Structure of Atom

Structure of Atom EARLIER EFFORTS TO REVEAL INTRODUCTION STRUCTURE OF ATOM STRUCTURE OF ATOM Rutherford's Bohr's Model Model CATHODE RAYS - DISCOVERY OF ELECTRON Wave mechanical Model Dalton’s concept of the indivisibility of the atom was completely discredited by a series of experimental evidences obtained by scientists. A number of new phenomena were brought to light and man’s idea about the natural world underwent a revolutionary change. the discovery of electricity and spectral phenomena opened the door for radical changes in approaches to experim entation. It was concluded that atoms are made of three particles : electrons, protons and neutrons. These particles are called the fundamental particles of matter. In 1859 Julius Plucker started the study of conduction of electricity through gases at low pressure (10–4atm) in a discharge tube When a high voltage of the order of 10,000 volts or more was impressed across the electrodes, some sort of invisible rays moved from the negative electrode to the positive electrode these rays are called as cathode rays. 37 Structure of Atom PROPERTIES OF CATHODE RAYS Cathode rays have the following properties. (i) Path of travelling is straight from the cathode with a very high velocity as it produces shadow of an object placed in its path. When this experiment is conducted, a faint red glow is observed on the wall behind the cathode. Since these rays originate from the anode, they are called anode rays. PROPERTIES OF ANODE RAYS : (ii) Cathode rays produce mechanical effects. If small  Anode rays travel along straight paths and hence light paddle wheel is placed between the electrodes, they cast shadows of object placed in their path. it rotates. This indicates that the cathode rays consist  They rotate a light paddle wheel placed in their of material particles. path. This shows that anode rays are made up of material particles. (iii) When electric and magnetic fields are applied to the cathode rays in the discharge tube. The rays are  They are deflected towards the negative plate of an electric field. This shows that these rays are positively deflected thus establishing that they consist of charged. charged particles. The direction of deflection showed  For different gases used in the discharge tube, the that cathode rays consist of negatively charged charge to mass ratio (e/m) of the positive particles particles called electrons. constituting the positive rays is different. When (iv) They produce a green glow when strike the glass hydrogen gas is taken in the discharge tube, the e/m wall beyond the anode. Light is emitted when they value obtained for the positive rays is found to be strike the zinc sulphide screen. maximum. Since the value of charge (e) on the positive particle obtained from different gases is the (v) Cathode rays penetrate through thin sheets of same, the value of m must be minimum for the positive aluminium and metals. particles obtained from hydrogen gas. Thus, the (vi) They affect the photographic plates positive particle obtained from hydrogen gas is the (vii) The ratio of charge(e) to mass(m) i.e. charge/mass lightest among all the positive particles obtained from is same for all cathode rays irrespective of the gas different gases. This particle is called the proton. used in the tube. e/m = 1.76 × 1011 Ckg–1 DISCOVERY OF NEUTRON : Thus, it can be concluded that electrons are basic Later, a need was felt for the presence of electrically constituent of all the atoms. neutral particles as one of the constituent of atom. These particles were discovered by Chadwick in PRODUCTION OFANODE RAYS (DISCOVERY 1932 by bombarding a thin sheet of Beryllium with OF PROTON) -particles, when electrically neutral particles having Goldstein (1886) repeated the experiment with a a mass slightly greater than that of the protons were discharge tube filled with a perforated cathode and emitted. He named these particles as neutrons. found that new type of rays came out through the 9 4 1  12 4 Be  2He  6 C  0n hole in the cathode. 38 Structure of Atom THE NUCLEUS : Electrons, protons & neutrons are the fundamental particles present in all atoms,(except hydrogen) Table : 1 Particles Symbol Electron –1e 0 or  Mass Charge 9.10939 x 10-31 kg 1H 1 1.6722 x 10–27 kg + 1.6022 x 10–19 Coulombs 1 0n 1.67493 x 10–27 kg neutral 1.00867 u 1 amu  1.66 × Goldstein Rutherford1907 + 4.803 x 10–10 esu 1.00727 u Neutron J.J. Thomson Stoney Lorentz 1887 – 4.803 × 10–10 esu 0.00054 u Proton – 1.6022 x l0–19 Coulombs Discoverer 0 10–27 James Chadwick 1932 kg ATOMIC MODELS : Observation : (A) Thomson’s Model of the Atom : 1. An atom is electrically neutral. It contains positive charges (due to the presence of protons ) as well as negative charges (due to the presence of electrons). 2. Hence, J.J. Thomson assumed that an atom is a uniform sphere of positive charges with electrons 3. embedded in it. Most of the -particles passed straight through the gold foil without suffering any deflection from their original path. A few of them were deflected through small angles, while a very few were deflected to a large extent. A very small percentage (1 in 20000) was deflected through angles ranging from 90° to 180°. Rutherford’s nuclear concept of the atom. (B) Rutherford’s Experiment : (i) The atom of an element consists of a small positively charged ‘nucleus’ which is situated at the centre of the atom and which carries almost the entire mass of the atom. (ii) The electrons are distributed in the empty space of the atom around the nucleus in different concentric circular paths, called orbits. (iii) The number of electrons in orbits is equal to the number of positive charges (protons) in the nucleus. Hence, the atom is electrically neutral. (iv) The volume of the nucleus is negligibly small as compared to the volume of the atom. (v) Most of the space in the atom is empty. 