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Mid Term_Sol - 1 Because of its particle nature light can...

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1. Because of its particle nature, light can exert pressure on a surface. Suppose a 5.0 W laser is focused on a circular area 100.0 μ m in radius. The energy of the laser light is 3.612 x 10 -19 J. Assume the photons travel perpendicular to the surface and are perfectly reflected, with their momenta changing sign after reflection from the surface. Pressure = Force/Area; 1W = 1 J/s; Force = Δ p/ Δ t where p = momentum and t = time; a.) What is the frequency of the laser light? Hz Js J h E h E 14 34 19 10 451 . 5 10 62618 . 6 10 612 . 3 ! = ! ! = = = " " # # b.) What is the pressure exerted by the light on the surface? The number of photons hitting the surface is s photons photon J Js x / 10 384275 . 1 ) ( 10 612 . 3 0 . 5 19 1 19 1 ! = ! = " " " The momentum change for each photon is c E h p p p p p p p p = = ! = ! = ! = " # # 1 1 1 2 1 2 2 So the pressure is 2 2 2 6 19 1 8 19 1 . 1 ) 10 100 ( 1 10 384275 . 1 10 3 10 612 . 3 2 ! ! ! ! = " " " " " " " " " = = Nm m s photons photons ms J A F P # 2. (20 Points) For each fragment below, draw the correct Lewis structure, labeling formal charges, any resonance structures, and average bond orders. a. N 3 -
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N N N N N N 2- N N N 2- average bond order = (2+1+3)/3 = 2 b. ClO 4 - (Only unique resonance structures should be drawn. Repetitive resonance structures should not be drawn out, merely noted) O Cl O O O O Cl O O O bond order = (2*3+1)/4=1.75 bond order = (2*2+2*1)/4=1.5 O Cl O O O bond order = (2+3*1)/4=1.25 2+ overall average = (1.75+1.5+1.25)/3 = 1.5 3. (20 points) If a bromide anion reacts with a cationic organic fragment, in many cases a carbon-bromine bond forms: CH 3 + + Br - CH 3 —Br In molecules where the cationic charge can be distributed via resonance, various C-Br bond-forming reactions are possible. H H H H H H H H H
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For this cationic molecular fragment we can simplify our analysis, and circle the letters of the carbons that could possibly be subject to such a reaction. ( i.e. which carbons can carry a positive charge? ) H H H H H H H H H A B C D E F G H I H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H A B C D E F G H I 4. Since the electron is trapped in the box, it must have zero amplitude at the edges. Mathematically, this means that 0 ) ( ) 0 ( = ! = ! L . a. In order to satisfy the boundary conditions, 0 ) 0 sin( ) 0 ( = ! = " k A and 0 ) sin( ) ( = ! = " L k A L . Since sin (x) = 0 for x = {0, π , 2 π , 3 π , … n π }, you can now solve for L n k ! = .
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b. ( ) ) sin( 8 ) ( ˆ 2 2 2 2 kx A dx d m h x H e ! " = # ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 8 ) ( 8 , using and ), ( 8 ) sin( 8 ) cos( 8 L m n h E x L m n h L n k x m k h kx A m k h kx A dx d m k h e e e e e = ! " = = " = = # = $ $ $ $ d. 2 2 2 2 2 2 2 2 1 2 8 3 8 1 8 2 L m h L m h L m h E E E e e e n = ! = ! = " Using L = 7.75 Å = 7.75 × 10 -10 m, m e = 9.10939 × 10 -31 kg, and h = 6.62607 × 10 -34 J · s, ! hc J E = " = # $ 19 10 0092 . 3 Using c = 2.99792458 × 10 8 m/s, we find that λ = 660. nm. Yes, this is consistent, because 660 nm is in the red spectral region. If red is absorbed, then green is reflected – which is what we see! 5. The Bohr radius of an orbital is 0.52 Å and we have shown that the radius is given by: r = n 2 h 2 4 "# 0 Ze 2 m e where m e is the electron mass, and the other symbols have their usual meaning.
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