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SF027 UNIT 12: ATOMIC STRUCTURE SF027 1 12.1 Early Models of the Atom 12.1.1 Thomson’s model of the atom { In 1898, Joseph John Thomson suggested a model of an atom that consists of homogenous positively charged spheres with tiny negatively charged electrons embedded throughout the sphere as shown in figure 12.1a. { The electrons much likes currants in a plum pudding. { This model of the atom is called ‘plum pudding’ model of the atom. positively charged electron sphere Fig. 12.1a 12.1.2 Rutherford’s model of the atom { In 1911, Ernest Rutherford performed a critical experiment that showed the Thomson’s model is not correct and proposed his new atomic model known as Rutherford’s planetary model of the atom as shown in figure 12.1b. SF027 2 SF027 { { nucleus electron { Fig. 12.1b { +Ze ‘plop’ plop’ -e { { According to this model, the atom was pictured as electrons orbiting around a central nucleus which concentrated of positive charge. The electrons are accelerating because their directions are constantly changing as they circle the nucleus. Based on Maxwell’s electromagnetic theory, an accelerating charge emits energy. Hence the electrons must emit the e.m. radiation as they revolve around the nucleus. As a result of the continuous loss of energy, the radii of the electron orbits will be decreased steadily. This would lead the electrons spiral and falls into the nucleus, hence the atom would collapse as shown in figure 12.1c. energy loss Fig. 12.1c SF027 3 12.2 Bohr’s Model of Hydrogen Atom { { SF027 In 1913, Neils Bohr proposed a new atomic model based on hydrogen atom. According to Bohr’s Model, he assumes that each electron moves in a circular orbit which is centred on the nucleus, the necessary centripetal force being provided by the electrostatic force of attraction between the positively charged nucleus and the negatively charged electron as shown in figure 12.2a. { On this basis he was able to show that the energy of an orbiting electron -e depends on the radius of its orbit. { This model has several features which r r are described by the postulates v FE (assumptions) stated below : +e 1. The electrons move only in certain circular orbits, called r STATIONARY STATES or ENERGY LEVELS. LEVELS When it is in one of these orbits, it does not radiate energy. energy Fig. 12.2a 4 SF027 2. The only permissible orbits are those in the discrete set for which the angular momentum of the electron L equals an integer times h/2π . Mathematically, nh and L = mvr 2π nh or mvr = mvr = nh 2π where r : radius of the orbit m : mass of the electron L= (12.2a) n : principal quantum number = 1,2 ,3 ,... h : planck constant h h = 2π 3. Emission or absorption of radiation occurs only when an electron makes a transition from one orbit to another. The frequency f of the emitted (absorbed) radiation is given by where SF027 ∆E = hf = E f − Ei E f : final energy state Ei : initial energy state (12.2b) 5 Note : z If ∆E ⇒ negative value ∆E ⇒ positive value z If 12.2.1 Bohr’s Radius in Hydrogen atom { Emission of e.m. e.m. radiation. Absorption of e.m. e.m. radiation. Consider one electron of charge –e and mass m moves in a circular orbit of radius r around a positively charged nucleus with a velocity v (Figure 12.2a). { The electrostatic force between electron and nucleus contributes the centripetal force as write in relation below : FE = Fc centripetal force electrostatic force { kq1q2 mv 2 where q1 = e ; q2 = e = r2 r2 e mv 2 = (12.2c) 4 