1. CHAPTER 3 ATOMIC STRUCTURE 3.1 Bohr’s Atomic Model 3.1 Bohr’s Atomic Model 3.2 Quantum Mechanical Model 3.2 Quantum Mechanical Model 3.3 Electronic Configuration 3.3 Electronic Configuration

2. Bohr’s Atomic Model At the end of this topic students should be able to:- 1) Describe the Bohr’s atomic postulates. Describe 2) Explain the existence of electron energy levels in an atom and calculate the energy of electron Explain 3) Differentiate between line spectrum and continuous spectrum. Differentiate 4) Perform calculation involving the Rydberg equation for Lyman, Balmer, Paschen, Brackett and Pfund series Perform calculation Perform calculation 5) Calculate the ionisation energy from Lyman series Calculate the ionisation energy Calculate the ionisation energy 6) Outline the weaknesses of Bohr’s atomic model. Outline the weaknesses Outline the weaknesses 7) State the de Broglie’s postulate and Heisenberg’s uncertainty principle State

3. In 1913, a young Dutch physicist, Niels Böhr proposed a theory of atom that shook the scientific world. The atomic model he described had electrons circling a central nucleus that contains positively charged protons. Böhr also proposed that these orbits can only occur at specifically “permitted” levels only according to the energy levels of the electron and explain successfully the lines in the hydrogen spectrum. BOHR’S ATOMIC MODELS

4. 1. Electron moves in circular orbits about the nucleus. In moving in the orbit, the electron does not radiate any energy and does not absorb any energy. Postulates H Nucleus (proton) H 1 1 BOHR’S ATOMIC MODELS

5. ii)The energy of an electron in a hydrogen atom is quantised, that is, the electron has only a fixed set of allowed orbits, called stationary states. n=1 n=2 n=3 H Nucleus (proton) [ orbit = stationary state = energy level = shell ] BOHR’S ATOMIC MODELS Postulates

6. 1. At ordinary conditions the electron is at the ground state (lowest level). If energy is supplied, electron absorbed the energy and is promoted from a lower energy level to a higher ones. (Electron is excited) 4.Electron at its excited states is unstable. It will fall back to lower energy level and released a specific amount of energy in the form of light. The energy of the photon equals the energy difference between levels. BOHR’S ATOMIC MODELS Postulates

7. Ground state the state in which the electrons have their lowest energy Excited state the state in which the electrons have shifted from a lower energy level to a higher energy level Energy level energy associated with a specific orbit or state Points to Remember

8. The energy of an electron in its level is given by: R H (Rydberg constant) or A = 2.18  10 -18 J. n (principal quantum number) = 1, 2, 3 ….  (integer) Note: n identifies the orbit of electron Energy is zero if electron is located infinitely far from nucleus Energy associated with forces of attraction are taken to be negative (thus, negative sign) THE BOHR ATOM

9. Radiant energy emitted when the electron moves from higher-energy state to lower-energy state is given by the difference in energy between energy levels: THE BOHR ATOM  E = E f - E i where Thus,

10. The amount of energy released by the electron is called a photon of energy. The amount of energy released by the electron is called a photon of energy. A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength. A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength. where; where; h (Planck’s constant) =6.63  10 -34 J s h (Planck’s constant) =6.63  10 -34 J s = frequency = frequency THE BOHR ATOM  E = h Where; c (speed of light) = 3.00  10 8 ms -1 Thus,

11. n =1n = 2n = 3n = 4 Electron is excited from lower to higher energy level. A specific amount of energy is absorbed  E = h = E 1 -E 3 (+ve) Electron falls from higher to lower energy level. A photon of energy is released.  E = h = E 3 -E 1 (-ve)

12. Energy level diagram for the hydrogen atom Potential energy n = 1 n = 2 n = 3 n = 4 n =  Energy released Energy absorbed

13. Exercises: 1) Calculate the energy of an electron in the second energy level of a hydrogen atom. (-5.448 x 10 -19 J) 1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom. 3) Calculate the energy change (J), that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom. (answer: 1.55 x 10 -19 J) (answer: 1.55 x 10 -19 J) 1) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3.

