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CHAPTER 2 ATOMIC STRUCTURE. 2.1 Bohr’s Atomic Model 2.2 Quantum Mechanical Model 2.3 Electronic Configuration. LECTURE 1. 2.1 Bohr’s Atomic Model. Bohr’s Atomic Model. At the end of this topic students should be able to:- Describe the Bohr’s atomic models.

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## CHAPTER 2 ATOMIC STRUCTURE

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**CHAPTER 2ATOMIC STRUCTURE**2.1 Bohr’s Atomic Model 2.2 Quantum Mechanical Model 2.3 Electronic Configuration**LECTURE 1**2.1 Bohr’s Atomic Model**Bohr’s Atomic Model**At the end of this topic students should be able to:- Describe the Bohr’s atomic models. Explain the existence of electron energy levels in an atom. Calculate the energy of electron using : En = - RH (1/n2) , RH = 2.18 x 10-18 J**BOHR’S ATOMIC MODELS**In 1913, a young Dutch physicist, Niels Bohr 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. Bohr also proposed that these orbits can only occur at specifically “permitted” levelsonly according to the energy levels of the electron and explain successfully the lines in the hydrogen spectrum.**H Nucleus (proton)**1 H 1 BOHR’S ATOMIC postulateS 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.**BOHR’S ATOMIC postulateS**2. 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. [ orbit = stationary state = energy level = shell ] n=1 H Nucleus (proton) n=3 n=2**BOHR’S ATOMIC POSTULATES**3. 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**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. 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. [ orbit = stationary state = energy level = shell ]**Energy levels in an atom**• 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**THE energy level**• The energy of an electron in its level is given by: RH (Rydberg constant) or A = 2.1810-18J. 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)**LECTURE 2**At the end of this topic students should be able to:- d) Describe the formation of line spectrum of hydrogen atom e) Calculate the energy change of an electron during transition. E = RH (1/n12 - 1/n22) , where RH = 2.18 x 10-18 J f) Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition E = h where = c/λ**Emission Spectra**Emission Spectra Line Spectra Continuous Spectra**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.**FORMATION OF CONTINUOUS 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 spectrum is continuous in that all wavelengths are presents and each colour merges into the next without a break.**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**FORMATION OF ATOMIC / LINE SPECTRUM**film prism 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**n = 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). n = 5 n = 4 Energy n = 3 n = 2 n = 1 When an electrical discharge is passed through a sample of hydrogen gas at low pressure, hydrogen molecules decompose to form hydrogen atoms.**FORMATION OF ATOMIC / LINE SPECTRUM**n = n = 6 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) n = 5 Energy n = 4 n = 3 n = 2 Emission of photon**FORMATION OF ATOMIC / LINE SPECTRUM**n = n = 5 n = 4 n = 3 Energy n = 2 n = 1 Emission of photon Line spectrum E Lyman Series**FORMATION OF ATOMIC / LINE SPECTRUM**n = n = 5 n = 4 n = 3 Energy n = 2 n = 1 Emission of photon Line spectrum E Lyman Series Balmer Series**Emission series of hydrogen atom**n = Pfund series n = 4 Brackett series n = 3 Paschen series n = 2 Balmerseries Lyman series n = 1**E** Exercise The following diagram depicts the line spectrum of hydrogen atom. Line A is the first line of the Lyman series. A B C D E Line spectrum 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? ii) the lowest frequency?**Exercise**W Y Line spectrum Describe the transitions of electrons that lead to the lines W, and Y, respectively. Balmer series Solution**Homework**Calculate En 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 • Lyman series • Paschen series**Significance of Atomic Spectra**• In Lyman series, the frequency of the convergence of spectral lines can be used to find the ionisation energy of hydrogen atom: IE = h • The frequency of the first line of the Lyman series > the frequency of the first line of the Balmer series. Line spectrum E Balmer Series Lyman Series**Exercise**E D C B A Line spectrum Paschen series Which of the line in the Paschen series corresponds to the longest wavelength of photon? Describe the transition that gives rise to the line. Solution**Energy calculation**• 