1. x z y   r but  vary 0 - 180 o & 0 - 360 o r 2 = x 2 + y 2 + z 2 x = r sin  cos  y = r sin  sin  z = r cos  r = (x 2 + y 2 + z 2 ) ½  = Acos(z/r)  = Asin{y/(rsin  )} dV = d  = r 2 sin  dr d  d  or ….. dV(shell) = dx dy dz = 4  r 2 sin  dr d  d  Polar Coordinates

2. CM - 2D – Particle on a ring “If a particle is confined to the xy plane, then it has angular momentum along the z axis. angular momentum: L z = mvr L z = x p y – y p x E = K + V = K = ½mv 2 (m/m) = p 2 /2m = L z 2 /2mr 2 = L z 2 /2I moment of inertia: I = mr 2

3. QM - 2D – Particle on a ring “If a particle is confined to the xy plane, then it has angular momentum along the z axis. Ĺ z = x p y – y p x = -iħ d( )/d  Ĥ  = Ĺ z 2 /2I  = -ħ 2 /2I d 2  /d  2  = E  E = m 2 ħ 2 /2I  = Ne im   = (2  ) -1/2 e im  N 2 ∫ 0 2  e -im  e im  d  = 1 quantum #, m (or m l ) = 0, ±1, ±2, ±3, ….

4. 2D – Particle on a ring  = (2  ) -1/2 e im  CM E = ½mv 2 = p 2 /2m = J 2 /2I (I = mr 2 ) QM E = L 2 /2I (I = mr 2 ) L = hr/ = m l ħ = hr/L E = (m l ħ) 2 /2I moon data m = 7.3 x 1022 kg v = 1,020 m s -1 r = 384000 km 1 revolution = 27.3 days What is I? What is E (CM)? What is J? What quantum state m l ?

5. Goal - Separate problem into 2 distinct one-particle problems. Two Particle Problems Particle 1:Mass = m 1 + m 2 Position = center of mass XYZ e.g. X = (m 1 x 1 + m 2 x 2 )/(m 1 + m 2 ) example: translational motion of diatomic molecule (O 2 ) Particle 2:Mass =  = m 1 m 2 /(m 1 + m 2 ) Position = position of smaller particle relative to the center of mass use x, y, z or (best) polar coordinates; r,  example: rotation of 2 bodies attached at fixed distance (O 2 )

6. x -1 +1 e.g. H 2 molecule m 1 = m 1 + m 2 ~ 2 amu Two Particle Problems x = (m 1 x 1 + m 2 x 2 )/(m 1 + m 2 ) x 1 =  m 2 = m 1 m 2 /(m 1 + m 2 ) = ½amu x cm = 0 r 2 = ½ bond distance = 0.60 Å e.g. H atom m 1 = m 1 + m 2 ~ 1 amu x = (m 1 x 1 + m 2 x 2 )/(m 1 + m 2 ) x -1 +1 x 1 ~  m 2 = m 1 m 2 /(m 1 + m 2 ) ~ m e x cm ~ -1 = -0.99891 r 2 ~ r = 0.529 Å m 2 = 9.04 x 10 -28 kg m 2 = 1.67 x 1010 -24 kg

7. V = 0 d = constant Internal motion is rotation only no vibration occurs RIGID ROTOR  rot

8. r 2 = x 2 + y 2 + z 2 x = r sin  cos  y = r sin  sin  z = r cos  r = (x 2 + y 2 + z 2 )  = Acos(z/r)  = Asin{y/(rsin  )} d  = r 2 dr sin  d  d  Ĥ  = -ħ 2 /2I {d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2  d 2 ( )/d  2 } & cot  cos  sin  Ĥ  = Ĺ 2 /2I + V ˆ 3D – rotations (fixed r) rigid rotor 0 - 360 o   r 0 – 180 o Ĺ 2 = -ħ 2 (d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2 d  d 2 ( )/d  2 ) L = {ℓ(ℓ+ 1)} ½ ħ Ĺ z = -iħ d( )/d  L z  = m ℓ ħ

