hw8_ans - Chem 113A HW8 Solutions 1(a Not a continuum(b Not...

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Unformatted text preview: Chem 113A HW8 Solutions 1. (a) Not a continuum (b) Not a continuum (c) Not a continuum. By looking at the potential energy function for the QHO, we see that V ( x ) → ∞ as x → ∞ . This is similar to the particle in a box, which also does not have a continuum of eigenstates. (d) Continuum of eigenstates as n → ∞ . (d) Continuum of eigenstates as n → ∞ . 2. Δ E = E f- E i =- e 2 8 πǫ a ( 1 n 2 f- 1 n 2 i ) =- 2 . 179 X 10 − 18 J * ( 1 3 2- 1 6 2 ) =- 1 . 816 X 10 − 19 J This means that 1 . 816 X 10 − 19 J of energy is released from the system in this transistion. The frequency and wavelength of this transition is then: ν = E h = 1 . 816 X 10- 19 J 6 . 626 X 10- 34 Js = 2 . 734 X 10 14 Hz λ = c ν = 3 . 00 X 10 8 m/s 2 . 734 X 10 14 Hz = 1 . 097 X 10 − 6 m 3. p x = radicalBig 3 4 π sinθcosφ and d xz = 1 √ w sinθcosθcosφ ˆ l z =- i planckover2pi1 ∂ ∂φ ˆ l z p x =- i planckover2pi1 ∂ ∂φ ( radicalBig 3 4 π sinθcosφ ) = i planckover2pi1 radicalBig 3 4 π sinθsinφ ˆ l z d xz =- i planckover2pi1 ∂ ∂φ ( 1 √ w sinθcosθcosφ ) = i planckover2pi1 1 √ w sinθcosθsinφ We can then see that p x and d xz are not eigenfunctions of the ˆ l z operator. 4. (a) Ψ 210 ( r,θ,φ ) = 1 4 √ 2 π ( 1 a ) 3 / 2 r a e − r/ 2 a cosθ < r > = integraltext 2 π dφ integraltext π dθ integraltext ∞ Ψ ∗ 210 r Ψ 210 r 2 sinθdr = 1 32 π ( 1 a ) 5 integraltext 2 π dφ integraltext π cos 2 θsinθdθ integraltext ∞ r 5 e − r/a dr = 1 32 π 1 a 5 (2 π )( 2 3 )( Γ(5+1) (1 /a ) 5+1 ) = 1 32 π 1 a 5 (2 π )( 2 3 )(120 a 6 ) = 5 a 1 (b) Ψ 210 ( r,θ,φ ) = 1 4 √ 2 π ( 1 a ) 3 / 2 r a e − r/ 2 a cosθ and Ψ 211 ( r,θ,φ ) = 1 8 √ π ( 1 a ) 3 / 2 r a e − r/ 2 a sinθe iφ integraltext 2 π dφ integraltext π dθ integraltext ∞ Ψ ∗ 210 Ψ 211 r 2 sinθdr...
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hw8_ans - Chem 113A HW8 Solutions 1(a Not a continuum(b Not...

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