πε0 r nh From the Bohr’s second postulate : mvr = 2π and k= 1 4 πε0 By taking square of both side of the equation, we get m2v 2 r 2 = SF027 n2h2 4π 2 (12.2d) 6 SF027 { By dividing eq. (12.2d) and (12.2c), thus n 2 h 2ε 0 e 2π n 2 h 2ε 0 r= ; n = 1,2,3,... me 2π mr = { (12.2e) which r are radii of the permissible orbits for the Bohr atom. Eq. (12.2e) can be written as 2 r = a0 n 2 ; n = 1,2,3,...with a0 = h ε0 me 2 π Where a0 is called the Bohr’ Bohr’s radius of hydrogen atom. It is defined as the radius of the lowest orbit or ground state (n=1) and is given by 2 r = a0 (1) = 2 (6.63 × 10 ) (8.85 × 10 ) (9.11 × 10 )(1.60 × 10 ) π − 34 − 31 −12 −19 2 a0 = 5.31 × 10 −11 m { The radii of the orbits associated with allowed orbits or states n 2,3,… are 4a0,9a0,…, hence the atomic radius is quantized. SF027 = 7 12.2.2 Discrete Energy in Hydrogen atom { The total energy E of the system is given by Kinetic energy of the electron E = K +U (12.2f) Potential energy of the electron The potential energy, U of the electron is 1 kq1q2 where q1 = e ; q2 = −e and k = 4πε0 r 2 e (12.2g) U =− 4 πε0 r U= and the kinetic energy, K of the electron is K= 1 2 mv 2 (12.2h) By substituting eq. (12.2c) into eq. (12.2h), thus 1 e2 2 4 πε0 r e2 K= 8 πε0 r K= SF027 (12.2i) 8 SF027 By substituting eq. (12.2g) and (12.2i) into eq. (12.2f), therefore the total energy E of the system is e2 + − 8πε 0 r 4πε 0 r e2 (12.2j) E=− 8 πε0 r Ze 2 In general, En = − 8 πε0 r Z : atomic number where E= { { e2 n=2 λ = 2πr1 Fig. 12.2b λ { SF027 n=3 2λ = 2πr2 Fig. 12.2c 3λ = 2πr3 Fig. 12.2d 9 λ n=4 n=5 4λ = 2πr4 5λ = 2πr5 Fig. 12.2e { (12.2k) From de Broglie’s relation, the electron is to be regarded as a wave, then its stable orbits in an atom are those satisfy the conditions of a standing (stationary) wave as shown in figures 12.2b, 12.2c, 12.2d , 12.2e and 12.2f. λ λ λ orbital n=1 SF027 Equation of discrete energy in hydrogen atom only Fig. 12.2f If there are n waves in the orbital and λ is wavelength of wave properties of electron therefore nλ = 2πrn (12.2l) th where rn : radius of the n orbit n = 1,2 ,3,... h Since λ = then eq. (12.2l) can be written as mv h n = 2πrn mv nh mvrn = Bohr’ Bohr’s second postulates 2π 10 SF027 { The energy level of hydrogen atom is given by n 2 h 2 ε0 Ze 2 where Z = 1 and r = me 2 π 8 πε0 r (hydrogen atom) (1)e 2 En = − n 2 h 2 ε0 8 πε0 2 me π me 4 me 4 and E = = 13.6 eV En = − 2 2 2 1 8 ε02 h 2 8 ε0 h n 13.6 eV E (12.2m) En = − 21 or En = − n2 n th where En : energy level of the n orbit(state) E1 : energy level of ground state n = 1,2 ,3,... En = − The negative sign means that work has to be done to remove the electron to infinity, where it is considered to have zero energy, i.e. the electron is bound to the atom. SF027 11 12.3 Energy Level of Hydrogen Atom { { The energies of the electron in an atom can have only certain values. This values are called the energy level of the atom. The energy level of hydrogen atom can be calculated by using eq. (12.2m) which is 13.6 eV En = − When n n2 ; n = 1,2,3,... = 1, the ground state (the state of the lowest energy level) level 13.6 eV E1 = − = −13.6 eV (1)2 13.6 eV = -3.39 eV When n = 2, the first excited state, E2 = − 2 2 13.6 eV n = 3, the second excited state,E3 = − = −1.51 eV 32 13.6 eV n = 4, the third excited state, E4 = − = -0 .85 eV 42 13.6 eV electron is completely n = ∞, E ∞ = − =0 2 removed from the atom. (∞ ) SF027 12 SF027 { Figure 12.3a shows diagrammatically the various energy levels in the hydrogen atom. n ∞ En (eV ) 0.0 Free