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Schrodinger Equation of Hydrogen Atom
- 1. The Hydrogen Atom By Saadia Shaukat 1
- 2. The Hydrogen Atom β’Introduction β’Schrodinger equation of Hydrogen atom β’Quantum number β’Magnetic Effect β’Example 2
- 3. Introduction A hydrogen atom is an atom of the chemical element hydrogen. The electrically neutral atom contains a single positively charged proton and a single negatively charged electron bound to the nucleus by the Coulomb force. General Symbol 1H β’Protons1 β’Neutrons 0 3
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- 5. Schrodinger Equation of Hydrogen Atom We know that Schrodinger equation: πππΏ πππ + πππΏ πππ + πππΏ πππ + ππ ππ ππ π¬ β π πΏ = π We can also write as: Laplacian ππ πΏ + ππ ππ ππ π¬ β π π³ = π 5
- 6. Continue.. We also know that: π = β πππ ππ πΊππ Here (r) is the distance between electron and radius. Hydrogen atom is a spherical symmetrical system. We canβt use cartesian co-ordinates, we use spherical co-ordinates. 6
- 7. Contiβ¦ Relation between cartesian and spherical co- ordinates is given as: π = πππβπ π π π = πππ§βπ ΰ΅ π² π± π = ππ + ππ + ππ 7
- 8. Contiβ¦. Schrodinger equation in spherical co-ordinates: π π π ππ ππ ππΏ ππ + π ππππππ π ππ½ πππΞΈ ππΏ ππ½ + π πππππππ½ ππ ππ½π + ππ π π ππ π¬ β π πΏ = π β¦ β¦ π we first separate the radial (r )and angular variables(ΞΈ,Ξ¦). For this we substitute: πΏ π, π½, π = πΉ π π π½, π β¦ . π By solving equation 2) we get: 8
- 9. Contiβ¦ ππΏ ππ = π ππΉ ππ β¦ . π ππΏ ππ½ = πΉ ππ ππ½ β¦ . π ππΏ ππ = πΉ ππ ππ Again differentiate we get: πππΏ πππ = πΉ πππ πππ β¦ . . π 9
- 10. Contiβ¦ By putting equ 3,4 and 5 in equ 1 we get: π ππ π ππ ππ π ππ ππ + π ππππππ π ππ½ πππΞΈπΉ ππ ππ½ + π πππππππ½ πΉ πππ ππ½π + ππ ππ ππ π¬ β π ππ = π Multiplying above expression by π«π ππ we get: ( π«π ππ ). π π«π π ππ« π«ππ ππ ππ« + π«π ππ . π π«ππ¬π’π§π π ππ π¬π’π§ΞΈπ ππ ππ + π«π ππ . π π«ππ¬π’π§ππ π πππ ππ½π + π«π ππ . ππππ π‘π π β π― ππ = π 10
- 11. Conti.. π π π ππ« π«π ππ ππ« + π π . π π¬π’π§π π ππ π¬π’π§ΞΈ ππ ππ + π π . π π¬π’π§ππ πππ ππ + π«π. ππππ π‘π π β π― = π And now we can write as: π πΉ π ππ ππ π π π π + π π . π ππππ π ππ½ πππΞΈ ππ ππ½ + π π . π πππππ½ πππ ππ + ππ. ππ ππ ππ π¬ β π = π π πΉ π ππ ππ π π π π + ππ. ππ ππ ππ π¬ β π = β π π π ππππ π ππ½ πππΞΈ ππ ππ½ + π πππππ½ πππ ππ 11
- 12. Conti.. Hence this equation can be correct only if both sides are equal to some constant. Let this constant be π π + π The radial equation is: π πΉ π ππ ππ π π π π + ππ. ππ ππ ππ π¬ β π = π π + π β¦ . π 12
- 13. Conti.. And angular equation is: β π π π ππππ π ππ½ πππΞΈ ππ ππ½ + π πππππ½ πππ πππ = π π + π β¦ . π Now again we have two variables there for π π½, π = πΈ π½ π(π) Put in above expression we get: β π πΈπ π ππππ π ππ½ ππππ½ ππ ππ½ π + π πππππ½ ππ π πππ πΈ = π π + π 13
- 14. Contiβ¦ Multiplying above expression with π ππ2ΞΈ we get: β π πΈ ππππ½ π π π½ ππππ½ π π π π½ β π π π ππ π ππ = π π + π πππππ½ Rearrange the equation we get: π πΈ ππππ½ π π π½ ππππ½ π π π π½ + π π + π πππππ½ = β π π π ππ π ππ 14
- 15. Conti.. Again two sides of the equation are function of different variables. They must be equal to some constant. β’ Let this constant be ππ π π πΈ ππππ½ π π π½ ππππ½ π πΈ π π½ + π π + π ππππ π½ = ππ π β¦β¦. 8 And β π π π π π π ππ = ππ π β¦ . π 15
- 16. Conti.. Equation 9 can be written as: π ππ π ππ = βπ ππ π π ππ π ππ + π ππ π = π β¦ ππ For more simplification we multiplying equation 8 with ΰ΅ πΈ ππππΞΈ we get: 16
- 17. Conti.. β’ πΈ πΊππππ½ π πΈ ππππ½ π π π½ ππππ½ π πΈ π π½ + π π + π ππππ π½ = ππ π After simplification we get: π ππππ½ π π π½ ππππ½ π πΈ π π½ + π π + π πΈ β ππ π πππππ½ =0 β¦..11 Multiplying equation 6 by ΰ΅ π π«π π π«π π π π ππ« π«π ππ ππ« + π«π. ππππ π‘π π β π― = π₯ π₯ + π 17
- 18. Conti.. πΉ ππ π πΉ π π π ππ π πΉ π π + ππ ππ ππ π¬ β π β π π + π ππ πΉ = 0 Thus we break down the Schrodinger equation of hydrogen atom into three ordinary differential equation, each having a single variable. We trying to find out acceptable solutions of these equation. 18