9.  m l = (2  ) -1/2 exp(im l  )  l,m l associated Legendre polynomials  0,0 = 2 -1/2  1,0 = (3/2) 1/2 cos   1,1 = (3/4) 1/2 sin   2,0 = (5/8) 1/2 (3cos 2   2,1 = (15/4) 1/2 sin  cos   2,2 = (15/16) 1/2 sin 2   rot =  m l  l,m l After solving Ĥ  = E rot .... where l = 0, 1, 2, … degeneracy = 2 l + 1 m l = - l, - l +1, - l +2,.... + l Ĥ  = -ħ 2 /2I {d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2  d 2 ( )/d  2 } L = {ℓ(ℓ+ 1)} ½ ħ Ĺ 2 = -ħ 2 (d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2 d  d 2 ( )/d  2 ) L = {ℓ(ℓ+ 1)} ½ ħ L z = m ℓ ħ Ĺ z = -iħ d( )/d  = m ℓ ħ E rot = E  + E  = { l ( l + 1)-m l 2 }(ħ 2 /2I) + m l 2 (ħ 2 /2I) = l ( l + 1)ħ 2 /2I I =  r 2

10.  m l = (2  ) -1/2 exp(im l  )  l,m l associated Legendre polynomials  0,0 = 2 -1/2  1,0 = (3/2) 1/2 cos   1,1 = (3/4) 1/2 sin   2,0 = (5/8) 1/2 (3cos 2   2,1 = (15/4) 1/2 sin  cos   2,2 = (15/16) 1/2 sin 2   rot =  m l  l,m l After solving Ĥ  = E rot .... where l = 0, 1, 2, … degeneracy = 2 l + 1 m l = - l, - l +1, - l +2,.... + l Ĥ  = -ħ 2 /2I {d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2  d 2 ( )/d  2 } E rot = E  + E  = { l ( l + 1)-m l 2 }(ħ 2 /2I) + m l 2 (ħ 2 /2I) = l ( l + 1)ħ 2 /2I I =  r 2 E   for E  = 0 since  0,0 = 2 -1/2 and E rot = 0 + 0 = 0 E   for E  = Ĥ  -ħ 2 /2  [d 2 ( )/d    + cot  d( )/d  ]  = E    -ħ 2 /2  (3/2) 1/2 [d 2 (cos  )/d    + cot  d(cos  )/d  ] = E   cot  cos  sin   -ħ 2 /2  (3/2) 1/2 [-1  - cos  sin  sin  ]  -ħ 2 /2  -2)  ħ 2 /   E   ħ 2 /  and E rot = ħ 2 /  + 0 = ħ 2 /  = ( +1)ħ 2 /2I

11. angular momentum l = 0 m ℓ = 0 L = 0 L z = 0 Ĺ 2 = -ħ 2 (d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2 d  d 2 ( )/d  2 ) L = {ℓ(ℓ+ 1)} ½ ħ Ĺ z = -iħ d( )/d  = m ℓ ħ  rot =   (spherical harmonics) … is an eigenfunction of the Ĺ z and Ĺ 2 operators but not Ĺ, Ĺ x nor Ĺ y. The overall angular momentum scalar value is determined by (Ĺ 2 ) ½. Note that this is not a diagram indicating where the electron may be found. Rather it is a vector diagram representing the limitations on the values of the angular momentum of the electron. The electron with l = 0 (an s orbital) Has no z-component to its angular momentum – it is not confined to a circle.

12. angular momentum l = 0, 1 L = 2 1/2 ħ L z = +1ħ L = 0 L z = 0 L = 2 1/2 ħ L z = -1ħ Ĺ 2 = -ħ 2 (d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2 d  d 2 ( )/d  2 ) L = {ℓ(ℓ+ 1)} ½ ħ Ĺ z = -iħ d( )/d  = m ℓ ħ  rot =   (spherical harmonics) … is an eigenfunction of the Ĺ z and Ĺ 2 operators but not Ĺ, Ĺ x nor Ĺ y. The overall angular momentum scalar value is determined by (Ĺ 2 ) ½.