39 DRAWBACKS OF RUTHERFORD’S MODEL: Structure of Atom Estimation of closest distance of approach (derivation) 1. This was not according to the classical theory of Ex. An -particle is projected from infinity with the electromagnetism proposed by maxwell. According velocity V0 towards the nucleus of an atom having to this theory, every accelerated charged particle atomic number equal to Z then find out (i) closest must emit radiations in the form of electromagnetic distance of approach (R) (ii) what is the velocity of waves and loses it total energy. the -particle at the distance R1 (R1 > R) from the nucleus. Since energy of electrons keep on decreasing, so radius of the circular orbits should also decrease and ultimately the electron should fall in nucleus. 2. It could not explain the line spectrum of H-atom. PROPERTIES OF CHARGE : 1. Q = ne ( charge is quantized) 2. Charge are of two types : (i) positive charge (ii) Negative Charge e = –1.6 x 10–19 Sol. From energy conservation P.E1 + KE1 = P.E2 + KE2  0+ 1 K(Ze)(2e) m V 2= +0 2   R p = + 1.6 x 10–19C 3. R= Same charge repel each other and opposite charges attract each other. Let velocity at R1 is V1. From energy conservation P.E1 + KE1 = P.E3 + KE3 Charge is a SCALAR Qty. and the force between the charges always acts along the line joining the charges. The magnitude of the force between the two charge placed at a distance ‘r’ is given by FE = 1 q1q2 4 0 r 2 (electrical force) 4. If two charge q1 and q2 are sepreated by distance r then the potential energy of the two charge system is given by. P.E. 5. 4KZe 2 This does not mean that a proton has a greater charge but it implies that the charge is equal and opposite. = 1 q1q2 4 0 r If a charged particle q is placed on a surface of potential V then the potential energy of the charge is q x V. m V 2  0+ (closest distance of approch) K(Ze)(2e) 1 1 m V2 = + m V12 R 2 2 1 SIZE OF THE NUCLEUS The volume of the nucleus is very small and is only a minute fraction of the total volume of the atom. Nucleus has a diameter of the order of 10–12 to 10– 13 cm and the atom has a diameter of the order of 10–8 cm. Thus, diameter (size) of the atom is 100,000 times the diameter of the nucleus. The radius of a nucleus is proportional to the cube root of the number of nucleons within it. R = R0 (A)1/3 cm where R0 can be 1.1 × 10–13 to 1.44 × 10–13 cm ; A = mass number ; R = Radius of the nucleus. Nucleus contains protons & neutrons except hydrogen atoms which does not contain neutron in the nucleus. 40  ATOMIC NUMBER (Z) AND MASS NUMBER (A)  Structure of Atom The atomic number (Z) and mass number (A) of an element ‘X’ are usually represented alongwith the symbol of the element as Atomic number (Z) of an element = Total number of protons present in the nucleus = Total number of electrons present in the atom   Atomic number is also known as proton number because the charge on the nucleus depends upon the number of protons. e.g. 1. As each of these particles has one unit mass on the atomic mass scale, therefore the sum of the number of protons and neutrons will be nearly equal to the mass of the atom. and 13 T and named as protium, deuterium (D) and tritium (T) respectively. Ordinary hydrogen is protium. 2. Isobars : Such atoms of different elements which have same mass numbers (and of course different atomic numbers) are called isobars 40 40 e.g. 18 Ar, 19 K, 40 20 Ca. Mass number of an element = No. of protons (Z) + No. of neutrons (n).  sotopes : Such atoms of the same element having 1 2 1H, 1 D 3.  and so on. same atomic number but different mass numbers are called isotopes. Since the electrons have negligible mass, the entire mass of the atom is mainly due to protons and neutrons only. Since these particles are present in the nucleus, therefore they are collectively called nucleons.  23 35 11 Na, 17 Cl The mass number of an element is nearly equal to the atomic mass of that element. However, the main 4. difference between the two is that mass number is always a whole number whereas atomic mass is usually not a whole number. Isotones : Such atoms of different elements which contain the same number of neutrons are called isotones 15 16 e.g. 14 6 C, 7 N, 8 O. Isoelectronic : The species (atoms or ions) containing the same number of electrons are called isoelectronic. For example, O2–, F–, Na+, Mg2+, Al3+, Ne all contain 10 electrons each and hence they are isoelectronic. Ex-1. Complete the following table : Particle Mass No. Atomic No. Protons Neutrons Electrons Nitrogen atom – – – 7 7 Calcium ion – 20 – 20 – Oxygen atom 16 8 – – – Bromide ion – – – 45 36 41 Structure of Atom But in the formation of bromide ion, one electron is gained by extra nuclear part according to equation Sol. For nitrogen atom. No. of electron = 7 (given) No. of neutrons = 7 (given)  No. of protons = Z = 7 Br + e–  Br – , But the composition of nucleus remains unchanged. ( atom is electrically neutral)  Atomic number = Z = 7 No. of protons in bromide ion = No. of electrons in bromine atom = 36 – 1 = 35 Mass No. (A) = No. of protons + No. of neutrons Atomic number (Z) = No. of protons = 35 = 7 + 7 = 14 Mass number (A) = No. of neutrons + No. of protons = 45 + 35 = 80. For calcium ion. No. of neutrons = 20 (Given) Atomic No. (Z) = 20 (Given)  No. of protons = Z = 20 ; No. of electrons in calcium atom = Z = 20 But in the formation of calcium ion, two electrons are lost from the extranuclear part according to the equation Ca  Ca2+ + 2e– but the composition of the nucleus remains unchanged. ELECTROMAGNETIC WAVE RADIATION The oscillating electrical/magnetic field are electromagnetic radiations. Experimentally, the direction of oscillations of electrical and magnetic field are prependicular to each other.  No. of electrons in calcium ion = 20 – 2 = 18 Mass number (A) = No. of protons + No. of neutrons = 20 + 20 = 40. For oxygen atom. Mass number (A) = No. of protons + No. of neutrons   E = Electric field, B = Magnetic field = 16 (Given Atomic No. (Z) = 8 (Given) No. of protons = Z = 8, No. of electrons = Z = 8 No. of neutrons = A – Z = 16 – 8 = 8 For bromide ion. No. of neutrons = 45 (given) No. of electrons = 36 (given) Direction of wave propogation. 