14. Emission Spectra Emission Spectra Continuous Spectra Line Spectra

15. Continuous Spectrum A spectrum consists all wavelength components (containing an unbroken sequence of frequencies) of the visible portion of the electromagnetic spectrum are present. It is produced by incandescent solids, liquids, and compressed gases.

16. Regions of the Electromagnetic Spectrum

17. When white light from incandescent lamp is passed through a slit then a prism, it separates into a spectrum. When white light from incandescent lamp is passed through a slit then a prism, it separates into a spectrum. The white light spread out into a rainbow of colours produces a continuous spectrum. The white light spread out into a rainbow of colours produces a continuous spectrum. The spectrum is continuous in that all wavelengths are presents and each colour merges into the next without a break. The spectrum is continuous in that all wavelengths are presents and each colour merges into the next without a break. FORMATION OF CONTINUOUS SPECTRUM

18. Line Spectrum (atomic spectrum) A spectrum consists of discontinuous & discrete lines produced by excited atoms and ions as the electrons fall back to a lower energy level. The radiation emitted is only at a specific wavelength or frequency. It means each line corresponds to a specific wavelength or frequency. Line spectrum are composed of only a few wavelengths giving a series of discrete line separated by blank areas

19. prism film The emitted light (photons) is then separated into its components by a prism. Each component is focused at a definite position, according to its wavelength and forms as an image on the photographic plate. The images are called spectral lines. FORMATION OF ATOMIC / LINE SPECTRUM

20. n = 1 n = 2 n = 3 n = 4 n = 5 n =  Energy When an electrical discharge is passed through a sample of hydrogen gas at low pressure, hydrogen molecules decompose to form hydrogen atoms. Radiant energy (a quantum of energy) absorbed by the atom (or electron) causes the electron to move from a lower-energy state to a higher-energy state. Hydrogen atom is said to be at excited state (very unstable).

21. FORMATION OF ATOMIC / LINE SPECTRUM Emission of photon n = 2 n = 3 n = 4 n = 5 n = 6 n =  Energy When the electrons fall back to lower energy levels, radiant energies (photons) are emitted in the form of light (electromagnetic radiation of a particular frequency or wavelength)

22. FORMATION OF ATOMIC / LINE SPECTRUM n = 1 n = 2 n = 3 n = 4 n = 5 n =  Lyman Series Emission of photon Line spectrum E Energy

23. FORMATION OF ATOMIC / LINE SPECTRUM n = 1 n = 2 n = 3 n = 4 n = 5 n =  Lyman Series Emission of photon Line spectrum Balmer Series E Energy

24. Emission series of hydrogen atom n = 1 n = 2 n = 3 n = 4 n =  Lyman series Balmer series Brackett series Paschen series Pfund series

25. Exercise: Complete the following table Seriesn1n1 n2n2 Spectrum region Lyman2,3,4,… 23,4,5,… Paschen4,5,6,…Infrared 45,6,7,…Infrared 56,7,8,…Infrared ultraviolet Visible Balmer Brackett Pfund 1 3

26. The following diagram depicts the line spectrum of hydrogen atom. Line A is the first line of the Lyman series. Exercise Line spectrum E Specify the increasing order of the radiant energy, frequency and wavelength of the emitted photon. Which of the line that corresponds to i) the shortest wavelength? i) the shortest wavelength? ii) the lowest frequency? ii) the lowest frequency? ABCDE Line E Line A

27. Describe the transitions of electrons that lead to the lines W, and Y, respectively. Solution Line spectrum WY Exercise Balmer series For W: transition of electron is from n=4 to n=2 For Y: electron shifts from n=7 to n=2

28. Homework Calculate E n for n = 1, 2, 3, and 4. Make a one- dimensional graph showing energy, at different values of n, increasing vertically. On this graph, indicate by vertical arrows transitions that lead to lines in a) Lyman series b) Paschen series

29. In Lyman series, the frequency of the convergence of spectral lines can be used to find the ionisation energy of hydrogen atom: In Lyman series, the frequency of the convergence of spectral lines can be used to find the ionisation energy of hydrogen atom: IE = h  IE = h  The frequency of the first line of the Lyman series > the frequency of the first line of the Balmer series. The frequency of the first line of the Lyman series > the frequency of the first line of the Balmer series. Significance of Atomic Spectra Lyman Series Line spectrum E Balmer Series 

30. Line spectrum ABCDE Exercise Paschen series Solution Which of the line in the Paschen series corresponds to the longest wavelength of photon? Describe the transition that gives rise to the line. Line A. The electron moves from n=4 to n=3.

