Difference between revisions of "Giancoli Physics (5th ed) Chapter 31"

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(/* 20. Suppose that the average power consumption, day and night, in a typical house is 300 W. What initial mass of U-235 would have to undergo fission to supply the electrical needs of such a house for a year? (Assume 200 MeV is released per fissi)
(/* 20. Suppose that the average power consumption, day and night, in a typical house is 300 W. What initial mass of U-235 would have to undergo fission to supply the electrical needs of such a house for a year? (Assume 200 MeV is released per fissi)
Line 137: Line 137:
  
 
P = E / t<br>
 
P = E / t<br>
−
300 W = (200 MeV)(1.60E-13 J/MeV)(n) / (3.16E7 s/yr)<br>
+
300 W = E/(3.16E7 s/yr)<br />
 +
E = 9.48E9 J<br>
 +
so the number of fissions you need is:<br>
 +
n = (9.48E9 J)/[(200E6 eV)(1.602E-19 J/eV)]<br>
 +
<br>
 
n = 2.96E20 fissions<br><br>
 
n = 2.96E20 fissions<br><br>
  
 
Using this number, knowing each fission requires one uranium atom, we can set up the equation:<br><br>
 
Using this number, knowing each fission requires one uranium atom, we can set up the equation:<br><br>
  
−
m = [(2.96E20 atoms) / (6.02E23 atoms/mol)](235 g/mol) = 4.82E5 g = '''482 kg'''
+
m = [(2.96E20 atoms) / (6.02E23 atoms/mol)](235 g/mol) = .116 g = '''482 kg'''
  
  

Revision as of 06:54, 12 March 2009

Main Page > Giancoli Physics (5th ed) Solutions > Giancoli Physics (5th ed) Chapter 31

Contents

Problems

1. Question

28/13Al
B- emitter
28/14Si

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2. Question

Solution goes here

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3. Question

Solution goes here

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4. Question

17.35 MeV is released

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5. Question

+ 18.000953 u

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6. Question

5.700 MeV is released

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7. Question

Solution goes here

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8. Question

Solution goes here

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9. Question

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10. Question

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11. Question

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12. Question

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13. Question

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14. Question

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15. Question

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16. Question

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17. Calculate the energy released in the fission reaction n + 92-U-235 -> 38-Sr-88 + 54-Xe-136 + 12n. Use Appendix F; assume that the initial KE of the neutron is small.

To calculate the released energy, we use the formula:

Q = [m(235U) + m(n) - 12m(n) - m(88Sr) - m(136Xe)]c2

Q = [(235.043924 u) - 11(1.008665 u) - (87.905618 u) - (135.90721 u)]c2(931.5 MeV / uc2) = 126.5 MeV


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18. Question

Solution goes here

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19. Question

Solution goes here

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20. Suppose that the average power consumption, day and night, in a typical house is 300 W. What initial mass of U-235 would have to undergo fission to supply the electrical needs of such a house for a year? (Assume 200 MeV is released per fission)

To find the number of fissions, we use:

P = E / t
300 W = E/(3.16E7 s/yr)
E = 9.48E9 J
so the number of fissions you need is:
n = (9.48E9 J)/[(200E6 eV)(1.602E-19 J/eV)]

n = 2.96E20 fissions

Using this number, knowing each fission requires one uranium atom, we can set up the equation:

m = [(2.96E20 atoms) / (6.02E23 atoms/mol)](235 g/mol) = .116 g = 482 kg


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21. Question

Solution goes here

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22. Question

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23. Question

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24. Question

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25. Question

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26. Question

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28. Question

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29. Question

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30. Question

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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

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38. Question

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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

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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

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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

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71. Question

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72. Question

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73. Question

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74. Question

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75. Question

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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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82. Question

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83. Question

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84. Question

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85. Question

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86. Question

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87. Question

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88. Question

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89. Question

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90. Question

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91. Question

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92. Question

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93. Question

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94. Question

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95. Question

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96. Question

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97. Question

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98. Question

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99. Question

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100. Question

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