42 SOME IMPORTANT CHARACTERISTICS OF A WAVE Structure of Atom Order of wavelength in Electromagnetic spectrum Cosmic rays <  – rays < X-rays < Ultraviolet rays < Visible < Infrared < Micro waves < Radio waves. PARTICLE NATURE OF ELECTROMAGNETIC RADIATION Planck's Quantum Theory Some of the experimental phenomenon such as diffraction and interference can be explained by the Wavelength of a wave is defined as the distance wave nature of the electromagnetic radiation. Howbetween any two consecutive crests or troughs. It is ever, following are some of the observations which represented by  (lambda) and is expressed in Å or could not be explained m or cm or nm (nanometer) or pm (picometer). (i) the nature of emission of radiation from hot bodies (black - body radiation) 1 Å = 10– 8 cm = 10–10 m (ii) ejection of electrons from metal surface when ra1 nm = 10– 9 m, 1 pm = 10–12 m diation strikes it (photoelectric effect) Frequency of a wave is defined as the number of BLACK BODY RADIATION waves passing through a point in one second. It is When solids are heated they emit radiation over a represented by  (nu) and is expressed in Hertz (Hz) wide range of wavelengths. or cycles/sec or simply sec–1 or s–1. The ideal body, which emits and absorbs all fre1 Hz = 1 cycle/sec quencies, is called a black body and the radiation emitted by such a body is called black body radiaVelocity of a wave is defined as the linear distance tion. The exact frequency distribution of the emitted travelled by the wave in one second. It is represented radiation (i.e., intensity versus frequency curve of by v and is expressed in cm/sec or m/sec (ms–1). the radiation) from a black body depends only on Amplitude of a wave is the height of the crest or the its temperature. depth of the trough. It is represented by ‘a’ and is expressed in the units of length. Wave number is defined as the number of waves present in 1 cm length. Evidently, it will be equal to the reciprocal of the wavelength. It is represented by  (read as nu bar).  1  If  is expressed in cm,  will have the units cm–1. Relationship between velocity, wavelength and frequency of a wave. As frequency is the number of waves passing through a point per second and  is the length of each wave, hence their product will give the velocity of the wave. Thus The above experimental results cannot be explained satisfactorily on the basis of the wave theory of light. Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete quantities and not in a continuous manner. v=× 43 QUANTUM THEORY OF LIGHT The smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation is called as quantum of light. According to Planck, the light energy coming out from any source is always an integral multiple of a smallest energy value called quantum of light. Let quantum of light be = E0(J), then total energy coming out is = nE0 (n = Integer) Quantum of light = Photon ( Packet or bundle of energy) Energy of one photon is given by E0 = h (- Frequency of light) h = 6.626 x 10–34 J-Sec (h - Planck const.) E0 = hc  Structure of Atom ONE ELECTRON VOLT (e.v.) : Energy gained by an electron when it is accelerated from rest through a potential difference of 1 volt. Note : Positive charge always moves from high potential to low potential and –ve charge always. moves from low potential to high potential if set free. From Energy conservation principle, P.E.i + K.E.i = P.E.f + K.E.f (– e) 0 + 0 = (– e)(1V) + (c - speed of light) ( - wavelength) K.E. = Order of magnitude of Eo = 10 34  10 8 10 10 = 10–16 J 1 mVf2 2 1 mVf2 = e(1 volt) 2 If a charge ‘q’ is accelerated through a potential dirrerence of ‘V’ volt then its kinetic energy will be increased by q.V. 1eV = 1.6 x 10–19 C x 1 volt  1eV = 1.6 x 10–19J Ex-2. Certain sun glasses having small of AgCl incorporated in the lenses, on expousure to light of PHOTOELECTRIC EFFECT When certain metals (for example Potassium, Ruappropriate wavelength turns to gray colour to reduce the glare following the reactions: bidium, Caesium etc.) were exposed to a beam of light electrons were ejected as shown in Fig. AgCl hv Ag(Gray) + Cl   If the heat of reaction for the decomposition of AgCl is 248 kJ mol–1, what maximum wavelength is needed to induce the desired process? Sol. Energy needed to change = 248 × 103 J/mol If photon is used for this purpose, then according to Einstein law one molecule absorbs one photon. Therefore,  = NA . hc  = 248 × 103 6.626  10 34  3.0  10 8  6.023  10 23 248  10 3 = 4.83 × 10–7 m 44 Structure of Atom The phenomenon is called Photoelectric effect. The results observed in this experiment were : (i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface, i.e., there is no time lag between the striking of light beam and the ejection of electrons from the metal surface. (ii) The number of electrons ejected is proportional to the intensity or brightness of light. (iii) For each metal, there is a characteristic minimum frequency, 0 (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency  > 0, the ejected electrons come out with certain kinetic energy. The kinetic energies of these electrons increase with the increase of frequency of the light used. When a photon of sufficient energy strikes an electron in the atom of the metal, it transfers its energy instantaneously to the electron during the collision and the electron is ejected without any time lag or delay. Greater the energy possessed by the photon, greater will be transfer of energy to the electron and greater the kinetic energy of the ejected electron. In other words, kinetic energy of the ejected electron is proportional to the frequency of the electromagnetic radiation. Since the striking photon has energy equal to h and the minimum energy required to