Physics 03, Fall 01 Modern Physics. Monday, Oct. 8 th, 01. Finish up examples for Ch. 8 Computer Exercise. Announcements: Take home Exam #1: Average 84.1, Average both 63.0 Quiz on Friday on Ch. 8 or Ch. 9 Submit a detailed outline of your paper by Monday, Oct. 9th
The General Equation was ψ (r ) = R(r)Θ(θ)Φ(φ) We have now solved for Θ(θ)Φ(φ) The solution for Θ(θ)Φ(φ) are Spherical Harmonics Y l m l (θ,φ) ψ (r ) = R(r)Y l m l (θ,φ) m l = 0,±1,±,... l m l l m l l
m d ( rr) + U dr eff ( rr) = E( rr) The final solution is ψ (r ) = R n,l (r)y l m l (θ,φ) a 0 = c ( ) ke mc
In General E n0 = m( ) ke n : Bohr model The Rydberg series are reproduced, but the physics is different. There are selection rules for optical excition or emission Δn any value Δl = ±1 Δm l = 0,±1 If there is an electron in the 3d state what lines will you observe?
m( ) ke E n0 = E nl > =E n0 = E 0 n n : Bohr model Which transitions are allowed Photon carries angular momentum Conservation of total angular momentum l f l i = 1 or Δl=±1 Optical Transition l f l i = 1 or Δl=±1
When we consider multi-electron systems the amplitude of the state at r=0 will become important. For small r all of the r /a R n0 states Ae For small r all of the r /a R n1 states Are For small r all of the R n states Ar r /a e If you are and l. one or more of these plots you should be able to determine n
I put an electron in the 4f level. It decays to the ground state, what are all possible emission energies? Δl ± 1, but it has to decay so from f d, must be 3d ΔE=13.6 1 9 1 16 = 0.66eV Next d p, must be p ΔE=13.6 1 4 1 9 = 1.85eV Next p s, must be 1s ΔE=13.6 1 1 4 = 10.eV What if this was singly ionized He? Can it go from the 4f 4d???
Ψ 1,0,0 = R 10 Y 0 0 = e r /a 0 πa 0 3 Ψ 1,0,0 = e r /a 0 πa 0 3 Radial Probability Distribu0on P(r)dr P(r)dr = Ψ 4πr dr P(r) = r R = rr(r) This is why we kept wrisng the radial equason in terms of rr(r)!
Calculate the electron-proton separation for Ψ 1,0,0 = e r /a 0 πa 0 3 Calculate the most probable position and compare to the average distance. The most probable position is found at the maximium of P(r). P 1s = 4r a 0 3 e r /a 0 dp 1s dr = 4r 3 a e r /a 0 r 0 a 0 r = a 0 + r = 0
Calculate the electron-proton separation for Ψ 1,0,0 = e r /a 0 The average postion is given by 0 r = rp(r)dr = 4 a 0 3 0 r 3 e r /a 0 dr πa 0 3 r = a 0 ( 4 3! ) = 3 a 0 Let z=r/a 0 r = a 0 4 0 z 3 e z dz definite integral z n e z dz = n! 0 r = a 0 Why?
P(r) = r R = rr(r) P 1,0 (r) = r e r /a 0 πa 0 3 P,0 (r) = r r a 0 8a 0 3 e r /a 0 P 3,0 (r) = r 1 r + 3a 0 7 7a 0 3 r a 0 e r /3a 0 P,1 (r) = r 4 e r /a 0 P 5 3, (r) = 8r 6 e r /3a0 7 4a 0 98415a 0
P(r) = r R = rr(r) Know how to answer quessons about the P(r). How many nodes is a give quantum state (n,l)? (1,0)=(1,s)=0 (,0)=(,s)=1 (3,0)=(3,s)= States n l 1 p states (1,1)=(1,p) doesn t exist (,1)=(,p)=0 (3,1)=(3,p)=1 States n l 1 d states: you do this.
