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 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?
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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