electron 5 4 3 − 0.54 − 0.85 − 1.51 4th excited state 3rd excited state 2nd excited state − 3.39 1st excited state Ionization energy is defined as the energy required by an electron in2 the ground state to escape completely from the attraction of the nucleus. An atom becomes ion. 1 Excitation energy is defined as the energy required by an electron that raises it to an excited state from its ground state. − 13.6 Ground state Fig. 12.3a SF027 { excited state is defined as the energy levels that higher than the ground state. is defined as the lowest stable energy state of an 13 atom. Example 1 : The electron in the hydrogen atom makes a transition from E2=-3.40 eV energy state to the ground state with E1= -13.6 eV. Calculate a. the energy in Joule, and b. the wavelength of the emitted photon. (Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1) Solution: a. The energy of the emitted photon is ∆E = E f − Ei ∆E = E1 − E2 ∆E = (− 13.6 ) − (− 3.40 ) ∆E = −10.2 (1.60 × 10 −19 ) ∆E = 1.63 × 10 −18 J b. The wavelength of the photon is hc λ hc λ= ∆E λ = 1.22 × 10 −7 m Negative sign means the energy is emitted. ∆E = SF027 14 SF027 { Example 2 : The lowest energy state for hydrogen atom is E1= -13.6 eV. Calculate the frequency of the photon required to ionize the atom. (Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1) Solution: An atom where its electron at ground state is raised to the zero energy level said to be ionized. The ionization energy is ∆E = E f − Ei ∆E = E∞ − E1 ∆E = (0 ) − (− 13.6 ) ∆E = 13.6 (1.60 × 10 −19 ) ∆E = 2.18 × 10 −18 J The frequency of the photon required to ionize the atom is ∆E = hf ∆E f = h f = 3.29 × 10 15 Hz SF027 15 { Example 3 : A hydrogen atom emits radiation of wavelengths 121.5 nm and 102.4 nm when the electrons make transitions from the 1st excited state and 2nd excited state respectively to the ground state. Calculate: a. the energy of a photon for each of the wavelengths above; b. the wavelength emitted by the photon when the electron makes a transition from the 2nd excited state to the 1st excited state. (Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1) Solution: λ1=121.5x10-9 m, λ2=102.4x10-9 m a. The energy of the photon due to transition from 1st excited state to the hc ground state is ∆E1 = λ1 ∆E1 = 1.64 × 10 −18 J The energy of the photon due to transition from 2nd excited state to the hc ground state is ∆E2 = λ2 ∆E2 = 1.94 × 10 −18 J SF027 16 SF027 b. ∆E3 ∆E1 ∆E2 2nd excited state 1st excited state Ground state ∆E3 = ∆E2 − ∆E1 ∆E3 = 3.00 × 10 −19 J Therefore the wavelength of the emitted photon due to the transition from 2nd excited state to the 1st excited state is hc λ3 λ3 = 6 .63 × 10 −7 m ∆E3 = { Example 4 : The electron of an excited hydrogen atom make a transition from the ground state to the 4th excited state. Determine the energy absorbs by the atom. (Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1) SF027 17 Solution: a. By applying the equation of energy level in hydrogen atom, 13.6 eV n2 13.6 eV Ground state, n = 1 E1 = − 12 E1 = 13.6 (1.60 × 10 −19 ) J E1 = 2.18 × 10 −18 J 13.6 eV E5 = − 4th excited state, n = 5 52 E5 = 0.54 (1.60 × 10 −19 ) J E5 = 8.64 × 10 −20 J En = − Hence the energy absorbs by the atom is ∆E = E f − Ei ∆E = E5 − E1 ∆E = 2.09 × 10 −18 J SF027 18 SF027 12.4 Emission Line Spectrum { { The emission lines correspond to photons of discrete energies that are emitted when excited atomic states in the gas make transitions back to lower energy levels. Figure 12.4a shows line spectra produced by emission in the visible range for hydrogen (H), mercury (Hg) and neon (Ne). Fig. 12.4a 12.4.1 Hydrogen Emission Line Spectrum { Emission processes in hydrogen give rise to series, which are sequences of lines corresponding to atomic transitions as shown in figure 12.4b. SF027 n ∞ 5 4 3 2 19 E n ( eV ) 0. 