1. It should be no surprise that the Schrödinger result exactly matches the Bohr result. Recall that Bohr was able to explain the spectrum of the H atom, exactly matching the best experimental data available at the time.

2. Explanation of the spectrum of the H atom

3. Explanation of the spectrum of the H atom
Energy Level Diagram for the H atom

4. Explanation of the spectrum of the H atom
Energy Level Diagram for the H atom
Consider two energy levels for the H atom, the upper one characterized by the quantum number nupper (abbreviated nu) and a lower level characterized by the quantum number nlower (abbreviated nl).

5. Two situations: (1) absorption and (2) emission.nu nl
absorption (photon absorbed)

6. nu nl
emission (photon emitted)

7. In both the absorption and emission processes
Recall so that

8. In both the absorption and emission processes
Recall so that k
From the formula En = we can n2
write :

9. and

10. and
Since we have

11. and
Since we have

12. and
Since we have

13. and
Since we have

14. That is

15. That is
Set

16. That is
Set
So that

17. That is
Set
So that
This is called the Rydberg formula and RH is called the Rydberg constant.

18. Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states.

19. Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states.
RH = cm-1 and k = x 10-18J

20. Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states.
RH = cm-1 and k = x 10-18J
Approach 1

21. Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states.
RH = cm-1 and k = x 10-18J
Approach 1

22. Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states.
RH = cm-1 and k = x 10-18J
Approach 1

23. = cm-1

24. = cm-1
Hence
= x10-5 cm
= nm

25. = cm-1
Hence
= x10-5 cm
= nm
Now use

26. = x 10-19J

27. = x 10-19J
Approach 2
Use k
En = - n2

28. So that
k k
E5 = E2 = -

29. So that
k k
E5 = E2 = -
= x J = x J

30. So that
k k
E5 = E2 = -
= x J = x J
Therefore = E5 - E2 = x J

31. So that
k k
E5 = E2 = -
= x J = x J
Therefore = E5 - E2 = x J
Now so

32. So that
k k
E5 = E2 = -
= x J = x J
Therefore = E5 - E2 = x J
Now so
= x 10-7 m

33. Definition: The ionization energy is the energy required to remove an electron from one mole of a substance in its ground state in the gas phase.

34. Definition: The ionization energy is the energy required to remove an electron from one mole of a substance in its ground state in the gas phase. For example, for substance X,
X(g) X(g)+ + e-

35. Problem: Calculate the ionization energy for the hydrogen atom.
(k = x 10-18J and NA = x 1023 mol-1).

36. Problem: Calculate the ionization energy for the hydrogen atom.
(k = x 10-18J and NA = x 1023 mol-1).
The process is:
H(g) H+(g) + e-

37. Problem: Calculate the ionization energy for the hydrogen atom.
(k = x 10-18J and NA = x 1023 mol-1).
The process is:
H(g) H+(g) + e-
The quantum numbers are nl = 1 and nu = (because the electron is completely removed from the atom).

38. Therefore = E1
= (-k)

39. Therefore = E1
= (-k)
= k
= x 10-18J

40. This corresponds to the energy to remove an electron from one atom.
Therefore = E1
= (-k)
= k
= x 10-18J
This corresponds to the energy to remove an electron from one atom.

41. Therefore = E1
= (-k)
= k
= x 10-18J
This corresponds to the energy to remove an electron from one atom. To get the ionization potential, we need the energy expended for 1 mole of atoms. Hence

42. Therefore = E1
= (-k)
= k
= x 10-18J
This corresponds to the energy to remove an electron from one atom. To get the ionization potential, we need the energy expended for 1 mole of atoms. Hence
Ionization
energy = x J ( x 1023 mol-1)
= x 106 J mol-1

43. Definition: The electron affinity is the energy required to add an electron to one mole of a substance in its ground state in the gas phase.

44. Definition: The electron affinity is the energy required to add an electron to one mole of a substance in its ground state in the gas phase. For example, for substance Y,
Y(g) + e Y(g)-

45. Exercise: Calculate the electron affinity for the hydrogen positive ion.

46. Exercise: Calculate the electron affinity for the hydrogen positive ion. (k = x 10-18J and NA = x 1023 mol-1).

47. Exercise: Calculate the electron affinity for the hydrogen positive ion. (k = x 10-18J and NA = x 1023 mol-1). (By convention the electron affinity is the difference between the energy of the product formed and the energy of the starting species, in this case E(H) – E(H+) ).

48. How does the Schrödinger eq. do for more complex systems, e. g
How does the Schrödinger eq. do for more complex systems, e.g. He atom, CO molecule, etcetera?

49. How does the Schrödinger eq. do for more complex systems, e. g
How does the Schrödinger eq. do for more complex systems, e.g. He atom, CO molecule, etcetera?
It turns out that the Schrödinger equation is too complex to solve exactly for systems with more than one electron. However, approximate solutions have been obtained which match up with experimental results.

50. How does the Schrödinger eq. do for more complex systems, e. g
How does the Schrödinger eq. do for more complex systems, e.g. He atom, CO molecule, etcetera?
It turns out that the Schrödinger equation is too complex to solve exactly for systems with more than one electron. However, approximate solutions have been obtained which match up with experimental results.
For example, some of the energy levels of the He atom have been determined to around 10 significant figures! The ground state energy level for He is known to over 40 significant figures!

51. It is generally accepted that the Schrödinger equation provides a very good description of the ground and excited states of atomic and molecular systems.

52. It is generally accepted that the Schrödinger equation provides a very good description of the ground and excited states of atomic and molecular systems.
The work of Schrödinger was one of the landmark breakthroughs in all of science.

53. The fact that the H atom can be solved exactly has turned out to be important for studying more complex atoms.

54. The fact that the H atom can be solved exactly has turned out to be important for studying more complex atoms. It is assumed the electronic behavior in many-electron atoms is not too different from that in the H atom.

55. The fact that the H atom can be solved exactly has turned out to be important for studying more complex atoms. It is assumed the electronic behavior in many-electron atoms is not too different from that in the H atom. So the results from the H atom can be used as a first approximation for describing the behavior of the electrons in more complex atoms.

56. The fact that the H atom can be solved exactly has turned out to be important for studying more complex atoms. It is assumed the electronic behavior in many-electron atoms is not too different from that in the H atom. So the results from the H atom can be used as a first approximation for describing the behavior of the electrons in more complex atoms.
The justification for this approach is that it works!

57. Some final thoughts on the Bohr model

58. Some final thoughts on the Bohr model
Obviously the result obtained by Bohr was very useful to Schrödinger.

59. Some final thoughts on the Bohr model
Obviously the result obtained by Bohr was very useful to Schrödinger.
Connection with Heisenberg Uncertainty Principle.

60. Some final thoughts on the Bohr model
Obviously the result obtained by Bohr was very useful to Schrödinger.
Connection with Heisenberg Uncertainty Principle.
According to Bohr, an electron is always circling around the nucleus in a well specified orbit.