorter the wavelength, the greater is the accuracy. But shorter wavelength means higher frequency and hence

higher energy. This high energy photon on striking the electron changes its speed as well as direction. But this is not true for macroscopic moving particle. Hence Heisenberg‟s uncertainty principle is not applicable to macroscopic particles.

Illustration 43:

Why electron cannot exist inside the nucleus according to Heisenberg‟s uncertainty principle?

Solution:

Diameter of the atomic nucleus is of the order of 10–15m

The maximum uncertainty in the position of electron is 10–15 _m. Mass of electron = 9.1 10–31 _kg. x. p = h 4 x (m. v) = h/4 v = h 1 4 x.m = 34 6.63 10 22 4 7 15 31 1 10 9.1 10 v = 5.80 1010 _ms–1

This value is much higher than the velocity of light and hence not possible.

What is the uncertainity in the position of electron, if uncertainity in its velocity is 0.0058 m/s? Solution: h x v 4 m 34 31 6.02 10 x 4 3.14 9.1 10 0.0058 x = 0.01 m Illustration 45:

What is the uncertainity in the position of a ball of mass 10 g and which is moving with a velocity of 100 m/s with 0.002 % uncertainity? Solution: h x v 4 m Now 100 0.002 v 0.002 m/ s 100 34 31 6.02 10 x 0.0289 m 4 3.14 9.1 10 0.002 Exercise 5:

(i) Two particles A and B are in motion. If the wavelength associated with particle A is 5 10–8 _{m, calculate the wavelength associated with} particle B if its momentum is half of A.

(ii) Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 kV.

(iii) Calculate the uncertainty in position of a particle when uncertainty in the momentum is

(a) 1 10–2 _{gm cm sec}–1 _{and (b) zero.} QUANTUM MECHANICAL MODEL OF ATOM

The atomic model which is based on the particle and wave nature of the electron is known as wave or quantum mechanical model of the atom. This was developed by Erwin Schrodinger in 1926. This model describes the electron as a three dimensional wave in the electronic field of positively charged nucleus. Schrodinger derived an equation which describes wave motion of an electron. The differential equation is

2 2 2 2

2 2 2 2

8 m

(E V) 0

x y z h

where x, y, z are certain coordinates of the electron, m = mass of the electron E = total energy of the electron. V = potential energy of the