CHAPTER 4, Levine, Quantum Chemistry, 5^th Ed.

4.12 For the ground state of the one-dimensional hrmonic oscillator, find the average value of the kinetic energy and of the potential energy. Verify that <T> = <V>.

y ₀ = (a /p )^1/4 e^{-(a x**2/2)}

T = -h²/(2m) d²/dx²

V = k x²/2 = 2p ^{2n 2}mx² = a ² h²x²/(2m)

dy ₀ /dx = (a /p )^1/4 (-a /2) (2x) e^{-(a x**2/2)}

d^2y ₀ /dx² = (a /p )^1/4 (-a ) d/dx [x e^{-(a x**2/2)}]

= (a /p )^1/4 (-a ) [e^{-(a x**2/2)} + x (-a /2) (2x) e^{-(a x**2/2)}]

= (a /p )^1/4 (-a ) [1 - a x²] e^{-(a x**2/2)}

<T> = ò _-¥ ¥ ^y ₀^* T y ₀ dx

= ò _-¥ ¥ (a /p )^1/4 e^{-(a x**2/2)} [-h²/(2m) d²/dx²] (a /p )^1/4 e^{-(a x**2/2)}

= (a /p )^1/2 (a ) [h²/(2m)] ò _-¥ ¥ e^{-a x**2} (1 - a x²) dx

= (a /p )^1/2(a ) [h²/(2m)][ ò _-¥ ¥ e^{-a x**2} dx - a ò _-¥ ¥ x²e^{-a x**2} dx]

= (a /p )^1/2a [h²/m][ ò ₀ ¥ e^{-a x**2} dx - a ò ₀ ¥ x² e^{-a x**2} dx]

= (a /p )^1/2a [h²/m][ (1/2) (p /a )^1/2 - a (1/4) (p /a ³)^1/2]

= (a /p )^1/2a [h²/m][ (1/2) (p /a )^1/2 - (1/4) (p /a )^1/2]

= a h²/(4m)

<V> = ò _-¥ ¥ ^y ₀^* V y ₀ dx

= ò _-¥ ¥ ^y ₀^* [h²/(2m)] (a ² x²) y ₀ dx

= a ²[h²/(2m)] (a /p )^{1/2 ò} _-¥ ¥ x² e^{-a x**2} dx

= a ²[h²/m] (a /p )^{1/2 ò} ₀ ¥ x² e^{-a x**2} dx

= a ²[h²/m] (a /p )^1/2 (1/4) (p /a ³)^1/2

NOTE: a h²/(4m) = [2 p n m/ h] [h²/(4m)] = hn /4

4.16 For the v = 1 harmonic oscillator state, find the most likely position(s) of the particle.

Most likely position occurs when the probability density is a maximum:

P = y ₁^{* y} ₁ = (4a ³/p )^1/2 x² e^{-a x**2}

Set dP/dx = 0 & find x:

dP/dx = 0 = 2x e^{-a x**2} + x² (-a 2x) e^{-a x**2}

= 1 - a x²

x = + 1/Ö a

21. Find <x> for the harmonic oscillator state with quantum number v.

<x> = ò _{-¥ ¥} ^{y *}x y dx

v = 0: y ₀ = c₀ e^{-a x**2/2} even function

<x> = ò _{-¥ ¥} ^y ₀ ^*x y ₀ dx

y ₀ ^*x y ₀ = (even) (odd) (even) = odd

<x> = ò _{-¥ ¥} odd dx = 0

v = 1: y ₁ = (4a ³/p )^1/4 x e^{-a x**2/2} odd function

<x> = ò _{-¥ ¥} ^y ₁ ^*x y ₁ dx = (4a ³/p )^1/2 ò _{-¥ ¥} e^{-a x**2} x³ dx

= (4a ³/p )^1/2 ò _{-¥ ¥} (even) (odd) dx

= (4a ³/p )^1/2 ò _{-¥ ¥} (odd) dx = 0

v = 2: y ₂ = (a /(4p ))^1/4 (2a x² -1) e^{-a x**2/2}

<x> = ò _{-¥ ¥} ^y ₂ ^*x y ₂ dx = (a /(4p ))^{1/2 ò} _{-¥ ¥} (2a x² -1)² x e^{-a x**2} dx

= (a /(4p ))^{1/2 ò} _{-¥ ¥} (even) (odd) (even) dx

= (a /(4p ))^{1/2 ò} _{-¥ ¥} (odd) dx = 0 & etc. for higher v

28. (A) The v = 0 ® band of LiH occurs at 1359 cm^-1. Calculate the ratio of the v = 1 to v = 0 populations at 25^oC and at 200^oC.

n = 1/ l = 1359 cm^-1

E_v = (v + 1/2) hn ; n = c/l = c n

E₀ = (0 + 1/2) hn = hn /2

E₁ = (1 + 1/2) hn = 3 hn /2

E₁ - E₀ = hn

N₁/N₀ = e^{-(E1 - E0)/kT} = e^{-hn /(kT)}; degeneracy = 1

T = 25^oC = 298K:

hn /(kT) = (6.63x10^-34 Js) (3.00x10⁸ m/s)

x (1359 cm^-1)/[1.38x10^-23J/K) (298K)] = 6.57

N₁/N₀ = e^-6.57 = 1.402x10^-3

T = 200^oC = 473K: hn /(kT) = 6.57 (298/473) = 4.14;

N₁/N₀ = e^-4.14 = 0.016 Higher relative population at higher T

4.27 (B) Do the same for ICl whose strongest infrared band occurs at 381 cm^-1.

T = 298K: hn /(kT) = 1.84, N₁/N₀ = e^-1.84 = 0.16

T = 473K: hn /(kT) = 1.16, N₁/N₀ = e^-1.16 = 0.31

Heavier molecule has higher relative population

4.24 (A) The infrared absorption spectrum of ¹H³⁵Cl has its strongest band at 8.65x10¹³ Hz. Calculate the force constant of the bond in this molecule.

n = (k/m )^1/2/2p ; k = m 4p ^{2n 2}

m = m_H m_Cl/(m_H + m_H) = (1.00) (35.0)/ (36.0) amu

=0.972 amu (1 g / 6.023x10²³ amu) = 0.161x10^-23 g

k = (0.161x10^-23 g) (4p ²) (8.65x10¹³ Hz)² 1 Hz = 1 s^-1

= 4.75x10⁵ g/s²

= 4.75x10⁵ dyn/cm (10^-8 cm/1 Angstrom)

= 4.75x10^-3 dyn/ Angstrom

= 4.75 millidyn/ Angstrom

(B) Find the approximate zero point vibrational energy of ¹H³⁵Cl.

E₀ = (0 + 1/2) hn = hn /2 = (6.63x10^-34 Js) (8.65x10¹³ Hz)/2

= 2.87x10^-20 J