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Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
Phy 310   chapter 5
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Phy 310 chapter 5

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  • 1. CHAPTER 5: Spectral Lines of Hydrogen (2 Hours) DR.ATAR @ UiTM.NS Dr Ahmad Taufek Abdul Rahman PHY310 Spectral Lines of Hydrogen School of Physics & Material Studies Faculty of Applied Sciences Universiti Teknologi MARA Malaysia Campus of Negeri Sembilan 72000 Kuala Pilah, NS 1
  • 2. Learning Outcome: 5.1 Bohr‟s atomic model (1 hour) At the end of this chapter, students should be able to:  Explain Bohr’s postulates of hydrogen atom. DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 2
  • 3. 5.1 Bohr‟s atomic model 5.1.1 Early models of atom Thomson’s model of 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 1. positively charged sphere electron Figure 1   The electrons much likes currants in a plum pudding. This model of the atom is called „plum pudding‟ model of the atom. DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 3
  • 4. Rutherford’s model of 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 2. nucleus electron Figure 2   According to Rutherford‟s 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. DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 4
  • 5.     Based on the wave theory, an accelerating charge emits energy. Hence the electrons must emit the EM 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 3. e +Ze „plop‟ energy loss Figure 3 DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 5
  • 6. 5.1.2 Bohr’s model of hydrogen atom   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 4. e  Fe +e  v Figure 4 r DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 6
  • 7.   On this basis he was able to show that the energy of an orbiting electron depends on the radius of its orbit. This model has several features which are described by the postulates (assumptions) stated below : 1. The electrons move only in certain circular orbits, called STATIONARY STATES or ENERGY LEVELS. When it is in one of these orbits, it does not radiate energy. 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 L 2 nh mvr  2 where DR.ATAR @ UiTM.NS and L  mvr (1) n : principalquantum number  1,2,3,... r : radius of theorbit m : mass of theelectron PHY310 Spectral Lines of Hydrogen 7
  • 8. 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 (2) E  hf  Ef  Ei where Note: E : change of energy h : Planck's constant Ef : final energy state Ei : initial energy state  If Ef > Ei Absorption of EM radiation  If Ef < Ei Emission of EM radiation DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 8
  • 9. Learning Outcome: 5.2 Energy level of hydrogen atom (1 hour) At the end of this chapter, students should be able to:  Derive Bohr’s radius and energy level in hydrogen atom.  Use 2 h  rn  n a0  n  2 2   4 mk e  2 and k e2  1  En    2 2 a0  n   Define ground state energy, excitation energy and ionisation energy. DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 9
  • 10. 5.2 Energy level of hydrogen atom 5.2.1 Bohr’s radius in hydrogen atom  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 as shown in Figure 11.3.  The electrostatic force between electron and nucleus contributes the centripetal force as write in the relation below: Fe  Fc centripetal force 1  Q1Q2  mv 2 and Q1  Q2  e  2  40  r  r e2 (3) mv 2  40 r electrostatic force DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 10
  • 11.  From the Bohr‟s second postulate: nh mvr  2 By taking square of both side of the equation, we get n2h2 m 2v 2 r 2  4 2  (4) By dividing the eqs. (11.4) and (11.3), thus  n2h2     4 2  2 2 2 m v r   2   e  mv 2    4 r  0   n 2 h 2 0 r me 2 DR.ATAR @ UiTM.NS and 1 0  4k PHY310 Spectral Lines of Hydrogen electrostatic constant 11
  • 12. n2h2 r me 2  1     4k  2 h2 rn  n   4 2 mk e2    ; n  1,2,3...   (5) which rn is radii of the permissible orbits for the Bohr‟s atom. Eq. (5) can also be written as rn  n a0 2 (6) and h2 a0  4 2 mk e2 where a0 is called the Bohr’s radius of hydrogen atom. DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 12