eject the electron is h0 (is also called work function, W0) then the difference in energy (h – h0) is transferred as the kinetic energy of the photoelectron. Following the conservation of energy principle, the kinetic energy of the ejected electron is given by the equation 1 h = h0 + me 2 2 where me is the mass of the electron and v is the velocity associated with the ejected electron. Ex-3. The threshold frequency 0 for a metal is 6 × 1014 s–1. Calculate the kinetic energy of an electron emitted when radiation of frequency  = 1.1 × 1015 s–1 hits the metal. Sol. BOHR’S ATOMIC MODEL It is based on quantum theory of light. Assumptions of Bohr’s model :  There are certain orbits around the nucleus such that if electron will be revolving in these orbit, then it not emit any electromagnetic radiation. These are called stationary orbit. The neccessary centripetal force is produced by attraction forces of nucleus. Ke 2 Z mv 2 = 2 r r  Angular momentum of the electron in these stationary orbit is always an integral multiple of nh mvr = 2  Electron can make jump from one stationary orbit to another stationary orbit by absorbing or emitting a photon of energy equal to difference in the energies of the stationary orbit. hc  = E E – difference in the energy of orbit MATHEMATICAL FORMS OF BOHR’S POSTULATES Calculation of the radius of the Bohr’s orbit : Suppose that an electron having mass ‘m’ and charge ‘e’ revolving around the nucleus of charge ‘Ze’ (Z is atomic number & e = charge) with a tangential/linear velocity of ‘v’. Further consider that ‘r’ is the radius of the orbit in which electron is revolving. According to Coulomb’s law, the electrostatic force of attraction (F) between the moving electron and nucleus is – 1 K.E. = meV2 = h ( – 0) 2  K.E. = (6.626 × 10–34) (1.1 × 1015 – 6 × 1014)  K.E. = (6.626 × 10–34) (5 × 1014) = 3.313 × 10–19 J h 2 F= KZe 2 r2 1 where : K = constant = 4 = 9 x 109 0 Nm2/C2 45 Structure of Atom and the centrifugal force F = CALCULATION OF VELOCITY OF mv 2 r AN ELECTRON IN BOHR’S ORBIT For the stable orbit of an electron both the forces are balanced. i.e Angular momentum of the revolving electron in nth orbit is given by KZe 2 mv 2 = 2 r r v2 = then nh mvr = 2 KZe 2 mr ......... (i) nh v = 2mr From the postulate of Bohr, nh 2 mvr =  v= nh 2mr n 2h 2 On squaring v2 = put the value of ‘r’ in the equation (iii) then, ........ (ii) 4 2m 2r 2 ......... (iii) v= v= nh  4 2mZe 2K 2mn2h2 Ze 2K nh on putting the values of , e-, h and K velocity of electron in nth x Figure-12 From equation (i) and (ii) v Z vn = 2.18 x 106 ; v  n 1 T, Time period of revolution of an electron in its orbit n 2h 2 KZe 2 = 2 22 4 m r mr = On solving, we will get r= Z m/sec; n orbit 2r v f, Frequency of revolution of an electron in its orbit n 2h 2 v 4  2mKZe 2 On putting the value of e , h , m, the radius of nth Bohr orbit is given by : 2 rn = 0.529 x The total energy of an electron revolving in a particular orbit is  22 nd + Sol. r1 (radius of 2 orbit of He ion) = 0.529  2   Å    32 rd +++ r2 (radius of 3 orbit of Be ion) = 0.529  4   Å   0.529  2 2 / 2 CALCULATION OF ENERGY OF AN ELECTRON n Å Z Ex.4 Calculate radius ratio for 2nd orbit of He+ ion & 3rd orbit of Be+++ ion. r1 = 2 r T.E. = K.E. + P.E. where : P.E. = Potential energy , K.E. = Kinetic energy, T.E. = Total energy The K.E. of an electron = 1 mv2 2 8 Therefore r = = 2 0.529  3 2 / 4 9 and the P.E. of an electron = – KZe 2 r 46 Structure of Atom Hence, T.E. = 1 KZe mv2 – 2 r 2 Ex-4 What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom ? KZe 2 mv2 = Sol. Since ni = 5 and nf = 2, this transition gives rise to a r 2 substituting the value of mv in the above equation : spectral line in the visible region of the Balmer series. KZe 2 mv 2 = 2 or r r we know that, KZe 2 KZe 2 KZe 2 – =– 2r r 2r T.E. = So, T.E. = – KZe 2r 2 substituting the value of ‘r’ in the equation of T.E. Then T.E. = – 42 Ze 2m 22 Z 2 e 4m K 2 KZe 2 x = – 2 n2h 2 n 2h 2 Thus, the total energy of an electron in nth orbit is given by 2 22 me 4 k 2  z  T.E. = En = – ... (iv)  n2  h2   Putting the value of m,e,h and  we get the expression of total energy Z2 En = – 13.6 2 eV / atom n n  T.E.  ; Z  T.E.  = – 2.18 × 10–18 T.E. = Z2 n2 J/atom 1 P.E. 2 T.E. = – K.E. Note : - The P.E. at the infinite = 0 The K.E. at the infinite = 0 CONCLUSION FROM EQUATION OF ENERGY (a) The negative sign of energy indicates that there is attraction between the negatively charged electron and positively charged nucleus. (b) All the quantities on R.H.S. in the energy equation [Eq. iv] are constant for an element having atomic number Z except ‘n’ which is an integer such as 1,2,3, etc . i.e. the energy of an electron is constant as long as the value of ‘n’ is kept constant. (c) The energy of an electron is inversely proportional to the square of ‘n’ with negative sign. E  1 = 2.18 × 10–18 J  5 2 – 1  22  = – 4.58 × 10–19 J It is an emission energy The frequency of the photon (taking energy in terms of magnitude) is given by  = 4.58  10 –19 J E = h 6.626  10 –34 Js = 6.91 × 1014 Hz = c  = 3.0  10 8 ms –1 6.91 1014 Hz = 434 nm FAILURES / LIMITATIONS OF BOHR’S THEORY (a) He could not explain the line spectra of atoms containing more than one electron. (b) He also could not explain the presence of multiple spectral lines. (c) He was unable to explain the splitting of spectral lines in magnetic field (Zeeman effect) and in electric field (Stark effect) (d) No conclusion was given for the principle of quantisation of angular momentum. (e) He was unable to explain the de-Broglie’s concept of dual nature of matter. (f) He could not explain Heisenberg’s uncertainty principle. ENERGY LEVEL DIAGRAM (i) Orbit of lowest energy is placed at the bottom, and all other orbits are placed above this. (ii) The gap between two orbits is proportional to the energy difference of the orbits. 