What are the possible Quantum numbers for the state n=1 state? Solution: n= 1 l = 0 m l = 0 Only one state Q numbers (1,0,0) 1s state Quantum Numbers Principal n= 1,, 3,... Orbital l = 0,1,,...(n-1) Magnetic m l l
What are the possible Quantum numbers for the state n= state? Solution: n= l = 0 m l = 0 n= l = 1 m l = 1 n= l = 1 m l = 0 n= l = 1 m l = +1 Quantum Numbers Principal n= 1,, 3,... Orbital l = 0,1,,...(n-1) Magnetic m l l Four states Q # s (,0,0), (,1, 1), (,1,0), (,1,+1) s and p
What are the possible Quantum numbers for the state n=3 state? Solution: n= 3 l = 0 m l = 0 n= 3 l = 1 m l = 1 n= 3 l = 1 m l = 0 n= 3 l = 1 m l = +1 n= 3 l = m l = n= 3 l = m l = 1 n= 3 l = m l = 0 n= 3 l = m l = +1 n= 3 l = m l = + Quantum Numbers Principal n= 1,, 3,... Orbital l = 0,1,,...(n-1) Magnetic m l l Nine states 3s, 3p, and 3d Can you do n=4?
Prove that Y 0 0 is a solution of the angular equation. 1 d dθ sinθ sinθ dθ dθ = l(l + 1) m Y 0 0 = 1 l sin θ Θ π l=0, m l = 0 1 d dθ sinθ sinθ dθ dθ = 0 dy 0 0 dθ = 0 0 = 0 : QED
Prove that Y 1 1 is a solution of the angular equation. 1 d dθ sinθ sinθ dθ dθ = l(l + 1) m Y 1 l sin θ Θ 1 = 1 l = 1, m l = 1 1 d dθ sinθ sinθ dθ dθ = 1 d d sinθ Asinθ sinθ dθ dθ = 1 sin θ Θ 1 sin θ Asinθ 1 d ( sinθ cosθ ) sinθ dθ = 1 sin θ sinθ Write Y 1 1 = Asinθ 1 sinθ cos θ sin θ + sinθ 1 sinθ = 0 3 π sinθeiφ
1 sinθ cos θ sin θ + sinθ 1 sinθ = 0 1 sinθ 1 sin θ + sinθ 1 sinθ = 0 1 1 sinθ + sinθ sinθ sinθ = 0 0 = 0 QED You do Y 1 0!
Prove that the R s wave funcson sassfies the radial Schrödinger EquaSon. Find A R s = A r d a 0 e r /a 0 dr d ( rr) = m ke n= and l=0 dr ( rr) = m ke l(l + 1) + E r mr (rr) ( ) 4 ( ) (rr) = 1 4α αr (rr) r + m ke α = mke = a 0 Use R 0 and do the left hand side. d ( ) = A 4 α dr rr 0 + 5r α r 4α 3 e r /α You do the right hand side. P(r)dr = 4πA r r a 0 e r /α dr = 3πα 3 A = 1 A = 0 0 1 3πα 3
Consider a parscle in a rigid rectangular box with sides a and b=a/. Find the lowest six energy levels. ψ(x, y)= b ψ(x, y)= sin k x x b ( ) sin k y y sin n xπx sin n yπy a a / k x = n xπ a : k y = n yπ : E= ( k) a m ( ): ψ=0 for x=a and y=a/ Is there any degeneracy? Can you draw P(r) plots? What is the wavelength. Lowest levels n x n y E/E 0 1 1 5 1 8 3 1 13 1 17 0 4 1 0 E=E x + E y = n xπ a m + n yπ a m = ma ( n x + 4n ) y = E ( 0 n x + 4n ) y
Show that the allowed energies of a mass M confined in a 3 D rectangular rigid box with sides a, b and c are give by. E= = M n x a + n y b + n z c ( ψ(x, y)= ) 3/ abc sin ( k xx) sin ( k y y)sin ( k z y): ψ =0 for x=a, y=b, z=c k x = n xπ a : k y = n yπ b : k z = n zπ c E= ( k) m E= ( k) M = π M n x a + n y b + n z c What is the wavelength for the,1,1 state? What are the energy levels? Is there degeneracy?
(a) If the H atom is in the n=5 and m= state how many different states are consistent with this informason? (b) Answer the same quesson in general for arbitrary values of n and m? Answer to (a): m l and l < n We know m= l < 5 = n So l=, 3, 4 Answer to (b): m l and l < n l can be m, m + 1,...,n 1 Which gives n- m possible states