13. angular momentum l = 2 L = 6 1/2 ħ L z = +2ħ L = 6 1/2 ħ L z = +1ħ L = 6 1/2 ħ L z = 0 L = 6 1/2 ħ L z = -1ħ L = 6 1/2 ħ L z = -2ħ

14. The Hydrogen atom a 3D, 2 particle problem.  =  N  e  N is translational motion of H atom  e is electron motion relative to nucleus  e (r,  )  =  (  )  (  ) R (r) “Radial function” Spherical harmonics

15. Force between two charges in vacuum F = q 1 q 2 /(4  o r 2 ) (N = kg m s -2 ) V = F r = q 1 q 2 /(4  o r) (J = kg m 2 s -2 ) Ĥ  R  - ħ 2 /2  [ 1/r 2 d(r 2 d  /dr)/dr + 1/(r 2 sin  ) d(sin  d  / d  )/d  + 1/(r²sin²  ) d²  /d  ²] - Ze 2 /(4  o r) = E  Hamiltonian: Ĥ  V  = E   e (r,  )  = R (r)  (  )  (  ) V = -Ze 2 /(4  o r) Applies to H atom and any H-like ion with only 1 e -.  ( R (r)  (  )  (  ) ) VV

16. Spherical harmonics  (  )  (  ) These solutions are identical to the rigid rotor model Ĥ  -ħ 2 /2  [1/r 2 d (r 2 d  / d r)/ d r + 1/(r 2 sin  ) d (sin  d  / d  )/ d  + 1/(r²sin²  ) d ²  / d  ²] - Ze 2 /(4  o r) reduces to …… E SH = ħ 2 l ( l + 1)/2  r 2 ĤR = -ħ 2 /2  [1/r 2 d(r 2 dR/dr)/dr + {ħ 2 l ( l +1)/2  r 2 - Ze 2 /(4  o r)}R Ĥ  (r,  )  = Ĥ(  ℓ,mℓ (  )  mℓ (  ) ) + Ĥ R (r)

17. Ĥ R = -ħ 2 /2  [1/r 2 d(r 2 dR/dr)/dr + {ħ 2 l ( l +1)/2  r 2 - Ze 2 /(4  o r)}R As the solutions to the  function already existed in the form of the associated Legendre Polynomials ….. An exact solution to the Hamiltonian for R exists using pre-existing mathematics ….. the associated Laguerre Polynomials  n,ℓ,mℓ (r,  )  =  ℓ,mℓ (  )  mℓ (  ) R n,ℓ (r) These solutions contain the quantum numbers (n and ℓ) such that …. n = 1, 2, 3, ….. (principle quantum #) ℓ = 0, 1, … n – 1 (angular momentum q# as in  ) m ℓ = -ℓ, -ℓ+1, … +ℓ (magnetic quantum number as in  and 

18. R(r) = radial function R = cst * (n-1) th order polynomial in r * e -Zr/2a a = 4  o ħ 2 /  e 2 (units = m) which happens to equal the Bohr orbit radii (0.529Å for H)  n,ℓ,mℓ (r,  )  =  ℓ,mℓ (  )  mℓ (  ) R n,ℓ (r) ĤR  -ħ 2 /2  [1/r 2 d(r 2 dR / d r)/ d r+ { ħ 2 l ( l +1)/2  r 2 - Ze 2 /(4  o r)}R = ER E  Z 2 e 4  /(8  o 2 h 2 n 2 ) eq 11.66