31. Calculate a) the wavelength in nm b) the frequency c) the energy that associated with the second line in the Balmer series of the hydrogen spectrum. Exercise Solution (a) Second line of Balmer series: the transition of electron is from n 2 =4 to n 1 =2 = RHRHRHRH11 n12n12n12n12 n22n22n22n22 1 = (1.097x10 7 m  1 ) 11 22222222 42424242 1 = x 1 10 9 m nm = 486 nm 4.86x10  7 m

32. Wavelength emitted by the transition of electron between two energy levels is calculated using Rydberg equation: Wavelength emitted by the transition of electron between two energy levels is calculated using Rydberg equation: Rydberg Equation R H = 1.097  10 7 m -1 = wavelength Since should have a positive value thus n 1 < n 2 where          2 2 2 1 H n 1 n 1 R λ 1

33. Calculate the wavelength, in nanometers of the spectrum of hydrogen corresponding to ni = 2 and nf = 4 in the Rydberg equation. Example Solution: Rydberg equation: 1/λ = RH (1/ni2 – 1/nf2) ni = 2 nf = 4 RH = 1.097 x 10m7 1/λ = RH (1/22 – 1/42) = RH(1/4-1/16) λ = 4.86m x 102 m = 486nm

34. Use the rydberg equation to calculate the wavelength of the spectral line of hydrogen atom that would result when an electron drops from the fourth orbit to the second orbit, then identified the series the line would be found. Example Solution: 1/λ = RH (1/n1 2 – 1/n2 2) n1 = 2 n2 = 4 1/λ = 1.097 x 107 (1/22 – 1/42) λ = 4.86 x 10-7 m = 486 nm *e dropped to the second orbit (n=2), >>> Balmer series

35. EXAMPLE 3 Calculate the wavelengths of the fourth line in the Balmer series of hydrogen. n 1 = 2 n 2 = 6 R H = 1.097 x 10 7 m -1 λ = 4.10 x 10 -7 m RH RH 2 6262 111 = λ

36. Different values of R H and its usage Different values of R H and its usage 1. 1. R H = 1.097  10 7 m -1 RH RH n2n2 1 n2n2 2 111 = λ R H = 2.18 x 10 -18 J n 1 < n 2

37. EXAMPLE 4 Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom. ΔE = 4.58 x 10 -19 J ΔE = (6.63  10 -34 Js)X(3.00  10 8 ms -1 ) RH RH n2n2 1 n2n2 2 111 = λ 5252 1.097 x 10 7 2 111 = λ 1 λ = 0.2303 X 10 7 m -1 X (0.2303 X 10 7 m -1 )

38. Calculate what is; i ) Wavelength ii ) Frequency iii ) Wave number of the last line of of the last line of hydrogen spectrum hydrogen spectrum in Lyman series in Lyman series Wave number = 1/wavelength Wave number = 1/wavelength EXERCISE For Lyman series; n 1 = 1 & n 2 = ∞ Ans: i. 9.116 x10 -8 m ii. 3.29 x10 15 s -1 iii. 1.0970 X 10 7 m -1

39. i) 1.097 x 10 7 1212 111 = λ ∞ 2 = 1.097 X 10 7 m -1 λ = 9.116 X 10 -8 m ii) V = c λ =3.00 X 10 8 ms -1 9.116 X 10 -8 m = 3.29 X 10 15 s -1 iii) Wave number = 1 = 1.097 X 10 7 m -1 λ SOLUTION

40. Defination : Ionization energy is the Defination : Ionization energy is the minimum energy required to remove one mole of electron from one mole of gaseous atom/ion. M (g)  M + (g) + e  H = +ve M (g)  M + (g) + e  H = +ve The hydrogen atom is said to be ionised when electron is removed from its ground state (n = 1) to n = . The hydrogen atom is said to be ionised when electron is removed from its ground state (n = 1) to n = . At n = , the potential energy of electron is zero, here the nucleus attractive force has no effect on the electron (electron is free from nucleus). At n = , the potential energy of electron is zero, here the nucleus attractive force has no effect on the electron (electron is free from nucleus). Ionization Energy