47 Structure of Atom (vi) Excitation Potential : Potential difference through which an electron must be accelerated from rest to so that its kinetic energy become equal to excitation energy of any state is called excitation potential of that state. Excitation potential of third state = excitation potential of second exicited state = secondexcitation potential = 12.09 V. DEFINITION VALID FOR SINGLE (vii) Binding Energy ‘or’ Seperation Energy : Energy required to move an electron from any state to n =  is called binding energy of that state. Binding energy of ground state = .E. of atom or on. ELECTRON SYSTEM : (i) Ground state : Lowest energy state of any atom or ion is called ground state of the atom It is n = 1. Ground state energy of H–atom = – 13.6 ev Ex-5 A single electron system has ionization energy 11180 Ground state energy of He+ on = – 54.4 ev kJ mol–1 . Find the number of protons in the nucleus (ii) Excited State : of the system. States of atom other than the ground state are called Z2 excited states : Sol. .E. = 2 × 21.69 × 10–19 J n n=2 first excited state 11180  10 3 Z2 n=3 second excited state –19 23 = 2 × 21.69 × 10 6.023  10 1 n=4 third excited state Ans. Z = 3 n=n+1 nth excited state (iii) Ionisation energy (E) : HYDROGEN SPECTRUM Minimum energy required to move an electron from Study of Emission and Absorption Spectra : ground state to An instrument used to separate the radiation of n =  is called ionisation energy of the atom or ion. different wavelengths (or frequencies) is called onisation energy of H–atom = 13.6 ev spectroscope or a spectrograph. Photograph (or the onisation energy of He+ ion = 54.4 ev pattern) of the emergent radiation recorded on the onisation energy of Li+2 ion = 122.4 ev film is called a spectrogram or simply a spectrum of (iv) Ionisation Potential (.P.) : the given radiation The branch or science dealing Potential difference through which a free electron with the study of spectra is called spectroscopy. must be accelerated from rest, such that its kinetic energy becomes equal to ionisation energy of the atom is called ionisation potential of the atom. .P. of H atom = 13.6 V, .P. of He+ on= 54.4 V (v) Excitation Energy : Energy required to move an electron from ground state of the atom to any other state of the atom is called excitation energy of that state. Excitation energy of 2nd state = excitation energy of 1st excited state = 1st excitation energy = 10.2 ev. 48 Structure of Atom Emission spectra : When the radiation emitted from some source e.g. from the sun or by passing electric discharge through a gas at low pressure or by heating some substance to high temperature etc, is passed directly through the prism and then received on the photographic plate, the spectrum obtained is called ‘Emission spectrum’. Depending upon the source of radiation, the emission spectra are mainly of two type : (a) Continuous spectra : When white light from any source such as sun, a bulb or any hot glowing body is analysed by passing through a prism it is observed that it splits up into seven different wide band of colours from violet to red. These colours are so continuous that each of them merges into the next. Hence the spectrum is called continuous spectrum. (ii) Line Spectrum : This is the ordered arrangement of lines of particular wavelength seperated by dark space eg. hydrogen spectrum. Line spectrum can be obtained from atoms. (b) Absorption spectra : When white light from any source is first passed through the solution or vapours of a chemical substance and then analysed by the spectroscope, it is observed that some dark lines are obtained in the continuous spectrum. These dark lines are supposed to result from the fact that when white light (containing radiations of many wavelengths) is passed through the chemical substance, radiations of certain wavelengths are absorbed, depending upon the nature of the element. (b) Discrete spectra : It is of two type (i) Band spectrum EMISSION SPECTRUM OF HYDROGEN Band spectrum contains colourful continuous bands sepearted by some dark space. Generally molecular spectrum are band spectrum When hydrogen gas at low pressure is taken in the discharge tube and the light emitted on passing electric discharge is examined with a spectroscope, the spectrum obtained is called the emission spectrum of hydrogen. 49 Structure of Atom LINE SPECTRUM OF HYDROGEN Line spectrum of hydrogen is observed due to excitation or de-excitation of electron from one stationary orbit to another stationary orbit Let electron make transition from n2 to n1 (n2 > n1) in a H-like sample Energy of emitted photon = (E)n2  n1 =  13.6Z 2 n22 2    13. 6Z –  n2    1   1 1 = 13.6Z2  n 2  n 2  2   1 Ex-6 Calculate the wavelength of a photon emitted when an electron in H- atom maker a transition from n = 2 to n=1 Wavelength of emitted photon hc  = (E) n 2 n1 = 1   1 hc  1 1 13.6Z 2  2  2 n n2  1 = (13 .6)z 2 hc 1  1 Sol.  = RZ2  2  2  n 2   n1      1 1   n 2 n 2 2  1     1 1  1    = R(1)2  2  2  2  1  1  = 4 3R or   3R 4 SPECTRA LINES OF HYDROGEN ATOM LYMAN SERIES Wave number, 1   1 1  =  = RZ 2  2  2   n1 n 2  R = Rydberg constant = 1.09678 × 107m–1 ; R 1.1 × 107 m–1 ; R = 13.6eV hc ; R ch = 13.6 eV * It is first spectral series of H. * It was found out in ultraviolet region in 1898 by Lyman. * It’s value of n1 = 1 and n2 = 2,3,4 where ‘n1’ is ground state and ‘n2’ is called excited state of electron present in a H - atom. 50 Structure of Atom * 1  1 1   1  = R × 12  2  2  1    1 = RH 12  n 2  where n2 > 1 always.  2   =R * For Lyman series, n12 The wavelength of marginal line = R for all series. H  longest = 1 So for lyman series  = R . H * st line of lyman series  2  BALMER SERIES : 1 nd line of lyman series = 3  1 Last line of lyman series =   1 [10.2 eV  (E) lyman  13.6 eV] 12400 12400  lyman   13 .6 10.2 * * * Aº * 12400 (E)min Series limit : t is the last line of any spectral series. R3 3R = 4 4  1  Aº   balmer  12400 Å 1. 9 1   1  2 2 3  2  1 1   1 =  =  = R × 12  2  2  2  1 =  1 = RH  2 2  n2  where n2 > 2 always.  