19. What is  n l m ?  100 = 1s = (  -½ ) (Z/a) 3/2 exp(-Zr/a)  200 = 2s = (32  -½ (Z/a) 3/2 {2-(Zr/a)} exp(-Zr/a)  210 = 2p z = (32  ) -½ (Z/a) 5/2 r exp(-Zr/2a) (cos  )  211 = (64  ) -½ (Z/a) 5/2 r exp(i  ) exp(-Zr/2a) sin   21-1 = (64  -½ ) (Z/a) 5/2 r exp(-i  ) exp(-Zr/2a) sin  2p x = 2 -1/2 (  211 +  21-1 )2p y = -i2 -1/2 (  211 -  21-1 ) If any 2 wave functions satisfy H and give the same E, then any linear combination of those 2 wave functions (renormalized) will also give the same E. 2p x = (32  -½ (Z/a) 5/2 r exp(-Zr/2a) sin  cos  2p y = (32  -½ (Z/a) 5/2 r exp(-Zr/2a) sin  sin  2p z = (32  -½ (Z/a) 5/2 r exp(-Zr/2a) cos  a = 4  o ħ 2 /  e 2 (units = m)

20. 1s 2s 3s H-atom – Radial Functions

21. 2p 3p 3d H-atom – Radial Functions

22. dV = dx dy dz = d  = 4  r 2 dr sin  d  d  dV = dx dy dz dV dV (cube) = dv x dv y dv z = ? dr dV = 4  (r+dr) 3 /3 – 4  r 3 /3 = 4  r 2 dr 4  r 3 /3 + 4  r 2 dr + 4  rdr 2 + 4  dr 3 /3

23. r2R2r2R2 1s 2s 3s Probability = r 2 R 2  R 2 Even though the radial function for s orbitals is maximal at r = 0 the r 2 term in d , drops the probability to 0 at the nucleus. r2R2r2R2 r/a

24. r2R2r2R2 1s 2s 3s  1s = (  -½ ) (1/a) 3/2 exp(-r/a) r/a r mp = r when d (r 2  1s )dr = 0 = a

25. r2R2r2R2 2p 3p 3d

26. Ĥ  -ħ 2 /2  [1/r 2 d (r 2 d  / d r)/ d r + 1/(r 2 sin  ) d (sin  d  / d  )/ d  + 1/(r²sin²  ) d ²  / d  ²] - Ze 2 /(4  o r) = E  and =  E =  Z 2 e 4  /(8  o 2 h 2 n 2 ) eq 11.66 eV

27. Probability = ∫  e *  e d  = 4  ∫ r (r+dr) r 2 R dr ∫  (  +d  ) sin  d  ∫  (  +d  ) d   e (r,  )  =  (  )  (  ) R (r) 1 = 4  ∫ 0 ∞ r 2 R dr ∫   sin  d  ∫   d  =11 1 prob = 4  ∫ r (r+dr) r 2 R dr (over all space)

28. Probability = ∫  e *  e d  Prob = 4  ∫ 0 0.1Å r 2 R 2 dr = 4  ∫ 0 0.189a r 2 R 2 dr 51 R = 2 (1/a) 3/2 exp(-r/a) R 2 = 4/a 3 exp(-2r/a) Prob = 16  a 3  ∫ 0 0.189a r 2 exp(-2r/a) dr ∫ r 2 exp(-2r/a) dr = (-ar 2 e -2r/a )/2 – (-a  r e -2r/a dr) = (-ar 2 e -2r/a )/2 – (-a a 2 /4 e -2r/a (-2r/a – 1) ] 0 0.189a  x m e cx dx = (x m e cx )/c – m/c  x (m-1) e cx dx  x 2 e cx dx = (x 2 e cx )/c – 2/c  x e cx dx  x e cx dx = e cx )/c 2 (cx – 1) c = -2/a Prob = 4  4  a 3  [(-ar 2 e -2r/a )/2 – (-a a 2 /4 e -2r/a (-2r/a – 1) ] 0 0.189a a (Å)Zr-Zr/aA2A2 e -2zr/a r 2 /b-2r/b 2 2/b 3 e -zr/a Prob 0.52911.00E-01-3.78E+0027.00.68518-2.65E-03-1.40E-02-3.70E-0216.80E-03 0.529101.00E-01-3.78E+0127020.50.02281-2.65E-04-1.40E-04-3.70E-0517.28E-01