  • 13.  The Bohr‟s radius is defined as the radius of the most stable (lowest) orbit or ground state (n=1) in the hydrogen atom and its value is a0  6.63 10  9.1110 9.00 10 1.60 10  34 2 4 2 31 a0  5.31 10  11 m OR 9 19 2 0.531 Å (angstrom) Unit conversion: 1 Å = 1.00 1010 m  The radii of the orbits associated with allowed orbits or states n = 2,3,… are 4a0,9a0,…, thus the orbit’s radii are quantized. DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 13
  • 14. 5.2.2 Energy level in hydrogen atom  is defined as a fixed energy corresponding to the orbits in which its electrons move around the nucleus. The energy levels of atoms are quantized.  The total energy level E of the hydrogen atom is given by  E U  K Potential energy of the electron  (7) Kinetic energy of the electron Potential energy U of the electron is given by kQ1Q2 U r k e2 U  2 n a0 DR.ATAR @ UiTM.NS where Q1  e; Q2  e and r  n 2 a0 nucleus electron (8) PHY310 Spectral Lines of Hydrogen 14
  • 15.  Kinetic energy K of the electron is given by 1 2 e2 K  mv but mv 2  2 40 r 1  e2    where 1  k and r  n 2 a0 K  2  40 r  40  1  ke2  (9) K  2  n a  2 0   Therefore the eq. (11.7) can be written as 2 2 ke 1  ke  En   2   2  n a 0 2  n a0    ke2  1  En    2 2 a0  n  DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen (10) 15
  • 16.   In general, the total energy level E for the atom is ke2  Z 2   2 En   2 a0  n    where Z : atomic number (11) Using numerical value of k, e and a0, thus the eq. (11.10) can 2 be written as 9.00  10 9 1.60  10 19  1  En Note:     11    2 n  2 5.31  10 2.17  10 18  1   eV 2  19 1.60  10 n  13.6 En   2 eV; n  1,2,3,... (12) n th where En : energy level of n state (orbit) Eqs. (10) and (12) are valid for energy level of the hydrogen atom. DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 16
  • 17.   The negative sign in the eq. (11.12) indicates that work has to be done to remove the electron from the bound of the atom to infinity, where it is considered to have zero energy. The energy levels of the hydrogen atom are when n=1, the ground state (the state of the lowest energy level) ; E1   13.6 1 2 eV  13.6 eV n=2, the first excited state; n=3, the second excited state; n=4, the third excited state; n=, the energy level is E   DR.ATAR @ UiTM.NS 13.6  2 eV  0 E2   13.6 E3   13.6 2 2 E4   eV  3.40 eV 3 2 13.6 4 2 eV  1.51 eV eV  0.85 eV electron is completely removed from the atom. PHY310 Spectral Lines of Hydrogen 17
  • 18.  Figure 4 shows diagrammatically the various energy levels in the hydrogen atom.  En (eV ) 0.0 Free electron 5 4 3 th  0.54 4rd excited state  0.85 3 excited state  1.51 2nd excited state n Figure 4 Ionization energy is defined as the energy required by an electron in the 2 ground state to escape completely from the attraction of the nucleus. An atom becomes ion. DR.ATAR @ UiTM.NS 1  3.40 1st excited state 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 PHY310 Spectral Lines of Hydrogen excited state is defined as the energy levels that higher than the ground state. is defined as the lowest stable energy state of an atom. 18
  • 19. Example 1 : The electron in the hydrogen atom makes a transition from the energy state of 0.54 eV to the energy state of 3.40 eV. Calculate the wavelength of the emitted photon. (Given the speed of light in the vacuum, c =3.00108 m s1 and Planck‟s constant, h =6.631034 J s) DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 19
  • 20. Solution : The change of the energy state in joule is given by Ei  0.54 eV; Ef  3.40 eV E  Ef  Ei E   3.40    0.54    2.86  1.60 10 19 E  4.58  10 19 J  Therefore the wavelength of the emitted photon is E  DR.ATAR @ UiTM.NS hc  4.58  10 19 6.63 10 3.00 10   34 8    4.34  10 m 7 PHY310 Spectral Lines of Hydrogen 20
  • 21. Example 2 : The lowest energy state for hydrogen atom is 13.6 eV. Determine the frequency of the photon required to ionize the atom. (Given the speed of light in the vacuum, c =3.00108 m s1 and Planck‟s constant, h =6.631034 J s) DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 21
  • 22. Solution : The ionization energy in joule is given by Ei  13.6 eV; Ef  E  0 E  Ef  Ei E  0   13.6   13.6 1.60 10 19 E  2.18  10 18 J  Therefore the frequency of the photon required to ionize the atom is E  hf 2.18 10 18  6.63 10 34 f   f  3.29 1015 Hz DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 22