- 19. Quantum Number Solution of the Equations Solution of F(Ο) equation 10 is: π = π¨πππππ A is constant. Since Ο repeats its value after 2Ο. Then: π(π + ππ ) = π(π) π¨ππππ(π+ππ ) = π¨πππππ 19
- 20. Conti.. ππππ(ππ ) = π πππ ππππ β ππππ ππππ = π πππ ππππ = π This can only happen when ππ is zero or positive or negative integer. β’ Hence ππ = π, Β±π, Β±π, Β±π, β¦ . β’ ππ is magnetic quantum number. 20
- 21. Conti.. Finding the solution of equation 11 β’ πΈπππ π½ = π΅πππ π·π ππ πππ(π½) π΅πππ is a constant. π·π ππ is associated Legendre polynomial . Solution is acceptable if π β₯ ππ Here l is orbital quantum number. 21
- 22. Conti... For solving equation 12 we must specify v(r). β’ In this case: π π = β πππ ππ πΊππ Then the solution is given as: πΉπ,π π = π΅π,ππ ΰ΅ βππ πππ πππ πππ π³π+π ππ+π πππ πππ π΅π,π is a constant. 22
- 23. Conti.. π³π+π ππ+π is associated Laguerre polynomial. Where β’ ππ = πππΊπ π πππ β’ This solution is only acceptable when E is positive or has one of the negative values π¬π (which correspond the bound states). π¬π = β πππππ ππΊπ πππ π ππ 23
- 24. Contiβ¦ Where n is an integer which must be equal to or greater than (l+1). π β₯ π + π i-e n=l+1,l+2,l+3,β¦.. Or l=0,1,2,3,β¦β¦.. (n-1) n is principle quantum number. 24
- 25. Magnetic effect on hydrogen atom The Dutch physicist Pieter Zeeman (and 1902 physics Nobelist) observed with a state of the art spectrometer of the time, (which we would now consider pretty crude) that many spectral lines split in a magnetic field into three (and more) spectral lines, one stays at the original position, the spacing of the other two depends linearly on the strength of the magnetic field. It is called the Zeeman effect. β’ A good theory of the hydrogen atom needs to explain this !! 25
- 26. Normal zeeman effect β’ Normal Zeeman effect, which is actually not observed with modern spectrometers, historically βnormalβ because of easy explanation β’ Model the electron in the H atom as a small permanent magnet. β’ Think of an electron as an orbiting circular current loop of I = dq / dt around some nucleus (that was not known to exist at that time). β’ The current loop has a magnetic moment ΞΌ and the period T = 2Οr / v. (donβt confuse this ΞΌ with reduced mass) β’ where L = mvr is the magnitude of the orbital β’ angular momentum for a circular path. 26
- 27. Normal zeeman effect β’ When there is no magnetic field to align them, doesnβt have a effect on total energy. In an external magnetic field a dipole has a potential energy 27
- 28. Normal zeeman Effect β’ If there is a magnetic field in direction z, it will act on the magnetic moment, this brings in an (extra) potential energy term into the Hamiltonian operator β’ So an external magnetic field should have an effect on atoms, spectral lines are βfingerprintβ characteristics of atoms, in an external magnetic field, each spectral line should be splitting into three lines, distance between two extra lines proportional to the strength of the magnetic field, tested by experiment, observed, part of Nobel prize 28
- 29. The Normal zeeman effect for Hydrogen β’ The potential energy due to the external magnetic field is quantized due to the magnetic quantum number mβ. β’ When a magnetic field is applied, the otherwise degenerate 2p level of atomic hydrogen is split into three different energy states with energy difference of ΞE = ΞΌBB Ξmβ. 29
- 30. Conti⦠⒠The larger B, the larger the splitting, if B is switched off suddenly, the three lines combine as if nothing ever happened, total intensity of lines remains constant in the splitting 30
- 31. Contiβ¦ πΌ = π πππ π³. π© = ππ© πππ π³π = βππππ β’ What is really observed with good spectrometers: there are a lot more lines in atomic spectra when they are in a magnetic field ! So called Anomalous Zeeman effect, which is the only one observed with good spectrometers !!. 31
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