Ce diaporama a bien été signalé.
Nous utilisons votre profil LinkedIn et vos données d’activité pour vous proposer des publicités personnalisées et pertinentes. Vous pouvez changer vos préférences de publicités à tout moment.
Prochain SlideShare
What to Upload to SlideShare
Suivant

Partager

Atomic structure

atomic structure presentation

Livres associés

Gratuit avec un essai de 30 jours de Scribd

Tout voir

Atomic structure

  1. 1. ATOMIC STRUCTURE Chemistry
  2. 2. Session Objectives 1. Dalton’s theory 2. Discovery of fundamental particles 3. Thomson’s model of an atom 4. Rutherford’s model 5. Concept of atomic number and mass number 6. Drawback of Rutherford’s model 7. Electromagnetic waves 8. Planck’s quantum theory 9. Bohr’s model
  3. 3. • All matter is composed of atoms. • All atoms of a given element are identical. Atom → Can not be cut. Indivisible and indestructible Dalton’s Theory – Atom Is Fundamental Particle Pre 1897
  4. 4. Cathode rays
  5. 5. Properties of Cathode Rays They are deflected from their path by electric and magnetic fields
  6. 6. Properties of cathode rays They are material particles as they produce mechanical motion in a small paddle wheel e/m ratio of cathode rays :1.758820 x 10 11 C kg -1
  7. 7. Anode rays + – Z n S C o a t in g P e r fo r a te d c a t h o d e H g a s in s id e a t lo w p r e s s u r e 2 Unlike cathode rays, the charge to mass ratio of canal rays on the nature of gas taken in the discharge tube. It is maximum when lightest gas hydrogen is taken which is equal to 9.58 x 107 Ckg-1
  8. 8. Properties of anode Rays They travel in straight line They are deflected by electric and magnetic field The nature of anode rays depends upon the nature of gas e/m ratio for anode rays is not constant It is maximum when lightest gas hydrogen is taken which is equal to 9.58 x 107 Ckg-1
  9. 9. Sub – atomic particles Subatomic particles Symbol Unit charge Unit mass Charge in Coulomb Mass in amu Proton Neutron Electron Negligible 1 1H 1 0n 0 1e− -19 +1.60 × 10 -19 +1.60 × 10 -4 5.489×10 1.008665 1.007825 -1 0 +1 1 1 0
  10. 10. Electron Positive sphere Thomson’s Model of an Atom
  11. 11. Rutherford experiment
  12. 12. Rutherford’s Experiment - Results A beam of α particles aimed at thin gold foil. • Most of the particles passed through. Most of the space is empty • A few came back Presence of concentrated mass at the centre • Others deflected at various angles Repulsion between two +vely charged particles “ Like firing shells at paper handkerchief with few of them coming back.” - Ernst Rutherford
  13. 13. Rutherford’s Model Atom consist of two parts: (a)Nucleus:Almost the whole mass of the atom is concentrated in this small region (b)Extra nuclear part:this is the space around the nucleus in which electrons are revolving at high speeds in fixed path
  14. 14. Concept of atomic mass and atomic number Atomic number(Z)=number of protons Mass number(A)=number of protons+number of neutrons Entire mass of the atom is concentrated at the centre
  15. 15. Concept of atomic number and mass number aN23 11 For example: Mass number=number of protons+number of neutrons =23 Atomic number =number of protons =11
  16. 16. Concept of atomic number and mass numberWe express weight of an atom in terms of atomic mass unit (a.m.u). Mass of a proton=Mass of neutron =1 a.m.u(approx) ∴ Mass number=Atomic weight (expressed in a.m.u)
  17. 17. Drawback of Rutherford’s model
  18. 18. Atom thus collapses. Drawback of Rutherford’s model
  19. 19. Electromagnetic waves  Light is an oscillating electro-magnetic field.  Oscillating electric field generates the magnetic field and vice-versa. Electric and magnetic fields are perpendicular to each other
  20. 20. Direction of propagation Electric field component Magnetic field component X Y Z Electric and magnetic fields are perpendicular to each other Electromagnetic waves
  21. 21. . λ (i) Wavelength: It is represented by Units: m, cm(10-2 m), nm(10-9 m), pm(10-12 m) or A0 (10-10 m). Characteristics of a wave λ Direction of Propogation of wave
  22. 22. (iv) Wave number: The number of waves present in 1 cm length. It is represented by . Its unit is cm-1 . 1 λ (iii) Velocity: The linear distance travelled by a crest or a trough in one second. Its unit is cm s-1 . (ii) Frequency: The number of waves passes through a given point in 1 second. It is represented by .ν Its unit is Hertz or second-1 . Characteristics of a wave
  23. 23. Electromagnetic spectrum
  24. 24. Radio city broadcasts on a frequency of 5,090 KHz.What is the wavelength of electromagnetic radiation emitted by the transmitter? Illustrative problem 1 c λ = ν 8 3 3 10 5090 10 × λ = × 2 0.589 10= × 58.9 m= s/m103isc 8 ×
