ps2s.5.112.fall06

ps2s.5.112.fall06 - Problem set#2 solutions for 5.112 Fall...

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Problem set #2 solutions for 5.112 Fall 2006 Page 1 of 12 5.112 PRINCIPLES OF CHEMICAL SCIENCE Problem Set #2 solutions This problem set was graded out of 71 points Question (out of 6 points) The transmitting antenna for a radio station is 7.00 km from your house. The frequency of the electromagnetic wave broadcast by this station is 535 kHz. The station builds a second transmitting antenna that broadcasts the same electromagnetic wave in phase with the original one. The new antenna is 8.12 km from your house. (a) Does constructive or destructive interference occur at the receiving antenna of your radio in your house? Show your calculations. Let us draw a sketch of what information is contained in the problem: Antenna #1 Antenna #2 7 . 0 0 k m 8 . 1 2 km The condition for constructive interference is r 2 r 1 = n λ . We can calculate the wavelength of the radio station from its frequency: λ = c ν = 2.9979 x10 8 m s -1 535 x 10 3 s -1 = 560.4 m We can now substitute it into the equation to see if constructive interference occurs: r 2 r 1 = n λ 8.12 x10 3 m 7.00 x10 3 m = n 560.4 m ( ) n = 1.120 x10 3 m 560.4 m = 1.999 = 2.00 Since n is a multiple of the wavelength, there is constructive interference. (b) Suppose a third antenna is built 9.52 km from your house. Does constructive or destructive interference occur at the receiving antenna of your radio in your house? We can then use the same approach to solve for the new antenna: r 2 r 1 = n λ 9.52 km 7.00 km = n 560.4 m ( ) n = 2.520 x 10 3 m 560.4 m = 4.497 = 4.50 Since n is a half-integer of the wavelength, there is destructive interference.
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Problem set #2 solutions for 5.112 Fall 2006 Page 2 of 12 Question 2 (out of 9 points) Sodium, which has a work function of 4.41 x 10 -19 J, is illuminated by light with a wavelength of 4383.4 Å. (a) With how much kinetic energy, momentum and velocity will each photoelectron emerge? We have to first determine how much energy the light has: 4383.4 A 0 = 4383.4 x 10 -10 m E = hc λ = (6.626076 x 10 34 J s)(2.99792x10 8 m s 1 ) 4383.4 x 10 -10 m = 4.53174 x 10 -19 J Since this energy is greater than the work function, a photoelectron will be ejected. We can calculate the electron’s kinetic energy by the following equation: KE = E i − Φ = 4.53174 x 10 -19 J - 4.41 x 10 -19 J = 0.122 x 10 -19 J =1.2 x 10 -20 J (correct sig figs) The momentum of the ejected electron can be calculated using this equation: E = p 2 2m e or E = 1 2 mv 2 p = E ( ) 2 ( ) m e ( ) = 1.22 x 10 -20 J ( ) 2 ( ) 9.109390 x 10 -31 kg ( ) = 1.49 x 10 -25 kg m s -1 = 1.5 x 10 -25 kg m s -1 (correct sig figs) Remember that an electron is a particle. We can now calculate the velocity of the ejected electron by the following equation: p = mv v = p m = 1.49 x 10 -25 kg m s -1 9.109390 x 10 -31 kg ( ) = 1.64 x 10 5 m s -1 = 1.6 x 10 5 m s -1 (correct sig figs) The ejected electron has KE of 1.2 x 10 -20 J, momentum of 1.5 x 10 -25 kg m s -1 and velocity of 1.6 x 10 5 m s -1 . (b) How many such photoelectrons will be produced by a burst of light with a total energy of 4.00 J?
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Problem set #2 solutions for 5.112 Fall 2006 Page 3 of 12 The energy is delivered in packets of 4.53174 x 10 -19 J, and with each packet (photon), one photoelectron is produced. The total # of photoelectrons produced is just the total amount of energy divided by the energy per photon: total number of photoelectrons produced =
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