Giancoli Physics (5th ed) Chapter 27

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Main Page > Giancoli Physics (5th ed) Solutions > Giancoli Physics (5th ed) Chapter 27

Contents

Problems

1. Question

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2. Question

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3. Question

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4. Question

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5. Question

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6. Question

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7. Question

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8. Question

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9. Question

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10. Question

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11. Question

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12. What is the energy range in eV, of photons in the visible spectrum, of wavelength 400 nm to 700 nm?

put these in J first
1.77 eV < E < 3.10 eV

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13. Question

4.23 x 10-7 eV

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14. Question

Solution goes here

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15. Question

2.4 x 1013 Hz
1.2 x 10-5 m

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16. Question

Solution goes here

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17. Question

400 nm

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18. Barium has a work function of 2.48 eV. What is the maximum kinetic energy of electrons if the metal is illuminated by light of wavelength 390 nm? What is their speed?

.70 eV - but also convert this to joules
5.0 x 105 m/s

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19. Question

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20. Question

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21. Question

0.62 eV

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22. Question

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23. Question

1.08 eV
no ejected electtrons

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24. Question

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25. Question

3.75 eV

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26. Question

1.52 MeV

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27. Question

1.9 x 10-24 kg*m/s
6.3 x 10-33 kg2

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28. Question

6.62 x 10-16 m

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29. Question

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30. Question

1.71 MeV
7.24 x 10-13 m

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31. Question

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32. Question

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33. Question

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34. Question

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35. Question

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36. Question

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37. Question

7.2 x 10-12 m

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38. Question

29 V

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39. Question

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40. Question

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41. Question

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42. Question

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43. Question

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44. Question

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45. Question

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46. Question

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47. Question

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48. Question

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49. Question

3.4 eV

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50. Question

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51. Question

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52. Question

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53. Question

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54. Question

(a) 486 nm
(b) 102 nm
(c) 434 nm

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55. Question

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56. Question

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57. Question

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58. Question

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59. Question

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60. Question

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61. Question

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62. Question

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63. Question

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64. Question

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65. Question

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66. Question

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67. Question

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68. Question

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69. Question

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70. Question

4.66 x 1026 photons/s

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71. Question

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72. A beam of red laser light (wavelength = 633 nm) hits a black wall and is fully absorbed. If this exerts a total force of F = 5.5 nN, how many photons per second are hitting the wall?

Use de Broglie's relation of momentum:
p = \frac{h}{\lambda}
And the relationship between force and momentum:
F = \frac{\Delta p}{\Delta t}

So if \Delta p\ is due to the photons hitting the surface and stopping, then \Delta p\ is just np\ where n\ is the number of photons, and p\ is their momentum

So we have
F = \frac{\Delta p}{\Delta t} = \frac{np}{\Delta t}


We are all set = the momentum of a 633nm photon is
p = \frac{h}{\lambda} = \frac{(6.626E-34\ Js)}{(633E-9\ m)} = 1.04676E-27 kg\ m/s
Plugging into

F =  \frac{np}{\Delta t}
So
(5.5E-9\ N) =  \frac{n(1.04676E-27 kg\ m/s)}{1\ s}

and n = 5.254E+18 photons per second
So this is just if the photons hit the surface and stop. What would happen if they bounced off? (i.e. the surface were white or silver - reflective?? Would this exert more force or less?)

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73. Question

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74. Question

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75. Question

4.7 x 10-14

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76. Question

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77. Question

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78. Question

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79. Question

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80. Question

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81. Question

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