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

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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. E = E f- E i =- e 2 8 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: = E h = 1 . 816 X 10- 19 J 6 . 626 X 10- 34 Js = 2 . 734 X 10 14 Hz = c = 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 sincos and d xz = 1 w sincoscos l z =- i planckover2pi1 l z p x =- i planckover2pi1 ( radicalBig 3 4 sincos ) = i planckover2pi1 radicalBig 3 4 sinsin l z d xz =- i planckover2pi1 ( 1 w sincoscos ) = i planckover2pi1 1 w sincossin We can then see that p x and d xz are not eigenfunctions of the l z operator. 4. (a) 210 ( r,, ) = 1 4 2 ( 1 a ) 3 / 2 r a e r/ 2 a cos < r > = integraltext 2 d integraltext d integraltext 210 r 210 r 2 sindr = 1 32 ( 1 a ) 5 integraltext 2 d integraltext cos 2 sind integraltext r 5 e r/a dr = 1 32 1 a 5 (2 )( 2 3 )( (5+1) (1 /a ) 5+1 ) = 1 32 1 a 5 (2 )( 2 3 )(120 a 6 ) = 5 a 1 (b) 210 ( r,, ) = 1 4 2 ( 1 a ) 3 / 2 r a e r/ 2 a cos and 211 ( r,, ) = 1 8 ( 1 a ) 3 / 2 r a e r/ 2 a sine i integraltext 2 d integraltext d integraltext 210 211 r 2 sindr...
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hw8_ans - Chem 113A HW8 Solutions 1. (a) Not a continuum...

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