Chem110MT1BookletSolutionsFall2018.pdf

Chem110MT1BookletSolutionsFall2018.pdf - Prep101 CHEM110...

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©Prep101 CHEM110 MT1 Booklet Solutions 1 Chapter 6 Questions: Q 6.1 Solution : λ c E h = E c λ h = ( )( ) J 10 x 3.90 ms 10 x 2.998 Js 10 x 6.626 λ 19 - 1 8 -34 = m 10 x 5.09 λ -7 = cm 10 x 5.09 λ -5 = λ = 509 nm Therefore, GREEN light will be emitted Q 6.2 Solution: We know the energy (499 kJ/mol). By using E = hν, we can solve for the frequency (ν). The frequency is related to wavelength (λ) by the equation ν = λ/c E = 499 kJ mol ( ) = 499 × 10 3 J mol ( ) 6.022 × 10 23 mol 1 ( ) = 8.286 × 10 19 J ( ) E = h ν ν = E h = 8.203 × 10 19 J ( ) 6.626 × 10 34 J s ( ) = 1.251 × 10 15 s 1 or Hz ( ) ν = c λ λ = c ν = 2.998 × 10 8 m s ( ) 1.251 × 10 15 s 1 ( ) = 2.397 × 10 7 m ( ) = 239.7 nm ( )
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©Prep101 CHEM110 MT1 Booklet Solutions 2 Q 6.3 Solution : E = h ν E = (6.626 x 10 - 34 Js)(4.35 x 10 14 s - 1 ) = 2.88 x 10 - 19 J Q 6.4 Solution : First calculate the energy of a photon: E = h ν E = 6.626 x 10 - 34 Js (1.50 x 10 14 s - 1 ) = 9.94 x 10 - 20 J Now, let “a” be the number of photons (factor) to contribute 30.1 J since we know that 1 photon emits 9.94 x 10 20 J of energy E(a) = 30.1 J 9.94 x 10 - 20 J (a) = 30.1 J a = 3.03 x 10 20 photons Q 6.5 Solution : Divide by Avagadro’s number to give the energy per photon J x10 4.28 mol 10 x 6.022 mol J 2.58 24 4 1 23 -1 = Wavelength can now be determined by dividing the product of the speed of light and Plank’s constant by the energy per photon. λ c E h = E c λ h = J 4.28x10 ) ms x10 Js)(2.998 (6.626x10 λ 24 - -1 8 -34 = λ = 0.0464 m or alternatively 4.64 x 10 - 2 m
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©Prep101 CHEM110 MT1 Booklet Solutions 3 Q 6.6 Solution : First convert the wavelength from μm → m 3.34 μm x (1 m/1E6 μm) = 3.34 x 10 - 6 m Now energy can be calculated by the standard equation: λ c E h = E = (6.626x10 - 34 Js)(2.998x10 8 ms - 1 )/(3.34x10 - 6 m) = 5.95 x 10 - 20 J Q 6.7 Solution: Amplitude is the maximum height of the wave above the centre line or the maximum depth below. Wavelength, λ is the distance of one cycle of the wave. It is the distance between the tops of two successive crests (or the bottoms of two troughs). Q 6.8 Solution : Correct answer: D B has a higher frequency because the distance between the peaks of the waves are smaller than A. B also has a higher energy because Energy is inversely proportional to wavelength. Since A has a longer wavelength, it will be of lower energy.
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©Prep101 CHEM110 MT1 Booklet Solutions 4 Q 6.9 Solution : Correct answer: A Speed of light = 3.0 x 10 8 m/sec. Speed of electrons = (3.0 x 10 8 m/sec)0.005 = 1.5 x 10 6 m/sec λ = h/mv = (6.626 x 10 - 34 Js)/[( 1.5 x 10 6 m/sec)(9.11 x 10 - 31 kg) = 4.8 x 10 - 10 m = 0.48 nm Q 6.10 Solution : ∆E = 2 f 2 i H 1 1 R n n R H = 2.179 × 10 18 J. ** your prof indicates that hcR H = 2.17 E - 18 J E = (2.179 × 10 18 J) 2 2 ) 2 ( 1 ) 5 ( 1 = - 4.58 × 10 - 19 J (EMISSION!) E = λ hc , therefore J s m Js E hc 19 8 34 10 58 . 4 ) / 10 0 . 3 )( 10 626 . 6 ( × × × = = λ = 4.34 E - 7 m 4.34 E - 7 m (1 E9 nm / 1 m) = 434 nm
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