2011SpChem1BUnit1PS1Solutions

2011SpChem1BUnit1PS1Solutions - Chemistry 1B Spring 2011...

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Unformatted text preview: Chemistry 1B Spring 2011 Unit 1, Problem Set 1 Answer Key 48. The distant galaxy called Cygnus A is one of the strongest sources of radio waves reaching Earth. The distance of this galaxy from Earth is 3x1024 m. How long (in years) does it take a radio wave of wavelength 10 m to reach the Earth? What is the frequency of this radio wave? Radio waves are a specific type of electromagnetic radiation and all electromagnetic radiation travels at the speed of light (at least in a vacuum). c = 2.998 ⋅ 10 8 m/s distance = time travelled velocity 3 ⋅ 10 24 m 1 year ⋅2 = 3 ⋅ 10 8 years 8 2.998 ⋅ 10 m/s 60 ⋅ 24 ⋅ 365 s Calculating the frequency of the radio wave: ν= ν= c λ 2.998 ⋅ 10 8 10m 1 ν = 3 ⋅ 10 7 s m s 50. Compare the energy (in Joules) carried by an x‐ray photon (wavelength λ = 0.20 nm) with that carried by an AM radio wave photon (λ = 200 m). Calculate the energy of 1.00 mol of each type of photon. What effect do you expect each type of radiation to have for inducing chemical reactions in the substances through which it passes? Calculation for the x‐ray photon: ν= ν= c λ 2.998 ⋅ 10 m/s 1 = 1.50 ⋅ 1018 −9 s 0.20 ⋅ 10 m E = hν 8 ⎛ m kg ⎞ ⎛ 1⎞ ⎟ ⋅ ⎜1.50 ⋅ 1018 ⎟ E = ⎜ 6.626 ⋅ 10 −34 ⎟ ⎜ s ⎠⎝ s⎠ ⎝ E = 9.9 ⋅ 10 −16 J 2 Calculation for the AM radio wave photon: ν= ν= c λ 2.998 ⋅ 10 m/s 1 = 1.50 ⋅ 10 6 200 m s E = hν 8 ⎛ m 2 kg ⎞ ⎛ 1⎞ ⎟ ⋅ ⎜1.50 ⋅ 10 6 ⎟ E = ⎜ 6.626 ⋅ 10 −34 ⎜ s ⎟⎝ s⎠ ⎠ ⎝ E = 9.9 ⋅ 10 − 28 J The AM radio wave will not strongly interact with matter through which it passes (molar energy of (9.9 ⋅10 ) units ⎞ J ⎛ −4 ). The x‐ray, however, will ionize matter through J ⋅ ⎜ 6.02 ⋅ 10 23 ⎟ = 6 ⋅ 10 mol ⎠ mol ⎝ units ⎞ J ⎛ 8 which it passes (molar energy of 9.9 ⋅ 10 −16 J ⋅ ⎜ 6.02 ⋅ 10 23 ). ⎟ = 6 ⋅ 10 mol ⎠ mol ⎝ − 28 ( ) 51. The maximum in Planck’s formula for the emission of blackbody radiation can be shown to occur at a wavelength λmax = 0.20 hc/kT. The radiation from the surface of the sun approximates that of a blackbody with λmax = 465 nm. What is the approximate surface temperature of the sun? hc kT hc λmax = 0.20 T = 0.20 kλ max ⎛ m 2 kg ⎞ ⎛ m⎞ ⎜ 6.626 ⋅ 10 −34 ⎟ ⋅ ⎜ 2.998 ⋅ 10 8 ⎟ ⎜ s ⎟⎝ s⎠ ⎠ T = 0.20 ⎝ ⎛ − 23 J ⎞ −9 ⎜1.381 ⋅ 10 ⎟ ⋅ 465 ⋅ 10 m K⎠ ⎝ T = 6.2 ⋅ 10 3 K ( ) 57. The energies of macroscopic