hw1_ans - Chem 113A HW1 Solutions 1. = h p = h mv =...

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Chem 113A HW1 Solutions 1. λ = h p = h mv = 6 . 626 10 - 34 Js (2 oz. )(135 mi/hr ) * 1 oz. 0 . 0283495 kg * 1 mi/hr 0 . 44704 m/s = 1 . 94 * 10 34 m 2. a) KE = - θ = 0 , ν = θ h = 4 . 65 eV 6 . 626 10 - 34 Js * 1 . 6 10 - 19 J 1 eV = 1 . 12 * 10 15 Hz b) i. KE = - θ = hc λ - θ = (6 . 626 10 - 34 Js )(3 . 00 10 8 m/s ) 200 10 - 9 m - (4 . 65 eV ) * 1 . 6 10 - 19 J 1 eV = 2 . 49 * 10 19 J ii. KE = hc λ - θ = (6 . 626 10 - 34 Js )(3 . 00 10 8 m/s ) 300 10 - 9 m - (4 . 65 eV ) * 1 . 6 10 - 19 J 1 eV = - 8 . 23 * 10 20 J The fnal kinetic energy is negative, meaning there was not enough energy to eject and electron. 3. a) λ = hc E = (6 . 626 10 - 34 Js )(3 . 00 10 8 m/s ) 4 . 085 10 - 19 J = 4 . 86 * 10 7 m n 1 = r 1 / ( 1 + 1 n 2 2 ) = r 1 / ( 1 (1 . 097 10 7 m - 1 )(4 . 86 10 - 7 m ) + 1 16 ) = 2 b) Balmer series. 4. i. ∂cos ( x ) sin (3 y ) ∂x = - sin ( x ) sin (3 y ) ∂cos ( x ) sin (3 y ) ∂y = 3 cos ( x ) cos (3 y ) 2 sin ( x ) ∂x = - sin ( x ) ii. (3 - 2 i )(5 + 6 i ) = 15 + 18 i - 10 i + 12 = 27 + 8 i (3 - 2 i )(3 - 2 i ) = (3 - 2 i )(3 + 2 i ) = 9 + 6 i - 6 i + 4 = 13 iii. i 5 e x dx = 5 e x + c i π π 5 sin ( x ) dx = - 5 cos ( x )
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This note was uploaded on 12/05/2011 for the course CHEM 113A taught by Professor Lin during the Fall '07 term at UCLA.

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hw1_ans - Chem 113A HW1 Solutions 1. = h p = h mv =...

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