29. What is an orbital? the size of the orbital depends on confidence level desired. e.g.....  0 r r 2 R dr *  0  sin   d  *      d  = 0.9 … says that the probability is 90% of finding e - within the distance r of the nucleus. r is different at different  &  angles for non s orbitals. The shape of an orbital is the volume enclosed by a surface of constant probability density =  |  2 | d  a one electron spatial wave function

30.  100 /1s(  -1/2  (Z/a) 3/2 exp(-Zr/a)  200 /2s(32  -1/2 (Z/a) 3/2 (2-Zr/a) exp(-Zr/2a)  21-1 (64  -1/2 (Z/a) 5/2 sin  exp(-i  ) r exp(-Zr/2a)  211 (64  -1/2 (Z/a) 5/2 sin  exp(i  ) r exp(-Zr/2a) 2p x (32  -1/2 (Z/a) 5/2 r exp(-Zr/2a) sin  cos  2p y (32  -1/2 (Z/a) 5/2 r exp(-Zr/2a) sin  sin   210 /2p z (32  -1/2 (Z/a) 5/2 r exp(-Zr/2a) cos  (2p x ) 2 (32  -1 (Z/a) 5 r 2 exp(-Zr/a) sin 2  cos 2  (2p y ) 2 (32  -1 (Z/a) 5 r 2 exp(-Zr/a) sin 2  sin 2  (2p z ) 2 (32  -1 (Z/a) 5 r 2 exp(-Zr/a) cos 2  Prob =  |  2 | d   0 r r 2 R dr *  0  sin   d  *      d  = 0.9

31. r/a Prob:  = 90 &  = 0 r22r22 2p x 2s 2p y & 2p z

32. r/a Prob:  = 90 &  = 90 r22r22 2p y 2s 2p x & 2p z

33. r/a Prob:  = 45 &  = 35 r22r22 2p y 2p x 2p z 2s 1s

34. angular momentum l = 0 m ℓ = 0 L = 0 L z = 0 Ĺ 2 = -ħ 2 (d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2 d  d 2 ( )/d  2 ) L = {ℓ(ℓ+ 1)} ½ ħ Ĺ z = -iħ d( )/d  = m ℓ ħ  rot =   (spherical harmonics) … is an eigenfunction of the Ĺ z and Ĺ 2 operators but not Ĺ, Ĺ x nor Ĺ y. The overall angular momentum scalar value is determined by (Ĺ 2 ) ½.

35. angular momentum l = 1 m ℓ = 0 L = √2ħ L z = 0 Ĺ 2 = -ħ 2 (d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2 d  d 2 ( )/d  2 ) L = {ℓ(ℓ+ 1)} ½ ħ Ĺ z = -iħ d( )/d  = m ℓ ħ Note that this is not a diagram indicating where the electron may be found. Rather it is a vector diagram representing the limitations on the values of the angular momentum of the electron. The electron with l = 0 (an s orbital) Has no z-component to its angular momentum – it is not confined to a circle.

36. angular momentum l = 0, 1 L = 2 1/2 ħ L z = +1ħ L = 0 L z = 0 L = 2 1/2 ħ L z = -1ħ Ĺ 2 = -ħ 2 (d 2 ( )/d  2 + cot  d( )/d  + 1/sin 2 d  d 2 ( )/d  2 ) L = {ℓ(ℓ+ 1)} ½ ħ Ĺ z = -iħ d( )/d  = m ℓ ħ  rot =   (spherical harmonics) … is an eigenfunction of the Ĺ z and Ĺ 2 operators but not Ĺ, Ĺ x nor Ĺ y. The overall angular momentum scalar value is determined by (Ĺ 2 ) ½.

37. angular momentum l = 2 L = 6 1/2 ħ L z = +2ħ L = 6 1/2 ħ L z = +1ħ L = 6 1/2 ħ L z = 0 L = 6 1/2 ħ L z = -1ħ L = 6 1/2 ħ L z = -2ħ