DK015 Chapter 2 : ATOMIC STRUCTURE 2.1 : Bohr’s Atomic Models CHEMISTRY UNIT KOLEJ MATRIKULASI KEDAH OBJECTIVES At the end of this lesson, students should be able to: describe Bohr’s atomic model. explain the existence of energy levels in an atom. calculate the energy of an electron using: 1 E n R H ( 2 ), n R H 2.18 x 10 -18 J calculate the energy change of an electron during transition. 1 ΔE R H ( 2 ni 1 - 2 ), nf where R H 2.18 x 10-18 J calculate the photon emitted by an electron that produces a particular wavelength during transition. History of Atomic Models The Development of Modern Atomic Theory 2.1 : Bohr’s Atomic Models 4 Bohr’s atomic postulate : 1 Electron moves in circular orbit about the nucleus. Each orbit has a different energy level/energy state. 2 In the specific energy level, the energy of electron is fixed in value or is quantised. The electron does not release or absorb any energy when it is in the ground state or at any particular energy level/orbit. This means that an electron moving in an orbit can have a certain amount of energy. (quantised = fixed/specific/definite) n=2 electron PYQ: ‘The energy of an electron in an atom is quantized.’ Describe the above statement. [3M] n=1 nucleus (proton) energy of electron at n=1 ≠ energy of electron at n=2 The energy of an electron in its level is given by: 1 E n RH 2 n where, RH (Rydberg constant)= 2.1810-18J. n (orbit/ energy level) = 1, 2, 3,…, (integer) NOTES: Energy is zero (E = 0) if electron is located infinitely (n = )far from nucleus. Energy associated with forces of attraction are taken to be negative (thus, negative sign) 3 At ordinary condition, the electron is at the ground state (lowest energy state). If energy is supplied, electron absorbed the energy and is promoted from lower energy level to a higher e nergy level (electron is excited) 4 Energy at excited state is unstable. It will fall back to lower energy level and released a specific amount of energy in the form of light or photon. The light have specific wavelength. The line spectrum is formed. The energy absorb / released during transition can be calculated by: 1 1 ΔE R H 2 2 ni n f Where, RH = Rydberg constant (2.18 x 10-18 J) n= principal quantum number (1, 2, 3, …, ) ni = position(n) of electron initially nf = position(n) of electron finally The amount of energy released by the electron during transition is called a photon of energy. A photon of energy is released in the form of light with appropriate frequency and wavelength. Therefore, the energy change (released/absorbed) can be related as: ΔE hυ where, h (Planck’s constant) =6.63 10-34 Js = frequency; c (speed of light) = 3.00108 ms-1 Photon = a packet of light energy equals to h Which of the following is an example when an electron absorbs energy? A.When it goes from the fourth energy level to the first. B.When an electron releases a photon. C.When it maintains constant energy and doesn't change his energy level. D.When it goes from the second energy level to the fourth. 1. The energy of an electron at particular orbit: 1 En RH 2 n 2. RH = 2.18 x 10-18 J The energy absorb / released during transition: 1 1 ΔE R H ni 2 n f 2 ΔE hυ Rydberg Equation Calculator ΔE h υ c υ A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength. Where: c υ Thus: hc ΔE Where: v = frequency λ = wavelength c = speed of light (3.00x108 ms-1) h = Plank’s constant (6.63 x 10-34 Js) 3. Find wavelength, λ : 1 1 ΔE R H 2 2 n1 n2 RH = Rydberg constant, =1.097 x 107 m-1 Using the Bohr Rydberg equation Calculation 1 Energy of electron at particular orbit 1 E n RH 2 n where, RH (Rydberg constant)= 2.1810-18J. n (orbit/ energy level) = 1, 2, 3,…, (integer) Example 1 Calculate the energy of electron that occupies n=2 Answer: Example 2 The energy of an electron at its orbit (n) is -1.36x10-19J. Determine the value of n. Answer: 1. Calculate the energy of electron that occupies n=6 2. The energy of an electron at its orbit (n) is -8.72x10-18J. Determine the value of n. Calculation 2 Energy released/absorbed by electron during transition 1 1 ΔE R H ni 2 n f 2 RH = Rydberg constant (2.18 x 10-18 J) n= principal quantum number (1, 2, 3, …, ) ni = position(n) of electron initially nf = position(n) of electron finally Example 1 Calculate the energy released when an electron falls from n=3 to n=2. Answer: Example 2 Calculate the energy required to promote an electron from the first energy level to the third energy level of a hydrogen atom. Answer: Calculate the energy released when an electron excited from n=1 to n=4. Calculation 3 Energy released/absorbed by electron during transition ΔE hυ h (Planck’s constant) =6.63 10-34 Js = frequency; c (speed of light) = 3.00108 ms-1 Example 1 Calculate the energy absorbed by an electron that produces a wavelength of 9.38 x 10-8 m. Answer: Example 2 An electron released 4.58x10-19 J of energy when falls from n=5 to n=2. Calculate the frequency and wavelength produced. Answer: 1. Calculate the energy released by an electron that produces a wavelength of 9.49 x 10-8 m. 2. An electron absorbed 2.09x10-18 J of energy when excited from n=1 to n=5. Calculate the frequency and wavelength produced. 