Exam 2 Solutions - Last Name First Name PSU ID Chem 452 Exam II Oct 15 2014 For a 1D harmonic oscillator 1 y2 2 2 4 = k m f x y=

# Exam 2 Solutions - Last Name First Name PSU ID Chem 452...

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Last Name: First Name: PSU ID#: Chem 452 – Exam II Oct 15 2014 For a 1D harmonic oscillator N v 2 = 1 α π 2 v v ! H v (y): Hermite polynomial H 0 = 1, H 1 = 2y, H 2 = 4y 2 -­‐2 … H v+1 -­‐ 2yH v + 2vH v-­‐1 =0 H v −∞ ( y ) H v ' ( y ) e y 2 dy = 0 if v ' v π 2 v v ! if v ' = v ' ( ) * ) For a 3D rotor only when l ” = l + 1 and m l ” = m l ’ + m ˆ L 2 = 2 Λ 2 = 2 1 sin 2 θ 2 ∂φ 2 + 1 sin θ ∂θ sin θ ∂θ # \$ % & ' ( For a classical rotor : Reduced mass : Trigonometric identity : sin 2 x + cos 2 x = 1 cos 2 θ a b sin θ d θ = 1 3 cos 3 θ " # \$ % & ' a b Coordinates conversion: x = rsinθcosϕ , y = rsinθsinϕ , z = rcosθ 2 = 2 x 2 + 2 y 2 + 2 y 2 = 2 r 2 + 2 r r + 1 r 2 Λ 2 , Λ 2 = 1 sin 2 θ 2 ∂φ 2 + 1 sin θ ∂θ sin θ ∂θ ψ v ( x ) = N v H v ( y ) e y 2 2 , y = x α , α = 2 k f m % & ' ' ( ) * * 1 4 E v = ( v + 1 2 ) ω , v = 0, 1, 2... ψ l , m l ( θ , φ ) = Y l , m l ( θ , φ ) E l = l ( l + 1) 2 2 I , I = mr 2 J = l ( l + 1) , J z = m l l = 0, 1, 2... m l = 0, ± 1, ± 2..., ± l Y l ', m l ' * Y l , m l d τ = Y l ', m l ' * Y l , m l 0 π 0 2 π sin θ d θ d φ = δ l ' l δ m l ' m l Y l ", m l " * Y l ', m l ' 0 π 0 2 π Y l , m l sin θ d θ d φ 0 ˆ l z = i ∂φ E = J 2 2 I I = mr 2 μ = m 1 m 2 m 1 + m 2 e cx dx a b = 1 c e cx [ ] a b sin cxdx a b = 1 c cos cx [ ] a b cos cxdx a b = 1 c sin cx [ ] a b d τ −∞ = r 2 dr sin θ d θ d φ 0 0 π 0 2 π h = 6.626*10 –34 Js = h/2 π k = 1.38*10 -­‐23 J/K m p = 1.67*10 -­‐27 kg m e = 9.109*10 –31 kg
Point Total Problem Possible Score Your Score 1 8 2 12 3 16 4 14 5 10 6 10 7 12 8 18 Total 100
Name: 1/7 1. (8 pts) Write the Hamiltonian operators for (a) a harmonic oscillator with mass m and spring constant k f and (b) an electron in a hydrogen atom separated by a fixed distance r (i.e. r is a constant). Hint: the restoration force exerted on a harmonic oscillator is linearly proportional to the displacement from the equilibrium position. The potential energy of a negative charge separated by r from a positive charge is -­‐1/r. ( a ) 2 2 m 2 x 2 + 1 2 k f x 2 ( b ) 2 2 μ r 2 1 sin 2 θ 2 ∂φ 2 + 1 sin θ ∂θ sin θ ∂θ # \$ % & ' ( 1 r ; μ = m e m p m e + m p m e 2. (12 pts) Sketch the first four (v = 0 ~ 3) wavefunctions (Ψ v ) and the probability densities (|Ψ v | 2 ) of a 1D harmonic oscillator and mark approximate classical turning points.

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