  • 23. Example 3 : For an electron in a hydrogen atom characterized by the principal quantum number n=2, calculate a. the orbital radius, b. the speed, c. the kinetic energy. (Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg; e=1.601019 C and k=9.00109 N m2 C2) DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 23
  • 24. Solution : a. The orbital radius of the electron in the hydrogen atom for n=2 level is given by n2   h2 2  rn  n  2  4 mke2    34 2  6.63  10 2 r2  2   4 2 9.11  10 31 9.00  10 9 1.60  10 19  r2  2.12 10 10 m  DR.ATAR @ UiTM.NS   PHY310 Spectral Lines of Hydrogen     2    24
  • 25. Solution : n  2 b. By applying the Bohr‟s 2nd postulate, thus nh mvrn  2 2h mvr2  2 6.63  10 34 9.11  10 31 v 2.12  10 10      6 v  1.09  10 m s 1 c. The kinetic energy of the orbiting electron is given by 1 2 K  mv 2 1  9.11 10 31 1.09 10 6 2 K  5.41  10 19 J  DR.ATAR @ UiTM.NS  PHY310 Spectral Lines of Hydrogen  2 25
  • 26. Example 4 : A hydrogen atom emits radiation of wavelengths 221.5 nm and 202.4 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 the speed of light in the vacuum, c =3.00108 m s1 and Planck‟s constant, h =6.631034 J s) DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 26
  • 27. Solution : a. The energy of the photon due to transition from 1st excited state to the ground state is 1  221.5 10 9 m; 2  202.4 10 9 m E1  DR.ATAR @ UiTM.NS hc 1 E1 6.63 10 3.00 10   34 8 221 .5  10 9 E1  8.98 10 19 J PHY310 Spectral Lines of Hydrogen 27
  • 28. Solution : 1  221.5 10 9 m; 2  202.4 10 9 m a. The energy of the photon due to transition from 2nd excited state to the ground state is E2 6.63 10 3.00 10   34 8 202.4  10 9 E2  9.83 10 19 J DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 28
  • 29. 1  221.5 10 9 m; 2  202.4 10 9 m Solution : b. ΔE3 2nd excited state 1st excited state ΔE1 ΔE2 E3  E2  E1 Ground state E3  9.83 10 19  8.98 10 19 E3  8.50 10 20 J Therefore the wavelength of the emitted photon due to the transition from 2nd excited state to the 1st excited state is E3  hc 3 DR.ATAR @ UiTM.NS 8.50  10  20 6.63 10 3.00 10   34 8  3  2.34 10 m 6 3 PHY310 Spectral Lines of Hydrogen 29
  • 30. Learning Outcome: 5.3 Line spectrum (1 hour) At the end of this chapter, students should be able to:  Explain the emission of line spectrum by using energy level diagram.  State the line series of hydrogen spectrum.  Use formula, E   hc 1 DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 30
  • 31. 5.3 Line spectrum   The emission lines correspond to the photons of discrete energies that are emitted when excited atomic states in the gas make transitions back to lower energy levels. Figure 11.5 shows line spectra produced by emission in the visible range for hydrogen (H), mercury (Hg) and neon (Ne). Figure 5 DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 31
  • 32. 5.3.1 Hydrogen emission line spectrum   Emission processes in hydrogen give rise to series, which are sequences of lines corresponding to atomic transitions. The series in the hydrogen emission line spectrum are  Lyman series involves electron transitions that end at the ground state of hydrogen atom. It is in the ultraviolet (UV) range.  Balmer series involves electron transitions that end at the 1st excited state of hydrogen atom. It is in the visible light range.  Paschen series involves electron transitions that end at the 2nd excited state of hydrogen atom. It is in the infrared (IR) range.  Brackett series involves electron transitions that end at the 3rd excited state of hydrogen atom. It is in the IR range.  Pfund series involves electron transitions that end at the 4th excited state of hydrogen atom. It is in the IR range. DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 32
  • 33.  Figure 6 shows diagrammatically the series of hydrogen emission line spectrum. En (eV ) 0.0 Free electron n  5 4 3  0.54 th Pfund series0.85 4rd excited state  3 excited state Brackett series  1.51 2nd excited state Paschen series 2  3.39 1st excited state Balmer series Figure 6 Lyman series 1 DR.ATAR @ UiTM.NS  13.6 Ground state PHY310 Spectral Lines of Hydrogen 33