2  (  ) 1st line = R ×1  Wave no of st line of Lyman series R × 12 4 3648 Å   balmer  6536 Å Lines of balmer series (for H atom) lies in the visible range. Ist line of balmer series = 3  2 last line of balmer series =   2 Last line of any spectral series is the shortest ( min) line.  4  1    4  1  12400 3.4 First line of any spectral series is the longest ( max) line. = 22 1.9  (E)balmer  3.4 eV.. All the lines of balmer series in H spectrum are not in the visible range. nfact only st 4 lines belongs to visible range. 12400 (E)max 1 It is the second series of H-spectrum. It was found out in 1892 in visible region by Balmer. It’s value of n1 = 2 and n2 = 3,4,5,............. The wavelength of marginal line of Balmer series n12 Shortest line : shortest wavelength line  shortest or  min = * * * * * = R = R = R H H H Longest line : longest wavelength line  longest or  max. = 12400 , shortest = 12400 ( E)21 E   1 (  ) last line = R  2 2  1   2  = 5R = 36 R 4 PASCHEN SERIES : (a) It is the third series of H - spectrum. (b) It was found out in infrared region by Paschen. (c) It’s value of n1 = 3 and n2 = 4,5,6 ........ (d) The wavelength of marginal line of Paschen series 4     3R    9 n12 32 = R = R = R . H H H Wave no of last line of Lyman series (e) 1   1 1 = RH  3 2  n 2  where n2 > 3 always.  2  51 Structure of Atom BRACKETT SERIES : NO. OF PHOTONS EMITTED BY (a) It is fourth series of H - spectrum. A SAMPLE OF H ATOM (b) It was found out in infrared region by Brackett. If an electron is in any higher state n = n and makes (c) It’s value of n1 = 4 and n2 = 5,6,7 .............. a transition to ground state, then total no. of different (d) The wavelength of marginal line of brackett series n  (n  1) 16 n12 42 photons emitted is equal to . 2 = R = R = R H H H If an electron is in any higher state n = n2 and makes  1 1 1 a transition to another excited state n = n1, then total (e)  = RH  4 2  n 2  where n2 > 4 always.  2  no. of different photons emitted is equal to PFUND SERIES : n ( n  1) , where n = n2 – n1 (a) It is fifth series of H- spectrum. 2 Note : In case of single isolated atom if electron (b) It was found out in infrared region by Pfund. (c) It’s value of n1 = 5 and n2 = 6,7,8 ............... where make transition from n th state to the ground state n1 is ground state and n2 is excited state. then max. number of spectral lines observed = (n-1) (d) The wavelength of marginal line of Pfund series n12 52 25 = R = R = R H H H (e) 1  1 1 = RH  5 2  n 2  where n2 > 5 always. 2   HUMPHRY SERIES : (a) It is the sixth series of H - spectrum. (b) It was found out in infrared region by Humphry. (c) It’s value of n1 = 6 and n2 = 7 , 8 , 9 ................... (d) The wavelength of marginal line of Humphry series 36 n12 62 = R = R = R H H H (e) 1   1 1 = RH  6 2  n 2  where n2 > 6.  2  Ex-7 Calculate wavelength for 2nd line of Balmer series of He+ ion 1 1 1 2 Sol.   R(2)  n 2  n2  2  1  n1 = 2 n2 = 4 1 1  1  R( 2 2 )  2  2   4  2 1 3R   4 4 = Ans. 3R Ex-8 If electron make transition from 7th excited state to 2nd state in H atom sample find the max. number of spectral lines observed. Sol. n = 8 – 2 = 6  6  1 7 spectral lines = 6  2  = 6× = 21   2 DUAL NATURE OF ELECTRON (DE-BROGLIE HYPOTHESIS) (a) Einstein had suggested that light can behave as a wave as well as like a particle i.e. it has dual character. (b) In 1924, de-Broglie proposed that an electron behaves both as a material particle and as a wave. (c) This proposed a new theory wave mechanical theory of matter. According to this theory, the electrons protons and even atom when in motion possess wave properties. (d) According to de-Broglie, the wavelength associated with a particle of mass m, moving with velocity v is given by the relation, = h mv where h is Planck’s constant 52 Structure of Atom (e) This can be derived as follows according to Planck’s equation. c = velocity of light If velocity of particle is zero then : dynamic mass = rest mass h.c E = h =  Rest mass of photon is zero that means photon is never at rest Energy of photon on the basis of Einstein’s mass energy relationship * E = mc2 or = h mc = mc2 or = h mc  = This was experimentally verified by Davisson and Germer by observing diffraction effects with an electron beam. m2v2 = 2emV 2m(K.E.) If a charge q is accelerated through a potential difference of ‘V’ volt from rest then K.E. of the charge is equal to : “ q.V” Let the electron is accelerated with a potential of V then the K.E. is 1 mv2 =eV 2 m.v. = 2m(K.E.) h Which is same as de - Broglie relation. h  = * If an electron is accelerated through a potential difference of ‘V’ volt from rest then :  =   150  2 =   Å  V   = * mvr = n × 2m( q.V ) mv = 2emV = p (momentum) = h h 2m e (eV ) 2emV If we associate Bohr’s theory with de - Broglie equation then 2r = n or = 2r n 1 From de-Broglie equation = 1 2 2 m v multiplied by mass on both side 2 m (K.E.) = Equating both we get h.c  1 mv2 2 K.E. = h mv therefore h 2r = mv n nh V = so, mvr = 2 m0 m = dynamic mass = 1   v  Å (V in volt) h 2 h mv h nh mv =  putting this in mvr = 2 2 c m0 = rest mass of particle 12.3 (on putting values of h, me and e)  2r   h nh r = 2    n     de Broglie wavelength depended on velocity 53 Structure of Atom * Ex-9 What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s–1 (NCERT) Sol. According to de Broglie equation = (6.626  10 –34 Js) h = mv (0.1 kg) (10 m s –1 ) = 6.626 × 10–34 m (J = kg m2 s–2). HEISENBERG’S UNCERTAINTY PRINCIPLE The exact position and momentum of a fast moving particle cannot be calculated precisely at the same moment of time. If x is the error in the measurement of position of the particle and if p is the error in the measurement of momentum of the particle, then: x . p where, h  4 or h p.x  4 P h t . x  t 4 P ( t = rate of change in momentum = F) h F.x.t  4  E . t  h 4 E  uncertainty in energy t  uncertainty in time  In terms of uncertainty in energy E, and uncertainty in time t, this principle is written as, h x . (mv)  4 x = uncertainty in position p = uncertainty in momentum (multiplied & divided by t) E.t  h 4 .  