  25. 25. Radiant energy is emitted or absorbed discontinuously in the form of quanta. Planck’s quantum theory : 34 c E h h hc Wavelength Frequency Wave number h Plank 's cons tant 6.626 10 Js− = ν = = ν λ λ = ν = ν = = = ×
  26. 26. Questions
  27. 27. The ratio of the energy of a photon of 2000 wavelength radiation to that of 6000 radiation is (a) ¼ (b) 4 (b)½ (d) 3 0 A 0 A Illustrative Problem 2 hc E h= ν = λ 1 2 1 2 hc hc E E= = λ λ 1 2 2 1 E E λ ∴ = λ 0 0 6000A 3 2000A = = Hence, answer is (d). Solution:
  28. 28. Bohr’s model Positively charged nucleus Negatively charged electrons Stationary Orbit + hν –hν
  29. 29. Bohr’s Postulates Retained key features of Rutherford’s model. Concept of stationary circular orbits. Quantization of angular momentum.nh mvr 2 = π Energy emitted/absorbed when electrons jump from one orbit to another. f iE E E h ∆ = − = ∆ν
  30. 30. Bohr’s model Bohr’s postulates energy of electron radius of various orbits velocity of electron
  31. 31. + r Nucleus electron 2 a 2 kZe F r = (i) Calculation of radius of Bohr orbit According to coulomb’s law
  32. 32. centrifugal force 2 c mu F r = (ii) F Fa c= 2 2 2 kZe mu r r ∴ = 2 2 kZe u mr = Calculation of radius of Bohr orbit nh mur 2 = π Bohr’s postulate nh u 2 mr = π 2 2 2 2 2 2 n h u 4 m r = π
  33. 33. For hydrogen Z=1 = × 2 0n r 0.529 A Z 2 2 kZe u mr = 2 2 2 2 2 2 n h u 4 m r = π 222 222 rm4 hn mr kZe π = 22 22 mkZe4 hn r π = For n=1,Z=1 , k = 9 109× Nm2 /C2 Calculation of radius of Bohr orbit
  34. 34. u is the velocity with which the electron revolves in an orbit nh mur 2 = π (1) 2 2 2 kZe mu rr = (2) Dividing (1) by (2),we get: Calculation of velocity of electron nh kZe2 u 2 π = u is in m/s
  35. 35. Number of revolutions per second velocity of electron circumference of an orbit = Calculation of number of revolutions
  36. 36. Total energy(T.E) P.E K.E= + 21 K.E mu 2 = 2 kZe P.E r = − 2 2 2 1 kZe T.E mu 2 r = − Calculation of energy of an electron
  37. 37. We know that 2 2 2 mu kZe r r = 2 2 kZe mu r = 2 2 kZe kZe E 2r r = − 2 kZe 2r = − Substituting the value of r we get 22 2422 n hn mkeZ2 E π −= P.E. = 2K.E. K.E. = -Total energy
  38. 38. Bohr’s model Bohr’s postulates 8 n Z v 2.18 10 cm /sec n = × 2 0 n n r 0.529 A Z = × 2 n 2 2 19 2 Z E 13.6 eVper atom n Z 21.8 10 J per atom n − = − × = − × ×
  39. 39. Quantum Mechanical Model of an Atom Schrodinger’s Equation: ∂2 /∂xψ 2  + ∂2 /∂yψ 2  + ∂2 /∂zψ 2  + 8π2 m/h2  ( E - v )  =ψ o  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 electron; h = Planck’s constant and ψ (psi) = wave function of the electron. Significance of ψ2 : ψ2 is a probability factor. It describes the probability of finding an electron within a small space. The space in which there is maximum probability of finding an electron is termed as orbital. The important point of the solution of the wave equation is that it provides a set of numbers called quantum numbers which describe energies of the electron in atoms, information about the shapes and orientations of the most probable distribution of electrons around nucleus.
  40. 40. Nodal point:   The point where there is zero probability of finding the electron is called nodal point. No. of radial nodes = n – l – 1 No. of angular nodes = l Total number of nodes = n – 1 Nodal plane: Nodal planes are the planes of zero probability of finding the electron. The number of such planes is also equal to l. 
  41. 41. λ = h/mv ……. De – Broglie’s Equation   Where mv = p, momentum of the particle   De- Broglie’s hypothesis Dual nature of Particles (PARTICLE AND WAVE CHARACTER OF MATTER AND RADIATION) E = hυ (According to the Planck’s quantum theory) E = mc2 (according to Einstein’s equation)
  42. 42. HEISENBERG’S UNCERTAINTY PRINCIPLE “It is impossible to measure simultaneously the position and momentum of a small microscopic moving particle with absolute accuracy or certainty”
  43. 43. 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): • main shell in which the electron resides • also tells the maximum number of electrons that a shell can accommodate is 2n2 , where n is the principal quantum number
  44. 44. Azimuthal (or) angular momentum quantum number (l): represents the number of subshells present in the main shell The orbital angular momentum of the electron is given as √l (l +1) h/2π or √l (l+1) nh/2Π for a particular value of ‘n’ (where h = Planck’s constant). For a given value of n values of possible l vary from 0 to n – 1
  45. 45. Magnetic Quantum Number (m): The magnetic quantum number determines the number of preferred orientations of the electron present in a subshell. m can assume all integral values between –l to +l including zero. Thus m can be –1, 0, +1 for l = 1. Total values of m associated with a particular value of l is given by 2l+ 1.