1. 2. Calculate the energy released by an electron that produces a wavelength of 9.49 x 10-8 m. An electron absorbed 2.09x10-18 J of energy when excited from n=1 to n=5. Calculate the frequency and wavelength produced. ans: v λ = = Calculation 4 Calculate: (a) The energy (in J) of an electron has when it occupies a level equivalent to the quantum number of n = 3 and n = 4. (b) The energy (in kJ/mol) of photon emitted when one mole of electron drops from the 4th energy level to the 3rd energy level. (c) The frequency (in s-1) and wavelength (in nm) of this photon. Answer: (a) Energy of an electron at n = 3 : Energy of an electron at n = 4 : b) For one electron: For 1 mol of electrons: 1 mol electrons = 6.02 x 1023 electron c) ∆E = – 1.06 x 10–19 J Calculate the energy of an electron in the hydrogen atom when n = 2, and when n = 6. Calculate the wavelength released when an electron moves from n = 6 to n = 2. If this line in the visible region of electromagnetic spectrum? (Note: visible light: 400 ~ 700 nm) 1 Calculate the energy of an electron in the En RH 2 hydrogen atom when n = 2, and when n = 6. n OBJECTIVES At the end of this lesson, students should be able to: describe the formation of line spectrum of hydrogen atom. perform calculations involving the Rydberg equation for Lyman, Balmer, Paschen, Brackett and Pfund series: 1 RH ( 1 n12 -n 1 2 ) 2 where RH 1.079 x 107 m-1 and n1 n2 calculate the ionisation energy of hydrogen atom from Ly man series. OBJECTIVES State the weakness of Bohr’s atomic model. State the dual nature of electron using de Broglie’s postulate and Heisenberg’s uncertainty principle. (calculation is not required) FORMATION LINE SPECTRUM HYDROGEN ATOMS 1 2 Spectrum is a series of lines or a set of colours. They are two types of spectrum: line spectrum, and c ontinuous spectrum. Line spectrum Continuous spectrum A spectrum that contains a series of diA spectrum that contains continuous screte lines. Each line corresponds to band of light radiation of all wavelen a specific wavelength. gths. Each line separated by blank area Continuous. No blank area in between. Source: emission spectrum of atom ( Source: white light, sunlight from gas discharge tube of hydrogen , etc.) How is the formation of line spectrum of hydrogen atom? When the electron of a hydrogen atom at its ground state absorbs s ufficient amount of energy, it will move to a higher energy level At higher energy evel, electron is unstable. It will falls back to a lower energy level. During the transition (falls form higher to lower energy level), energy will be released in a form of light at certain wavelength and frequency. Since energy is quantised (fix in value), line spectrum is produced. n=4 n=3 n=2 n=1 nucleus 3 The electromagnetic spectrum: 4 There is several emission series of lines obtained during the transition of electrons in hydrogen spectrum 5 6 The series in the hydrogen line spectrum are found in the visible region, infrared region and ultraviolet region. These spectral lines are the result of the transition of excited electron to the lower energy level. series Spectrum region Lyman Ultraviolet Balmer Visible Paschen Infrared Brackett Infrared Pfund Infrared 7 Balmer series is the only one visible to the unaided eye. It contains four coloured line against a black background. Line number 1st line 2nd line 3rd line 4th line Colour in Balmer series Wavelength (nm) red 656 Blue-green 486 indigo 434 violet 410 7 The possible energy levels in the hydrogen atom and the transitions of electrons that produce the lines in the hydrogen spectrum: 7 Lines produced caused by the transition of electrons from higher energy level to lower energy level. eeee- e- n= n=7 n=6 Series Line Lyman Transition of e- n=5 1st n=2 to n=1 n=4 2nd n=3 to n=1 3rd n=4 to n=1 4th n=5 to n=1 5th n=6 to n=1 n=3 n=2 n=1 Continuum limit ΔE h h c E,, eeee- n= n=7 n=6 n=5 Series Line Balmer Transition of e- 1st n=3 to n=2 n=4 2nd n=4 to n=2 n=3 3rd n=5 to n=2 4th n=6 to n=2 n=2 n=1 Continuum limit ΔE h h c E,, eeee- n= n=7 n=6 n=5 n=4 n=3 n=2 n=1 Series Paschen Line Transition of e- 1st n=4 to n=3 2nd n=5 to n=3 3rd n=6 to n=3 4th n=7 to n=3 Continuum limit ΔE h h c E,, eee- n= n=7 n=6 n=5 n=4 n=3 Series Brackett Line Transition of e- 1st n=5 to n=4 2nd n=6 to n=4 3rd n=7 to n=4 n=2 n=1 Continuum limit ΔE h h c E,, ee- n= n=7 n=6 Series n=5 Line Transition of e- n=4 1st n=6 to n=5 n=3 2nd n=7 to n=5 Pfund n=2 n=1 Continuum limit ΔE h h c E,, 8 Each line in the hydrogen spectrum has a specific wavelength. 