Heisenberg replaced the concept of definite orbits by the concept of probability. h = Plank’s constant m = mass of the particle v = uncertainty in velocity Ex-10 A golf ball has a mass of 40 g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position. If the momentum of the particle is measured precisely. (NCERT) i.e. p  0 then x   . If the position of a particle is measured precisely, i.e. x  0 then p   . 2 This is because of a principle of optics that if a light Sol. The uncertainty in the speed is 2%, i.e., 45 × 100 of wavelength ‘’ is used to locate the position of a = 0.9 m s–1. particle then maximum error in the position Using the equation measurement will be   i.e. x =   If x  0 ;  But,  p= x 0 h   p   So, to make x  0 ,   0 a photon of very high energy is used to locate it.  When this photon will collide with the electron then momentum of electron will get changed by a large amount. = h 4 m v = 6.626  10 –34 4  3.14  40  10 – 3 (0.9 m s –1 ) = 1.46 × 10–33 m This is nearly ~ 1018 times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for large particles, the uncertainty principle sets no meaningful limit to the precision of measurements. 54 Structure of Atom ORBITAL An orbital may be defined as the region of space around the nucleus where the probability of finding an electron is maximum (90% to 95%) Orbitals do not define a definite path for the electron, rather they define only the probability of the electron being in various regions of space around the nucles. Difference between orbit and orbitals Orbit 1. It is well defined circular path followed by revolving electrons around the nucleus 2. It represents planar motion of electron Orbitals 1. It is the region of space around the nucleus where electron is most likely to be found 2. It represents 3 dimensional motion of an electron around the nucleus. 3. The maximum no. of electron in an 2 orbits is 2n where n stands for no. 3. Orbitals can not accomodate more than 2 electrons. of orbit. 4. Orbits are circular in shape. 4. Orbitals have different shape e.g. s-orbital is spherical, p - orbital is dumb- bell shaped. 5. Orbit are non directional in character. Hence, they cannot explain shape of molecules 6. Concept of well defined orbit is against Heisenberg’s uncentainty principle. 5. Orbitals (except s-orbital) have directional character. Hence, they can account for the shape of molecules. 6. Concept of orbitals is in accordance with Heisenberg’s principle 18. SHAPE OF THE ORBITALS Shape of the orbitals are related to the solutions of Schrodinger wave equation, and gives the space in which the probability of finding an electron is maximum. s- orbital : Shape  spherical s- orbital is non directional and it is closest to the nucleus, having lowest energy. s-orbital can accomodate maximum no. of two electrons. 55 Structure of Atom p-orbital : Shape  dumb bell Dumb bell shape consists of two lobes which are separated by a region of zero probability called node. d - orbital can accomodate maximum no. of 10 electrons. f - orbital : Shape  leaf like f - orbital can accomodate maximum no. of 14 electrons. p - orbital can accomodate maximum no. of six electrons. d - Orbital : Shape  double dumb bell QUANTUM NUMBERS The set of four numbers required to define an electron completely in an atom are called quantum numbers. The first three have been derived from Schrodinger wave equation. (i) Principal quantum number (n) : (Proposed by Bohr) It describes the size of the electron wave and the total energy of the electron. It has integral values 1, 2, 3, 4 ...., etc., and is denoted by K, L, M, N. ..., etc. * Number of subshell present in nth shell = n n subshell 1 s 2 s, p 3 s, p, d 4 s, p, d, f * Number of orbitals present in nth shell = n2 . * The maximum number of electrons which can be present in a principal energy shell is equal to 2n2. No energy shell in the atoms of known elements possesses more than 32 electrons. * nh Angular momentum of any orbit = 2 56 Structure of Atom (ii) Azimuthal quantum number () : (Proposed by Sommerfield) * It describes the shape of electron cloud and the number of subshells in a shell. * It can have values from 0 to (n – 1) * value of  subshell * 0 s 1 p 2 d 3 f Number of orbitals in a subshell = 2 + 1 eh Spin magnetic moment s = 2 mc = * n (n  2 ) or  B.M. (n = no. of unpaired electrons) It represents the value of spin angular momentum which is equal to * s( s  1) h 2 s( s  1) Maximum spin of atom = 1 x No. of unpaired 2 electron. ELECTRONIC CONFIGURATION * Maximum number of electrons in particular subshell Pauli’s exclusion principle : = 2 × (2 + 1) No two electrons in an atom can have the same set of all the four quantum numbers, i.e., an orbital cannot h * Orbital angular momentum L = 2 ( 1) have more than 2 electrons because three quantum numbers (principal, azimuthal and magnetic) at the h   most may be same but the fourth must be different, =   (  1)   2     i.e., spins must be in opposite directions. i.e.Orbital angularmomentum of s orbital = 0, Orbital Aufbau principle : h Aufbau is a German word meaning building up. The angular momentum of p orbital = 2 , 2 electrons are filled in various orbitals in order of their increasing energies. An orbital of lowest energy is h Orbital angular momentum of d orbital = 3 filled first. The sequence of orbitals in order of their 2 increasing energy is : (iii) Magnetic quantum number (m) : 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, (Proposed by Linde) 6p, 7s, 5f, 6d, .... It describes the orientations of the subshells. It can The energy of the orbitals is governed by (n + ) have values from – to +  including zero, i.e., total rule. (2 + 1) values. Each value corresponds to an orbital. s-subshell has one orbital, p-subshell three orbitals ‘n +  RULE (p x , p y and p z ), d-subshell five orbitals The relative order of energies of various sub-shell in a multi electron atom can be predicated with the (d xy , dyz , dzx , dx 2  y 2 , dz 2 ) and f-subshell has seven help of ‘n + ’ rule orbitals. The total number of orbitals present in a  The sub-shell with lower value of (n + ) has main energy level is ‘n2’. lower energy and it should be filled first. (iv) Spin quantum number (s) : eg. 3d 4s (Proposed by Samuel Goldsmit & Uhlenbeck) (n +) = 3 + 2 (n +) = 4 + 0 It describes the spin of the electron. It has values =5 =4 +1/2 and –1/2. signifies clockwise spinning and Since, (n + ) value of 3d is more than 4s therefore, anticlockwise spinning. 