PROBLEM SET SOLUTIONS CHAPTER 8, Levine, Quantum ... - Njit

PROBLEM SET SOLUTIONS CHAPTER 8, Levine, Quantum Chemistry, 5th d. 8.27(a) or the ground state of the hydrogen atom use the aussian trial function φ = exp [-cr2/ (a02)]. ind the optimum value of c and the percent error in the energy.

(b) Multiply the trial function by the Spherical armonic Y20, and the minimize the variational integral. This yields an upper bound to the energy of which hydrogen atom state?

- c is variable parameter. respect to c:

ind c by minimizing with

∂ /∂c = 0. or the

atom,

= -h2/(2µ2) ∇2 - Z (e')2/r,

∇2 = ∂2/∂r2 + (2/r) ∂/∂r - L2/(r2h2)

L2 φ = L2 exp [-cr2/ (a02)] = 0 since φ has no angular dependence. µ ~ me → a0 = h2/[ me(e')2] Let c' = c/ a02 → φ = exp [-c'r2] φ = -h2/(2me 2) {-6c'r2 exp [-c'r2] + 4(c')2r2 exp [-c'r2]} - {Z (e')2/r} exp [-c'r2] ∫φ* φ dτ = - (h2/me ) (3/16) (π/c'')1/2 - (e')2/[2 c''], c'' = 2c' ∫φ*φ dτ = π1/2/[4c''3/2] = ∫φ* φ dτ / ∫φ*φ dτ = h23 c''/(4 me) - 2(e')2(c''/π)1/2 ∂ /∂c = (∂ /∂c'') (∂c''/∂c) = (∂ /∂c'') (2/a02) = 0 if (∂ /∂c'') = 0 (∂ /∂c'') = h23/(4me) - 2(e')2(1/2)/ (c''π)-1/2 → c'' = (e')4 me2 16/[π h49] = 2c/a02 → c = a02(e')4me216/[2π h49] a0 = h2/[ me(e')2] → c = 8/(9π) = ∫φ* φ dτ / ∫φ*φ dτ >

1

(n = 1 ground state of

Use the value for c to evaluate :

atom)

= - a02(e')2(9π/64) /{ a03(27π2/256)} = -0.424 (e')2/a0 or

atom, 1

n

= -Z2 me(e')4/[2n2 h2]

= - me(e')4/[2h2]

% error =  -

1

/  1x 100 = 0.424 - 0.500/(0.500)x100

= 15% 8.14 (b) φ = exp [-cr2/ (a02)] Y20 L2 φ = exp [-cr2/ (a02)] L2 Y20 = exp [-cr2/ (a02)] [2 (2+1) h2]Y20 = 6 h2 φ = -h2/(2µ2) ∇2 - Z (e')2/r, ∇2 = ∂2/∂r2 + (2/r) ∂/∂r - L2/(r2h2) φ = -h2/(2me ) {-6c'r2 exp [-c'r2] Y20 + 4(c')2r2 exp [-c'r2] Y20 - (6/r2) exp [-c'r2] Y20} - {Z (e')2/r} exp [-c'r2] Y20 ∫φ* φ dτ = (h2/me ) (3/16) (π/c'')1/2 - (e')2/[2 c''] (from Part a) + 6h2/(2me ) ∫0∞ exp [-c'r2] dr, (c'' = 2c' = 2c/ a02)

= (h2/me ) (27/16) (π/c'')1/2 - (e')2/[2 c''] Note: ∫02πdφ ∫0πdθ sin θ Y20* Y20 = 1 ∫φ*φ dτ = π1/2/[4c''3/2] (from Part a because Spherical armonics are normalized = ∫φ* φ dτ/∫φ*φ dτ = {(h2/me ) (27/16) (π/c'')1/2 - (e')2/[2 c'']}/{π1/2/[4c''3/2]} = h227c''/(4 me) - 2(e')2(c''/π)1/2 ∂ /∂c = (∂ /∂c'') (∂c''/∂c) = (∂ /∂c'') (2/a02) = 0 if (∂ /∂c'') = 0 (∂ /∂c'') = h227/(4me) - 2(e')2(1/2)/(c''π)-1/2 → c'' = (e')4 me2 16/[π h4272] = 2c/a02 → c = (a02/2)(e')4me216/[π h4272] a0 = h2/[ me(e')2] → c = 8/(272π) = ∫φ* φ dτ / ∫φ*φ dτ >

?

n=?

or c = 8/(272π), ∫φ* φ dτ = (272/64)πa02(e')2 or c = 8/(272π), ∫φ*φ dτ = - π2 a03273/256 = ∫φ* φ dτ / ∫φ*φ dτ = - (0.148/π) (e')2/ a0

n

= -Z2 me(e')4/[2n2 h2] = - [0.500 (e')2/ a0]/n2, Z = 1

0.500/ n2 ~ 0.148/π if n = 3. So have upper bound to n = 3 state.

8.28

pply the linear variation function φ = c1x2(L - x) + c2x(L-x)2

0