9 The wavelength can be calculated by using the Rydberg equation: 1 1 RH 2 2 n2 n1 1 Where = wavelength RH = Rydberg constant = 1.097 x 107 m-1 n1 and n2 = 1, 2, 3, …, (principal quantum numbers) Since should have a positive value thus n1 < n2 Example Calculate the wavelength of the indigo line in the Balmer series. Solution: Indigo line 3rd line in Balmer series n=5 1 1 RH 2 2 n n 2 1 1 ; n1 n2 to n=2 1 x 10-9 m nm Try!!! 1. Calculate the wavelength of the continuum limit in the Paschen series. 2. For the Lyman series, calculate the wavelength when electron falls from n=4 to n=1. Changing the Energy Ultraviolet Infrared Visible • The further they fall, more energy is releas ed and the higher the frequency. • The orbitals also have different energies in side energy levels. ΔE hc ΔE hv Example 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 ∆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? ii) the lowest frequency? 53 Example E D C B Line spectrum A 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. Example W Line spectrum Y Balmer series Describe the transitions of electrons that lead to the lines W, and Y, respectively. Compare the difference: 1 n2i E = RH 1 = RH 1 n2f 1 n12 RH = 2.18 x 10–18 J 1 n22 RH = 1.097 x 107 m-1 n1 can be ni or nf as long as n1 < n2 to get positive value for Both equations can be used to calculate wavelength () and frequency () of any line of H atom emission series Remember! RH values and units are different in both equations ! EXAMPLE-1 Use the Rydberg equation to calculate the wavelength (in nm) of the forth line in the Balmer series of Hydrogen spectrum . ANS: 410.2 nm ANS.EXAMPLE-1 Balmer series, electron: higher level to n = 2 1st line: n: 3 to 2 ; So, 4th line: n: 6(n2) to 2(n1) By using Rydberg equation: Therefore, = 4.102 x 10-7 m = 410.2 nm EXAMPLE-2 An electron in the n = 5 level of an H atom emits a photon of wavelength 1281 nm. To what energy level does the electron move? ANS:n = 3 ANS.EXAMPLE-2 Electron transition: ni = 5 nf = ? By using Rydberg equation: EXAMPLE-3 Calculate what is; For Lyman series; n1 = 1 i ) Wavelength & n2 = ∞ ii ) Frequency iii ) Wave number of the last line of hydrogen spectrum in Lyman series Wave number = 1/wavelength The minimum energy required to remove one mole of electrons from one mole of the gaseous atoms or ions. M (g) → M+ (g) + e- ΔH1= +ve kJ Ionization Energy is the energy required to cause an electron to escape from the atom. The energy levels are a bit like steps of a ladder that the electrons can climb, if an electron is given enough energy to climb to the last energy level the electron will escape from the atom and becomes a positive ion. The energy needed for this jump is called IONIZATION ENERGY EXAMPLE Calculate the ionization energy of H atom in kJmol-1. Ans: EXAMPLE Electron transition: ni = 1 (ground state) nf = ∞ (complete removal) E = RH 1 n2 i 1 n2 f = 2.18 x 10-18 J x 1 12 1 ∞2 = 2.18 x 10-18 J (for removal of one electron form one H atom) EXAMPLE EXAMPLE –– 14 14 – 14 Ans: Ans: EXAMPLE EXAMPLE 6.02 x 1023 electrons 6.02 x 1023 H atoms For removal of 1 mol electrons from 1 mol H atom: E = 2.18 x 10-18 J x 6.02 x 1023 electrons H 1 electron = 1.3124 x 106 J/mol = 1312.4 kJ/mol 1 mol Bohr’s theory has a number of weaknesses. Bohr’s atomic model • Unable to explain the line spectrum of atoms or ions cont aining more than one electron (such as helium). • Electron is restricted to move in a certain distance around the nucleus of an atom. • Unable to explain the extra lines formed in the hydrogen spectrum. • Unable to explain the dual nature of electrons. (A) de Broglie’s postulate • in 1924, Louis de Broglie, a French physicist, suggested that light and matter appear to have dual natures, that is, both light and matter are wave-like as well as particle-like. • de Broglie suggested an equation that allows the calculation of the wavelength of an electron or a particle with mass (m), moving at velocity (): h λ m where, h = Planck’s constant (J s) m = particle mass (kg) = velocity (m/s) = wavelength of a matter wave (B) Heisenberg’s uncertainty principle • Bohr’s theory: the electron was thought of as orbiting the nucleus. This would mean that at any moment, we would know both the precise position and precise speed of the electron. This is not allowed in quantum mechanics. • In 1927, Werner Heisenberg states in his principle that: “It is impossible to know simultaneously both the momentum, p (defined as mass times velocity) and the position of a particle with certain”. This principle is known as Heisenberg’s uncertainty principle. • Mathematically, his principle stated as: h x p 4 where x= uncertainty in measuring the position p=uncertainty in measuring the momentum = mv h = Planck’s constant Bohr’s Atomic Models 9M ANS: https://image.ibb.co/eu8bmQ/tysc2.jpg