0 Free electron − 0. 54 4th excited state *Pfund series − 0. 85 3rd excited state *Brackett series − 1. 51 2nd excited state Paschen series involves transitions that end with the Infrared range 2nd excited state. − 3. 39 1st excited state Balmer series involves transitions ending with the Visible light range 1st excited state of hydrogen. Lyman series Ultraviolet involves transitions that end with the range ground state of hydrogen. − 13 . 6 Ground state 1 Fig. 12.4b *Brackett series involves transitions that end with the 3rd excited state. Infrared *Pfund series involves transitions that end with the range SF027 20 Simulation 4th excited state. SF027 { Figure 12.4c shows “permitted” orbits of an electron in the Bohr model of a hydrogen atom. Fig. 12.4c : not to scale { Figure 12.4d illustrates the lines spectrum of Balmer series for hydrogen atom. SF027 21 Fig. 12.4d 12.4.2 Wavelength of Emission Line Spectrum for Hydrogen atom { { If an electron makes a transition from an outer orbit (ni) to an inner orbit (nf), energy is radiated. The energy radiated can be calculated by using equation below: ∆E = E f − Ei = En f − Eni Emission of e.m. e.m. radiation (you can ignore it) { ∆E = − me4 1 1 − 2 2 2 2 8 εo h n f ni (12.4a) Hence the wavelength of the photon emitted (energy radiated in form of e.m. radiation) is shown below. Since SF027 me4 me4 ∆E = − 2 2 2 − − 2 2 2 8ε h n 8 εo h ni o f ∆E = hc then eq. (12.4a) can be written as λ hc me4 1 1 = 2 2 2− 2 λ 8 εo h n f ni 22 SF027 1 λ = me4 1 1 − 2 2 2 3 8 εo ch n f ni 1 1 = RH 2 − 2 n λ ni f 1 where { RH : Rydberd constant n f : final value of n ni : initial value of n Note: z For Lyman series (nf=1) : z For Balmer series (nf=2) : z For Paschen series (nf=3) : SF027 z For Brackett series (nf=4) : z For Pfund series (nf=5) : (12.4b) RH = me4 = 1.0974 × 107 m −1 2 3 8ε 0 ch 1 1 = RH 2 − 2 λ 1 ni 1 1 1 = RH 2 − 2 λ 2 ni 1 1 1 = RH 2 − 2 λ 3 ni 1 23 1 1 = RH 2 − 2 λ ni 4 1 1 1 = RH 2 − 2 λ 5 ni 1 To calculate the shortest wavelength in any line series, ni=∞. Example 5 : Determine the wavelength for a line spectrum in Lyman series when the electron makes a transition from n=4 level. (Given Rydberg constant ,RH = 1.0974 x 107 m s-1) z { Solution: ni=4 By applying the equation of wavelength for Lyman series (nf=1), thus 1 1 = RH 2 − 2 λ 1 ni 1 1 1 = 1.0974 × 107 2 − 2 λ 1 4 −8 λ = 9.72 × 10 m 1 ( SF027 ) 24 SF027 { Example 6 : The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate at energy level n=2 as shown in the diagram below. E (eV ) n n ∞ 6 5 4 3 0. 0 − 0. 38 − 0. 54 − 0. 85 − 1. 51 2 − 3. 40 Calculate a. the longest-wavelength, and b. the shortest-wavelength photon emitted in this series. (Given Rydberg constant ,RH = 1.0974 x 107 m s-1) Solution: nf=2 The equation of wavelength for Balmer series is 1 1 = RH 2 − 2 λ 2 ni 1 SF027 25 a. The longest-wavelength photon results from the transition from n = 3 to n = 2 (Balmer series). Thus 1 λmax 1 1 = RH 2 − 2 2 3 ∆Emin = E f − Ei = E2 − E3 ∆Emin = (− 3.40 ) − (− 1.51) ∆Emin = −1.89 × 1.60 × 10 −19 ∆Emin = 3.02 × 10 −19 J hc = 3.02 × 10 −19 J or λmax = 6.56 × 10 −7 m λmax b. The shortest-wavelength photon results from the transition from n = ∞ to n = 2 (Balmer series). Thus 