  46. 46. Spin Quantum Number (s): 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 .↑ ↓ It helps to explain the magnetic properties of the substances.
  47. 47. Shapes of Orbitals S- and p- orbitals S- orbital – spherical in shape P- orbital – Dumb-bell in shape
  48. 48. Shape of d- orbitals d-orbitals – Double dumb-bell shape
  49. 49. Questions
  50. 50. Illustrative Problem 3 The energy of the electron in the second and third Bohr orbits of the hydrogen atom is -5.42 X 10-12 and –2.41 X 10-12 respectively. Calculate the wavelength of the emitted radiation, when the electron drops from third to second orbit.
  51. 51. Solution 2 1E E E∆ = − According to Planck’s quantum theory 12 12 5.42 10 ( 2.41 10 )ergs− − = − × − − × 12 3.01 10 ergs.− = − × E h= υ c E h= λ 27 h 6.6 10 ergs− = ×
  52. 52. Solution 5 6.6 10 cm− λ = × 3 0 6.6 10 A= × 0 8 1A 10 cm− = 27 8 12 6.62 10 3 10 3.01 10 − − × × × = × c h E ∴ =λ
  53. 53. Class Test
  54. 54. Class Exercise - 1 Which of the following fundamental particles are present in the nucleus of an atom? (a) Alpha particles and protons (b) Protons and neutrons (c) Protons and electrons (d) Electrons, protons and neutrons Solution The nucleus of an atom is positively charged and almost the entire mass of the atom is concentrated in it. Hence, it contains protons and neutrons. Hence, answer is (b).
  55. 55. Class Exercise - 2 The mass of the proton is (a) 1.672 × 10–24 g (b) 1.672 × 10–25 g (c) 1.672 × 1025 g (d) 1.672 × 1026 g Solution Hence, answer is (a). The mass of the proton is 1.672 × 10–24 g
  56. 56. Class Exercise -3 Which of the following is not true in case of an electron? (a) It is a fundamental particle (b) It has wave nature (c) Its motion is affected by magnetic field (d) It emits energy while moving in orbits Solution Hence, answer is (d). An electron does not emit energy while moving in orbit. This is so because if it would have done that it would have eventually fallen into the nucleus and the atom would have collapsed.
  57. 57. Class Exercise - 4 Positive charge of an atom is (a) concentrated in the nucleus (b) revolves around the nucleus (c) scattered all over the atom (d) None of these Solution Hence, answer is (a). Positive charge of an atom is present entirely in the nucleus.
  58. 58. Class Exercise - 6 Why only very few a-particles are deflected back on hitting a thin gold foil? Solution Due to the presence of a very small centre in which the entire mass is concentrated.
  59. 59. Class Exercise - 5 Calculate and compare the energies of two radiations which have wavelengths 6000Å and 4000Å (h = 6.6 x 10-34 J s, c = 3 x 108 m s-1 ) Solution − − − × × × × × = = λ × 34 8 1 1 7 hc 6.6 10 Js 3 10 ms E 6 10 m = 3.3 x 10-19 J − − − × × × = × 34 8 1 2 7 6.6 10 Js 3 10 ms E 4 10 m = 4.9 x 10-19 J 19 1 19 2 E 3.3 10 J E 4.95 10 J − − × = × = 0.666 : 1
  60. 60. Class Exercise - 7 Explain why cathode rays are produced only when the pressure in the discharge tube is very low. Solution This is happened because at higher pressure no electric current flows through the tube as gases are poor conductor of electricity.
  61. 61. Class Exercise - 8 If a neutron is introduced into the nucleus of an atom, it would result in the change of (a) number of electrons (b) atomic number (c) atomic weight (d) chemical nature of the atom Solution Hence, answer is (c). Neutrons contribute in a major way to the weight of the nucleus, thus addition of neutron would result in increase in the atomic weight.
  62. 62. Class Exercise - 9 The concept of stationary orbits lies in the fact that (a) Electrons are stationary (b) No change in energy takes place in stationary orbit (c) Electrons gain kinetic energy (d) Energy goes on increasing Solution Hence, answer is (c). When an electron revolves in a stationary orbit, no energy change takes place. Energy is emitted or absorbed only when the electron jumps from one stationary orbit to another.
  63. 63. Class Exercise - 10 What is the energy possessed by 1 mole of photons of radiations of frequency 10 × 1014 Hz? Solution E = h ν E = 6.6 × 10–34 × 10 × 1014 E = 66 × 10–20 = 6.6 × 10–19 joules ∴ energy of 1 mole of photons = 6.6 × 10–19 × 6.023 × 1023 = 39.7518 × 104 = 397.518 kJ/mol
  64. 64. Class test 1.The radius of hydrogen atom in ground state is 5.3x10–11 m. It will have a radius of 4.77A after colliding with an electron. The principal quantum number of the atom in the excited state is (a) 2 (b) 4 (c)3 (d)5 2 n o 2 10 11 2 n Since r r x Z n 4.77 10 5.3 10 (for hydrogen atom) 1 n 9 n 3 − − = × = × × = = Solution Hence, answer is (c).
  65. 65. Classification of Elements
  • BinduMallikarjun

