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### hw09

Course: PHY 107, Fall 2009
School: Ill. Chicago
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Word Count: 951

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9: Week Suggested Problems P27.53 What wavelength photon would be required to ionize a hydrogem atom in the ground state and give the ejected electron a kinetic energy of 10.0 eV? S27.53 The energy of the photon is hf = E ion + KE = 13.6 eV + 10.0 eV = 23.6 eV. We find the wavelength from = (1.24 x 10 eV-nm)/hf = (1.24 x 10 eV-nm)/(23.6 eV) = 3 3 52.5 nm....

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9: Week Suggested Problems P27.53 What wavelength photon would be required to ionize a hydrogem atom in the ground state and give the ejected electron a kinetic energy of 10.0 eV? S27.53 The energy of the photon is hf = E ion + KE = 13.6 eV + 10.0 eV = 23.6 eV. We find the wavelength from = (1.24 x 10 eV-nm)/hf = (1.24 x 10 eV-nm)/(23.6 eV) = 3 3 52.5 nm. --------------------------------------------------------------------------------------------------------------------P27.57 Construct the energy-level diagram for the He ion (see Fig. 27-32). + 0 3.4 6.0 13.6 Continuum S27.57 Singly ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the nucleus. The square of the product of the positive and negative n= n=5 n=4 n=3 n=2 We can use the results for hydrogen, if we replace e2 by Z e : En = Z (13.6 eV)/n 2 2 2 2 2 2 = 2 (13.6 eV)/n = (54.4 eV)/n . 2 Energy (eV) 54.4 charges appears in the energy term for the energy levels. n=1 ------------------------------------------------------------------------------------------------------------------P27.61 Is the use of nonrelativistic formulas justified in the Bohr atom? To check, calculate the electron's speed, v, in terms of c for the ground state of hydrogen, and then calculate (1 2 2 1/2 v /c ) . S27.61 We find the velocity from the quantum condition: mvrn = nh/(2); and for n = 1, we have (9.11 x 10 31 kg)v(0.529 x 10 6 10 m) = (1)(6.63 x 10 -3 34 J-s)/(2), which gives v = 2.18 x 10 m/s = 7.3 x 10 c Due 21 March 2000 P108S00: Week 9 Homework Page 1 of 4 The relativistic factor is [1 (v/c) ] 2 1/2 1 (v/c)2 = 1 2.7 x 10 . 5 Because this is essentially 1, the use of nonrelativistic formulas is justified. -------------------------------------------------------------------------------------------------------------------P27.65 Suppose a particle of mass m is confined to a one-dimensional box of width L. According to quantum theory, the particle's wave [with = h/(mv)] is a standing wave with nodes at the edges of the box. (a) Show the possible modes of vibration on a diagram. (b) 2 2 2 Show that the kinetic energy of the particle has quantized energies given by KE = n h /(8mL ), where n is an integer. (c) Calculate the ground-state energy (n = 1) for an electron confined -10 to a box of width 0.50 x 10 m wide? (d) What is the ground-state energy, and speed, of a baseball (m = 140 g) in a box 0.50 m wide? (e) An electron confined to a box has a groundstate-energy of 20 eV. What is the width of the box? S27.65 (a) (b) From the diagram we see that the wavelengths are given by n = 2L/n, (n = 1, 2, 3, ) . so the momentum is pn = h/ n = nh/(2L), (n = 1, 2, 3, ) . Thus the kinetic energy is 2 2 2 2 KEn = p n /(2m) = n h /(8mL ), (n = 1, 2, 3, ) . 3 2 1 (c) Because the potential energy is zero inside the box, the total energy is the kinetic energy. For the ground state energy, we get 2 2 2 En = KEn = n h /(8mL ); and for n = 1, 2 34 2 31 10 2 E1 = (1) (6.63 x 10 J-s) /[8(9.11 x 10 kg)(0.50 x 10 m) ] 17 = 2.41 X 10 J = 150 eV. (d) For the baseball, we get 2 2 2 E1 = KE1 = n h /(8mL ) 34 2 2 66 = (1)2(6.63 x 10 J-s) /8(0.140 kg)(0.50 m) = 1.6 x 10 J. We find the speed from 2 KE1 = mv ; 66 2 33 1.6 x 10 J = (0.140 kg)v , which gives v = 4.8 x 10 m/s. (e) We find the width of the box from 2 En = n2 h2/(8mL ); and for n = 1, 19 31 2 (20 eV)(1.60 x 10 J/eV) = (1)2(6.63 1034 x J-s)2/[8(9.11 10 kg)L ], 10 which gives L = 1.4 x 10 m = 0.14 nm. -----------------------------------------------------------------------------------------------------------------Due 21 March 2000 P108S00: Week 9 Homework 0 L Page 2 of 4 Week 9: Assigned Problems to Hand In P27.40 What is the wavelength of an electron of energy (a) 10 eV, (b) 100 eV, (c) 1.0 keV? 2 S27.40 Because all the energies are much less than m0c , we can use KE = p2/(2m0), so = h/p = h/[2m0(KE)]1/2 = hc/[2m0c (KE)] . (a) = hc/[2m0c (KE)] 2 2 1/2 1/2 2 1/2 = (1.24 x 10 eV-nm)/[2(0.511 x 10 eV)(10 eV)] 3 6 3 6 1/2 = 0.39 nm. = 0.12 nm. (b) = hc/[2m0c (KE)] (c) = hc/[2m0c (KE)] = 0.039 nm. 2 = (1.24 x 10 eV-nm)/[2(0.511 x 10 eV)(100 eV)] 3 6 3 1/2 1/2 = (1.24 x 10 eV-nm)/[2(0.511 x 10 eV)(1.0 x 10 eV)]1/2 ------------------------------------------------------------------------------------------------------------------P27.72 A beam of red laser light ( = 633 nm) hits a black wall and is fully absorbed. If this exerts a total force F = 5.5 nN on the wall, how many photons per second are hitting the wall? S27.72 Suppose that N red photons hit the wall in time t. The impulse on the wall is due to the change in momentum of the photons: F (t) = p = Np = Nnh/, or N/t = F/h = (5.5 x 10 N)(633 x 10 9 9 34 18 m)/(6.63 x 10 J-s) = 5.3 x 10 photons/s. ------------------------------------------------------------------------------------------------------------------P27.80 Electrons accelerated by a potential difference of 12.3 V pass through a gas of hydrogen atoms at room temperature. What wavelength of light will be emitted? S27.80 The potential difference produces a kinetic energy of 12.3 eV, so it is possible to provide this much energy to the hydrogen atom through collisions. From the ground state, the maximum energy of the atom is 13.6 eV + 12.3 eV = 1.3 eV. From Fig. 27-32, the energy level diagram, we see that this total energy value means the atom could be excited to the n = 3 state, but not the n = 4 state. Thus, the possible ...

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