1 λmin 1 1 = RH 2 − 2 2 ∞ λmin = 3.65 × 10 −7 m SF027 or ∆Emax ∆Emax ∆Emax ∆Emax hc λmin = E f − Ei = E2 − E∞ = (− 3.40 ) − (0 ) = −3.40 × 1.60 × 10 −19 = 5.44 × 10 −19 J = 5.44 × 10 −19 J 26 SF027 12.5 Many Electrons Atomic Model 12.5.1 Limitations of Bohr’s Model { The Bohr’s theory z predicts successfully the energy levels of the hydrogen atom but fails to explain the energy levels of more complex atoms. z can explain the spectrum for hydrogen atom but some details of the spectrum cannot be explained especially when the atom is placed in a magnetic field. z cannot explain the Zeeman effect. { Zeeman effect is defined as the splitting of spectral lines when the radiating atoms are placed in a magnetic field. { Figure 12.5a shows the zeeman effect. 2p Energy Levels Transitions 1s No magnetic field Magnetic field Fig. 12.5a SF027 Spectra 27 12.5.2 Utilises 4 quantum numbers: { Principal quantum number, n: Specifies the main energy level and corresponds to the number n in the Bohr atomic model where n = 1,2,3,... { Orbital (azimuthal (azimuthal)) quantum Number, l : Specifies the angular momentum of electron, where l = 0,1,2,..., (n − 1) For n = 1, l = 0 For n = 2, l = 0 and 1 For n = 3, l = 0,1 and 2 { Magnetic quantum number, ml : Specifies the orientation of the electron orbitals in a magnetic field and plays an important part in explaining the Zeeman effect E.g : a. l SF027 ml = −l ,−(l − 1),...,0,1,..., (l − 1),+l = 0, ml = 0 b. l = 1, ml = -1,0,+1 c. l = 2, ml = -2,-1,0,+1,+2 28 SF027 Spin quantum number, ms: Electron spins when put in a magnetic field and have a magnetic moment because of this spin. Its value is given by Specify whether the spin aligned 1 ms = ± with, or counter to, an applied 2 magnetic field. 12.5.3 Pauli Exclusion Principle { States “No No two electrons in an atom can have the same set of four quantum numbers.” numbers. { It is needed in describing the electron configuration of multi-electron atoms. { Example 7 : Write down the set of quantum number for electron with principal quantum number n=2. Solution: { When n=2; l = 0,1; ml=0,-1,0,+1; ms = ±½, ±½, ±½, ±½ The following set of quantum numbers are, therefore, possible : (2,0,0,+½) ; (2,0,0,-½) (2,1,-1, +½); (2,1,-1, -½) (2,1,0, +½); (2,1,0, -½) (2,1,1, +½); (2,1,1, -½) SF027 or in table form : { Table 12.5a SF027 There are 8 electron states with n = 2 and these form the L shell 29 n 2 l 0 1 ml 0 -1 0 +1 m s +½ - ½ +½ - ½ +½ - ½ +½ - ½ Note : z Table 12.5a shows the electron quantum states in the first four shell. Number of electron n l ml Subshell 1 0 0 1s 2 2 2 0 1 0 -1,0,1 2s 2p 2 6 3 0 0 3s 2 3 1 -1,0,1 3p 6 3 2 -2,-1,0,1,2 3d 10 4 0 0 4s 2 4 1 -1,0,1 4p 6 4 2 -2,-1,0,1,2 4d 10 4 3 -3,-2,-1,0,1,2,3 4f 14 Shell K 8 L 18 M 32 N 30 SF027 z The total number of electrons in all the shells equal to Z, the atomic number. Hence the electron configuration for various elements can be written as shown below in table 12.5b. Element Atomic Number (Z) He 2 1s2 Be 4 1s22s2 C 6 1s22s22p2 O 8 1s22s22p4 Ne 10 1s22s22p6 Mg 12 1s22s22p63s2 Si 14 1s22s22p63s23p2 S 16 1s22s22p63s23p4 Ar 18 1s22s22p63s23p6 Ca 20 1s22s22p63s23p64s2 Electron Configuration Table 12.5b SF027 31 z Figure 12.5b shows the energy levels for various subshell. 4p Energy level 3d 4s 3p 3s Maximum electron in shell 2p n = 2 2s n = 1 1s Fig. 12.5b { SF027 Example 8 : (exercise) Write down the set of quantum number of electron in form of table for nitrogen atom with atomic number, Z=14. 32 SF027 THE END… Next Unit… UNIT 13 : X-rays SF027 33