    Sep. 22, 2021
  • JanakiRamayya4

    Dec. 18, 2020
  • ShereenCqc

    Oct. 18, 2020
  • Vkgh1

    Sep. 23, 2020
  • SasmitaPatro5

    Sep. 20, 2020
  • nagaa786

    Aug. 30, 2020
  • KathirKing2

    Aug. 17, 2020
  • KathirKing2

    Aug. 17, 2020
  • KathirKing2

    Aug. 17, 2020
  • KathirKing2

    Aug. 17, 2020
  • KathirKing2

    Aug. 17, 2020
  • patelheemanshu

    Jul. 16, 2020
  • mlkkalim

    Jun. 21, 2020
  • MiraParmekar

    Jun. 4, 2020
  • dhnalingam

    Jun. 3, 2020
  • sravani1014

    May. 19, 2020
  • RashiGupta172

    Apr. 1, 2020
  • BhagyashreePriyadarshini1

    Jan. 4, 2020
  • Thongz

    Nov. 23, 2019
  • ISHWARSHARMA19

    Aug. 10, 2018

atomic structure presentation

Vues

Nombre de vues

959

Sur Slideshare

0

À partir des intégrations

0

Nombre d'intégrations

22

Actions

Téléchargements

0

Partages

0

Commentaires

0

Mentions J'aime

21

×