  • 34.  Figure 7 shows “permitted” orbits of an electron in the Bohr model of a hydrogen atom. Figure 7: not to scale DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 34
  • 35. 5.3.2 Wavelength of hydrogen emission line spectrum   If an electron makes a transition from an outer orbit of level ni to an inner orbit of level nf, thus the energy is radiated. The energy radiated in form of EM radiation (photon) where the wavelength is given by E   E   hc hc 1  (13) From the Bohr‟s 3rd postulate, the eq. (11.13) can be written as 1 1   Enf  Eni  hc where and DR.ATAR @ UiTM.NS ke2  1    Enf   2a0  nf 2    ke2  1    Eni   2  2a0  ni   PHY310 Spectral Lines of Hydrogen 35
  • 36. 1  ke2  1   ke2  1         2a0  nf 2   2a0  ni 2         1 ke2  1 1      2 hc 2a0  nf 2 ni    k e2  1 1  ke2    2  and  RH n 2 n  2hca0  f 2hca0 i  1    hc  1 1   RH  2  2  n  ni   f  1 where DR.ATAR @ UiTM.NS (14) RH : Rydberd' s constant 1.097107 m1 nf : final value of n ni : initial value of n PHY310 Spectral Lines of Hydrogen 36
  • 37. Note:  For the hydrogen line spectrum,  Lyman series( nf=1 )  Balmer series( nf=2 ) 1 1   RH  2  2  1 n   i    1 1 1   RH  2  2  2  ni    1 1 1   RH  2  2  3  ni     1 1 1   RH  2  2  4  ni     1 1 1   RH  2  2  5  ni    1   Brackett series( nf=4 )   Paschen series( nf=3 ) Pfund series( nf=5 ) To calculate the shortest wavelength in any series, take ni= . DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 37
  • 38. 5.3.3 Limitation of Bohr’s model of hydrogen atom  The Bohr‟s model of hydrogen atom  predicts successfully the energy levels of the hydrogen atom but fails to explain the energy levels of more complex atoms.  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.  cannot explain the Zeeman effect (Figure 11.7).  Zeeman effect is defined as the splitting of spectral lines when the radiating atoms are placed in a magnetic field. 2 1 Transitio ns No magnetic field DR.ATAR @ UiTM.NS Energy Levels Magnetic field Figure 7 Spectra PHY310 Spectral Lines of Hydrogen 38
  • 39. Example 5 : The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate at energy level n=2 as shown in the Figure 8.  6 5 4 3 En (eV ) 0 .0  0.38  0.54  0.85  1.51 2  3.40 n Figure 8 Calculate a. the longest wavelength, and b. the shortest wavelength of the photon emitted in this series. (Given the speed of light in the vacuum c =3.00108 m s1 ,Planck‟s constant h =6.631034 J s and Rydberg‟s constant RH = 1.097  107 m1) DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 39
  • 40. Solution : nf  2 a. The longest wavelength of the photon results due to the electron transition from n = 3 to n = 2 (Balmer series). Thus  1 1 1  E  RH  2  2  OR  n  ni   hc  f  1 Ef  Ei 1 1 7  1   1.097  10  2  2   hc max 3  2 E 2  E3 1 max  6.56 10 7 m  max hc  3.40   1.51  1.60 10 19  6.63  10 34 3.00  108 1        max  6.58 10 7 m DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 40
  • 41. Solution : n  2 f b. The shortest wavelength of the photon results due to the electron transition from n =  to n = 2 (Balmer series). Thus 1   Ef  Ei hc 1 min   E 2  E hc  3.40   0  1.60 10 19 6.63 10 34 min  3.66 10 7 m   3.00 10  8  1 1  OR  RH  2  2  n  ni   f  1 1  7  1  1.097  10  2  2  min   2 1   min  3.65 10 7 m DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 41
  • 42. Example 6 : Determine the wavelength for a line spectrum in Lyman series when the electron makes a transition from n=3 level. (Given Rydberg‟s constant ,RH = 1.097  107 m1) Solution : ni  3 ; nf  1 By applying the equation of wavelength for Lyman series, thus 1 1   RH  2  2  1 n   i   1 7  1  1.097  10  2  2  1 3  1     1.03  10 7 m DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 42
  • 43. Exercise 5.1 : Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, e=1.601019 C and RH =1.097107 m1 1. A hydrogen atom in its ground state is excited to the n =5 level. It then makes a transition directly to the n =2 level before returning to the ground state. What are the wavelengths of the emitted photons? ANS. : 4.34107 m; 1.22107 m 2. Show that the speeds of an electron in the Bohr orbits are given ( to two significant figures) by vn DR.ATAR @ UiTM.NS 2.2 10  6 m s 1  n PHY310 Spectral Lines of Hydrogen 43

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