41. n 1 = 1, n 2 = ∞ ∆E = R H (1/n 1 2 – 1/n 2 2 ) ∆E = R H (1/n 1 2 – 1/n 2 2 ) = 2.18 X 10 -18 (1/1 2 – 1/ ∞ 2 ) = 2.18 X 10 -18 (1/1 2 – 1/ ∞ 2 ) = 2.18 X 10 -18 (1 – 0) = 2.18 X 10 -18 (1 – 0) = 2.18 X 10 -18 J = 2.18 X 10 -18 J Ionisation energy Ionisation energy = 2.18 X 10 -18 x 6.02 X 10 23 J mol -1 =1.312 x 10 6 J mol -1 = 2.18 X 10 -18 x 6.02 X 10 23 J mol -1 =1.312 x 10 6 J mol -1 = 1312 kJ mol -1 = 1312 kJ mol -1 Example

42. 1 1 st line Convergent limit Finding ionisation energy experimentally:  Ionisation energy is determined by detecting the wavelength of the convergence point

43. 10.97 10.66 10.52 10.27 9.74 8.22 10.97 10.66 10.52 10.27 9.74 8.22 wave number (x10 6 m -1 ) The Lyman series of the spectrum of hydrogen is shown above. Calculate the ionisation energy of hydrogen from the spectrum. Example

44. ΔE = hc/ λ =h x c / λ = h x c x wave no. = 6.626 x 10 -34 J s x 3 x 10 8 m s -1 x 10.97x 10 6 m - 1 = 218.06 x 10 -20 J = 2.18 x 10 -18 J Ionisation energy = 2.18 X 10 -18 x 6.02 X 10 23 J mol -1 = 2.18 X 10 -18 x 6.02 X 10 23 J mol -1 =1.312 x 10 6 J mol -1 =1.312 x 10 6 J mol -1 = 1312 kJ mol -1 = 1312 kJ mol -1 Solution

45. Compute the ionisation energy of hydrogen atom in kJ mol  1. Exercise Solution n 1 = 1 n 2 = 1= RHRHRHRH 1 n12n12n12n12 n22n22n22n22 EEEE = 2.18x10  18 J 11 12121212 2222 = 6.02x10 23 1 atom mol J kJ x x1000 1 = 1312 kJ mol  1 1 atom

46. The weakness of Bohr’s Theory Bohr was successful in introducing the idea of quantum energy and in explaining the lines of hydrogen spectrum. Bohr was successful in introducing the idea of quantum energy and in explaining the lines of hydrogen spectrum. His theory could not be extended to predict the energy levels and spectra of atoms and ions with more than one electron. His theory could not be extended to predict the energy levels and spectra of atoms and ions with more than one electron. His theory can only explain the hydrogen spectrum or ions contain one electron eg He +, Li 2+. His theory can only explain the hydrogen spectrum or ions contain one electron eg He +, Li 2+. Modern quantum mechanics retain Bohr’s concept of discrete energy states and energy involved during transition of electrons but totally reject the circular orbits he introduced. Modern quantum mechanics retain Bohr’s concept of discrete energy states and energy involved during transition of electrons but totally reject the circular orbits he introduced.

47. Davisson & Germer observed the diffraction of electrons when a beam of electrons was directed at a nickel crystal. Diffraction patterns produced by scattering electrons from crystals are very similar to those produced by scattering X-rays from crystals. This experiment demonstrated that electrons do indeed possess wavelike properties. Thus, can the ‘position’ of a wave be specified??? Point to Ponder

48. de Broglie’s Postulate In 1924 Louis de Broglie proposed that not only light but all matter has a dual nature and possesses both wave and corpuscular properties. De Broglie deduced that the particle and wave properties are related by the expression: h = Planck constant (J s) m = particle mass (kg)  = velocity (m/s) = wavelength of a matter wave = wavelength of a matter wave = h mm

49. Heisenberg’s Uncertainty Principle It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain. Stated mathematically, where  x = uncertainty in measuring the position  p = uncertainty in measuring the momentum  p = uncertainty in measuring the momentum =  mv =  mv h = Planck constant h = Planck constant h 44  x  p 