4s will be filled before 3d. 57 Structure of Atom  If two sub-shell has same value of (n + ) then the sub-shell with lower value of n has lower energy and it should be filled first. 3d eg. (n +) = 3 + 2 =5 4p Ex-11 Write the electronic configuration and find the no. of unpaired electrons as well as total spin for the following atoms : =4+1 (1) 6 C (2) 8 =5 (3) 15 P (4) 21 (5) 26 Fe (6) 10 3d is filled before 4p. MEMORY MAP O Sc Ne Sol. (1) 6C  1s2, 2s2, 2p2 No. of unpaired electrons  2. Total spin = 2 2 or 2 2 (2) 8O  1s2, 2s2, 2p4 No. of unpaired electrons = 2  Total spin = (3) 15 P 1s2, 2s2, 2p6, 3s2, 3p3 No. of unpaired electrons = 3  HUND’S RULE No electron pairing takes place in the orbitals in a sub - shell until each orbital is occupied by one electron with parallel spin. Exactly half filled and fully filled orbitals make the atoms more stable, Total spin = (4) 2 2 or 2 2 3 3 or 2 2 Sc  1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d1 21 or [Ar] 4s2 3d1 i.e., p3, p6, d5, d10, f7 and f14 configuration are most stable. [Ar] 3d1 4s2  No. of unpaired electrons = 1  Total spin = 1 1 or 2 2 58 Structure of Atom 2 2 6 2 6 2 6 2 6 (v) 26Fe  1s , 2s 2p , 3s , 3p , 4s 3d or [Ar] 4s , 3d EXCEPTIONS (1) Cr = [Ar] 4s2, 3d4 (Not correct) 24 [Ar] 4s1, 3d5 (correct : as d5 structure is more stable than d4 structure) No. of unpaired electrons = 4 Total spin =  4 4 or 2 2 (2) (correct : as d10 structure is more stable than d9 structure). (vi) 10Ne  1s2, 2s2 2p6 THE SCHRODINGER EQUATION No. of unpaired electrons = 0 Total spin = 0 Ex-12 Write down the four quantum numbers for fifth and sixth electrons of carbon atom. Sol. 6C : 1s2, 2s2 2p2 fifth electron : n = 2 s=+ 1 2 or – s=+ 1 2 or – =1 m = – 1 or +1 1 2 sixth electron : n=2 =1 m=0 1 2 Ex-13 Calculate total spin, magnetic moment for the atoms having at. no. 7, 24 and 36. Sol. The electronic configuration are 1s2, 2s2 2p3 unpaired electron = 3 7 N: : 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1 unpaired electron = 6 24Cr : 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6 unpaired electron = 0 36Kr  Total spin for an atom = ± 1/2 × no. of unpaired electron For 7N, it is = ± 3/2 ; For 36Kr, it is = 0 For 24Cr, it is = ± 3 ; Also magnetic moment = n (n  2) For 7N, it is = 15 ; For 36Kr, it is = 0 Cu = [Ar] 4s1, 3d10 29 For 24Cr, it is = 48 ; The de Broglie wave relation is the basis for predicting the behavior of freely moving particles. Shortly after it was proposed, Erwin Schrodinger demonstrated that the de Broglie expression could be generalized so as to apply to bound particles such as electrons in atoms. The heart of Schrodinger's theory is that the allowed energies of physical systems can be found by solving an equation which so resembles the equations of classical wave theory that it is called the wave equation. For the motion of one particle in one (the x) direction, the Schrodinger wave equation is – h2 2 d2 8 m dx 2 + V = E . The "knowns" in this equation are m, the mass of the particle, and V, its potential energy expressed as a function of x. The "unknowns" to be found by solving the equation are E, the quantized or allowed energies of the particle, and , which is called the wave function. The quantity d2/dx2 represents the rate of change of d/dx, the rate of change of . When this equation is applied to real systems such as the hydrogen atom, it is found that it cannot be solved unless E takes on certain values which are related by integers. Thus quantized energy and quantum numbers are an automatic consequence of the Schrodinger theory, and do not have to be tacked on to Newtonian mechanics as was done by Bohr. What is  ? By itself, it has no physical meaning. However, the square of the absolute value of , 2, does have an important physical interpretation. 59 Structure of Atom It is a mathematical expression of how the probability of finding a particle varies from place to place. Thus the exact trajectories of Newtonian mechanics and the Bohr theory do not appear in the results of the Schrodinger quantum mechanics; this, according to the Uncertainty Principle, is as it should be. A similar analysis is possible for the other 2pfunctions. The px function has the yz-plane as an angular node, since the function is proportional sin  cos  and cos  = 0 everywhere in the yz-plane. The maximum values of 1 for sin  and cos  occur along the positive x-axis. The py function, proportional to the sin  sin , vanishes in the xz-plane, where sin  = 0, and has a maximum along the positive y-axis, where both sin  and sin  are unity. The 3d-orbitals of the hydrogen atom. Note the relation between the labeling of the d-orbitals and their orientations in space. (Adapted from K. B. Harvey and G. B. Porter, An Introduction to Physical Inorganic Chemistry. Reading, Mass.: AddisonWesley, 1963.) FIG. The 2p-orbitals of the hydrogen atom. (Adapted from K. B. Harvey and G. B. Porter, An Introduction to Physical Inorganic Chemistry. Reading, Mass.: Addison-Wesley, 1963.) Radial probability density for some orbitais of the hydrogen atom. Ordinate is proportional to 4r2 R2, and all distributions are to the same scale. (i) total nodes = n – 1, (ii) angular nodes = , (iii) radial nodes = n –  – 1. 60 Structure of Atom 61