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: where E = Ef - Ei Thus,**Energy calculation**• 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. where; h (Planck’s constant) =6.63 10-34 J s = frequency E = h Where; c (speed of light) = 3.00108 ms-1 Thus,**Electron is excited from lower to higher**energy level. A specific amount of energy is absorbed E = h = E1-E3 (+ve) n =1 n = 2 n = 3 n = 4 Electron falls from higher to lower energy level . A photon of energy is released. E = h = E3-E1 (-ve)**Energy level diagram for the hydrogen atom**n = n = 4 n = 3 Potential energy n = 2 Energy absorbed Energy released n = 1**Exercises:**• Calculate the energy of an electron in the second energy level of a hydrogen atom. (-5.448 x 10-19 J) • 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 10-19J) • Calculate the frequency and wavelength (nm) of the radiation emitted in question 3.**LECTURE 3**At the end of this topic students should be able to:- g) Perform calculations involving the Rydberg equation for Lyman, Balmer, Paschen, Brackett and Pfund series . 1/ λ = RH (1/n12 - 1/n22) , where RH = 1.097 x 107 m-1 and n1<n2 • Calculate the ionisation energy of hydrogen atom from Lyman series. • State the weaknesses of Bohr’s atomic model. • State the dual nature of electron using the Broglie’s postulate and Hesseinberg’s uncertainty principle.**æ**ö 1 1 1 ç = - R ç H 2 2 λ n n è ø 1 2 Rydberg Equation • Wavelength emitted by the transition of electron between two energy levels is calculated using Rydberg equation: where • RH = 1.097 107 m-1 • = wavelength Since should have a positive value thus n1 < n2**Example**Calculate the wavelength, in nanometers of the spectrum of hydrogen corresponding to ni = 2 and nf = 4 in the Rydberg equation. 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**Example**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. Solution: 1/λ = RH (1/n12 – 1/n22) 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**1**1 1 RH = 22 62 λ EXAMPLE 3 Calculate the wavelengths of the fourth line in the Balmer series of hydrogen. n1 = 2 n2 = 6 RH = 1.097 x 107m-1 λ= 4.10 x 10-7 m**1**1 1 RH = λ n2 1 n2 2 Different values of RH and its usage • RH = 1.097 107 m-1 n1 < n2 RH = 2.18 x 10-18 J**1**1 1 RH = λ n2 1 1 1 1 = 1.097 x 107 22 52 λ n2 1 2 = 0.2303 X 107 m-1 λ 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 = (6.63 10-34Js)X(3.00108 ms-1) X (0.2303 X 107 m-1) ΔE = 4.58 x 10-19 J**EXERCISE**Calculate what is; i ) Wavelength ii ) Frequency iii ) Wave number of the last line of hydrogen spectrum in Lyman series Wave number = 1/wavelength For Lyman series; n1 = 1 & n2 = ∞ Ans: 9.116 x10-8m 3.29 x1015 s-1 1.0970 X 107 m-1**Ionization Energy**• Definition : Ionization energy is the minimum energyrequired to remove one mole of electron from one mole of gaseous atom/ion. 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 = . • 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).**Example**n1 = 1, n2 = ∞ ∆E = RH (1/n12 – 1/n22) = 2.18 X 10 -18 (1/12 – 1/ ∞2) = 2.18 X 10 -18 (1 – 0) = 2.18 X 10 -18 J Ionisation energy = 2.18 X 10 -18x 6.02 X 1023J mol-1 =1.312 x 106 J mol-1 = 1312 kJ mol-1**Finding ionisation energy experimentally:**Convergent limit 1 st line 1 Ionisation energy is determined by detecting the wavelength of the convergence point**Example**10.97 10.66 10.52 10.27 9.74 8.22 wave number (x106 m-1) The Lyman series of the spectrum of hydrogen is shown above. Calculate the ionisation energy of hydrogen from the spectrum.**Solution**ΔE = hc/λ =h x c / λ = h x c x wave no. = 6.626 x 10-34 J s x 3 x 108 m s-1 x 10.97x 106 m-1= 218.06x 10-20 J = 2.18 x 10-18J Ionisation energy = 2.18 X 10 -18x 6.02 X 1023 J mol-1 =1.312 x 106 J mol-1 = 1312 kJ mol-1**Exercise**Compute the ionisation energy of hydrogen atom in kJ mol1. Solution J**The weakness of Bohr’s Theory**• His theory could not be extended to predict the energy levels and spectra of atoms and ions with more than one electron. It only can explain the hydrogen spectrum or ions contain one electron eg He+, Li2+. • Electrons are restricted to orbit the nucleus at certain fixed distances • It cannot explain for the dual nature of electron • It cannot explain for the extra lines formed in the hydrogen spectrum.**Point to Ponder**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. Thisexperiment demonstrated that electrons do indeed possess wavelike properties. Thus, can the ‘position’ of a wave be specified???**h** = m 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

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