objects, as well as those of microscopic objects, are quantized, but the effects of the quantization are not seen because the difference in energy between adjacent states is so small. Apply Bohr’s quantization of angular momentum to the revolution of Earth (mass 6.0 x1024 kg), which moves with a velocity of 3.0 x 104 m/s in a circular orbit (radius 1.5 x 1011 m) about the sun. The sun can be treated as fixed. Calculate the value of the quantum number n for the present state of the Earth‐sun system. What would be the effect of an increase in n by 1? L = me vr = nh 2π 2πme vr =n h m⎞ ⎛ 2π ⋅ 6.0 ⋅ 10 24 kg ⋅ ⎜ 3.0 ⋅ 10 4 ⎟ ⋅ 1.5 ⋅ 1011 m s⎠ ⎝ =n 2 −34 m kg 6.626 ⋅ 10 s n = 2.6 ⋅ 10 74 ( ) ( ) Increasing n by 1 would result in a negligible change in Earth’s angular momentum since 1 <<< n. 66. A particle of mass m is placed in a three‐dimensional rectangular box with edge lengths 2L, L and L. Inside the box the potential energy is zero, and outside it is infinite; therefore, the wave function goes smoothly to zero at the sides of the box. Calculate the energies and give the quantum numbers of the ground state and the first five excited states (or sets of states of equal energy) for the particle in the box. E nx ,n y ,nz E nx ,n y ,nz 2 2 h 2 ⎛ n x n y n z2 ⎞ ⎜ = + +⎟ 8m ⎜ L2 L2 L2 ⎟ y z⎠ ⎝x 2 2 n y n z2 h 2 ⎛ nx ⎜ = + + 8m ⎜ (2 L )2 L2 L2 ⎝ E nx ,n y ,nz = 2 h 2 ⎛ nx ⎜ 8m ⎜ (2 L )2 ⎝ ⎞ ⎟ ⎟ ⎠ 2 ny n2 ⎞ + 2+ z⎟ L L2 ⎟ ⎠ The ground state and first five excited states of the system are as follows, x y z Ground State 1 1 1 First Excited State 2 1 1 Second Excited State 3 1 1 Third Excited State 1 2 1 1 1 2 Fourth Excited State 4 1 1 Fifth Excited State 3 2 1 3 1 2 h 2 ⎛ 12 12 12 ⎞ 9h 2 ⎜ + 2 + 2⎟= 8m ⎜ (2 L )2 L L ⎟ 32mL2 ⎝ ⎠ h 2 ⎛ 22 12 12 ⎞ 12h 2 ⎜ + 2 + 2⎟= E2,1,1 = 8m ⎜ (2 L )2 L L ⎟ 32mL2 ⎝ ⎠ h 2 ⎛ 32 12 12 ⎞ 17h 2 ⎜ + 2 + 2⎟= E3,1,1 = 8m ⎜ (2 L )2 L L ⎟ 32mL2 ⎝ ⎠ 2⎛ 2 2 2⎞ h⎜1 2 1 21h 2 + 2 + 2⎟= E1, 2,1 = E1,1, 2 = 8m ⎜ (2 L )2 L L ⎟ 32mL2 ⎝ ⎠ h2 ⎛ 42 12 12 ⎞ 24h 2 ⎜ + 2 + 2⎟= E 4,1,1 = E 2, 2,1 = E 2,1, 2 = 8m ⎜ (2 L )2 L L ⎟ 32mL2 ⎝ ⎠ 2⎛ 2 2 2⎞ h⎜3 1 2 ⎟ 29h 2 + + = E3, 2,1 = E3,1, 2 = 8m ⎜ (2 L )2 L2 L2 ⎟ 32mL2 ⎝ ⎠ E1,1,1 = ...
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This note was uploaded on 09/10/2011 for the course CHEM 1B taught by Professor Pines during the Spring '08 term at University of California, Berkeley.

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