Atomic Structure: Bohr’s Correspondence Principle.pptx

Bohr’s Correspondence Principle

1. Lecture 2 Mphy 350 Atomic and Biomolecules spectra By Dr.Fatemah Alkallas

2. Bohr’s Correspondence Principle Dr.Fatemah Alkallas Bohr’s Correspondence Principle states that quantum mechanics is in agreement with classical physics when the energy differences between quantized levels are very small 1

3. To valid that we will prove that classical frequency V c for the electron is equal to quantum ν q frequency when n is very large Dr.Fatemah Alkallas 2

4. Dr.Fatemah Alkallas First at large n : Determine Determine Comparing between and   3

5. Using the angular momentum law to determine   Dr.Fatemah Alkallas 4 so

6. we know that the the angular velocity is And that Dr.Fatemah Alkallas 5

7. make the two velocities equal and put the angular frequency in the equation find: We know that the orbital radius is Dr.Fatemah Alkallas 6

8. Put r in equation 2 Dr.Fatemah Alkallas 7 We also know that :

9. Put R in 2 Dr.Fatemah Alkallas 8 Equation 4 is the classical frequency

10. To determine the quantum frequency Dr.Fatemah Alkallas 9 so

11. Using: At a very large n we there will not be a different between n i and n f then when transition happened between two successive states n i and n f Dr.Fatemah Alkallas 10

12. Put these approximations in equation 5 From 4 and 6 : Dr.Fatemah Alkallas 11

13. This true only if: 1- n very large 2- the transition is between successive states If not means for example . Then the is twice . Dependence on   Dr.Fatemah Alkallas 12

14. Experimental validation of individual energy levels existence. Dr.Fatemah Alkallas 13

15. Discrete Spectrum Bohr’s Postulates Predicted a Discrete Spectrum Consistent with the spectrum of hydrogen Direct proof found in measurements by James Frank and Gustav Hertz in 1914 Results consistent with spectra Dr.Fatemah Alkallas 14

16. The Franck-Hertz Experiment The heated filament ejects electrons into the tube, which can be either evacuated (vacuum) or filled with Hg vapor A variable (positive) accelerating voltage, V A , is applied to the grid (a wire mesh) The electrons acquire kinetic energy K = e V A There is (negative) retarding voltage ( V r ) between grid and collector Only electrons having enough [kinetic] energy will overcome this potential and reach the Collector and contribute to current measured with an Electrometer Dr.Fatemah Alkallas 15

17. The Franck-Hertz Experiment If V r > V A no electrons can reach the Collector, so no current would be measured. If V r < V A , then, if the tube is highly evacuated , most of the electrons would reach the Collector, and have energy | e| ( V A - V r ) Dr.Fatemah Alkallas 16

18. The Franck-Hertz Experiment If the tube contains some gas, the electrons can loose energy via collisions with the gas atoms Such collisions are inelastic, i.e. electrons lose energy, which is transferred to internal energy of atoms in the gas Dr.Fatemah Alkallas 17 Thus, even in the case when V r < V A , it is possible that the electrons would not be able to reach Collector, and contribute to the current, Why?

19. The Franck-Hertz Experiment Franck and Hertz observed the Collector current as a function of V A ( > V r ) when tube was filled with various gases (result for mercury gas is shown here) At first, the current increased as was expected for a typical vacuum tubes, but at ~4.9 V current suddenly dropped Then, the increase resumed until 9.8 V, an so on Dr.Fatemah Alkallas 18 V A

20. The Franck-Hertz Experiment This occurs only if the electrons undergo inelastic collisions Thus, when V A = 4.9 × n Volts ( n = 1, 2, 3, …) the electrons undergo inelastic collisions with Hg atoms! In inelastic collision kinetic energy of electrons becomes internal energy of Hg atoms – Hg atoms absorb the energy of electrons! The current drops because fewer electrons reach the Collector Dr.Fatemah Alkallas 19 V A

21. The Franck-Hertz Experiment Why do we see the drop only at specific voltages? If distribution of energy levels of Hg atom is continuous, then kinetic energy should be transferred to Hg atoms regardless of the energy of electrons However, If we assume that Hg spectrum is discrete, then only when electrons reach certain energies do they undergo inelastic collision with Hg atoms Thus energy spectrum of Hg atoms is such that an electron energy level lies ~4.9 eV above the ground state Dr.Fatemah Alkallas 20

22. The Franck-Hertz Experiment Explained Why did Frank and Hertz not observe dips in the current at other voltage? Their experiment was not sufficiently sensitive Since as soon as electrons gain energy of 4.9 eV they transfer it to Hg atoms, only small fraction of the electrons could have higher energies, (e.g., 6.65 eV), making it difficult to observe current dips associated with other (higher) voltages Dr.Fatemah Alkallas 21

23. We can observe that the voltage difference between two peaks in the plot is equal to first exited state energy And the wavelength when the exited atom return to ground state is Dr.Fatemah Alkallas 22

24. To calculate the needed voltage to make the excitation process Dr.Fatemah Alkallas 23 This match with the experimental results

25. Hg Atom Dr.Fatemah Alkallas 24 الجهد v 4.9 6.7 10.4 التيار I

26. Example. 1 In H atom, Valid = when: And when   Dr.Fatemah Alkallas 25

27. In large n , and in small in use the equations: Dr.Fatemah Alkallas 26

28. Solution First, in small n Dr.Fatemah Alkallas 27

29. Find Dr.Fatemah Alkallas 28

30. so Dr.Fatemah Alkallas 29

31. Second at large n We found: So The difference is very small about 0.3% in large n , but is 62% in small n Dr.Fatemah Alkallas 30

32. Example 2 If the ground state for the Hg atom is -10.4 ev , determine the first and second excited states energy Dr.Fatemah Alkallas 31

33. Hg Atom Dr.Fatemah Alkallas 32 الجهد v 4.9 6.7 10.4 التيار I

34. solution The energy voltage to make the first level excitation is E=4.9 e v So E i -E f =E E i first exited state energy E f. the ground state energy Dr.Fatemah Alkallas 33

35. E i -E f =4.9 E i =4.9+E f =4.9+(-10.9) E i =-5.5 eV So -5.5 ev is first exited state energy Same calculations when the energy voltage to second exited state is 6.7 ev Give -3.7 ev Second exited state energy Dr.Fatemah Alkallas 34