Giancoli Physics (5th ed) Chapter 30

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Main Page > Giancoli Physics (5th ed) Solutions > Giancoli Physics (5th ed) Chapter 30

Contents

Problems

1. Question

3727 MeV/c2

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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. Question

2.22 MeV

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13. Question

7.48 MeV

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14. Question

32.0 MeV
5.33 MeV

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15. Question

Solution goes here

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16. Question

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17. Question

(a) 12.1 MeV
(b) 15.7 MeV
repulsive electric force

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18. Question

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19. Question

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20. Question

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21. Question

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22. Question

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23. Question

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24. Question

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25. Question

228/90Th
228.02883 u

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26. Question

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27. Question

Solution goes here

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28. Question

6.12 MeV
0.27 meV

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29. Question

Solution goes here

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30. Question

5.41 mEv

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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

3.0 h

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36. Question

4.91 x 10-18 s-1

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37. The decay constant of a given nucleus is 5.4 x 10-3 s-1. What is its half-life

T1/2 = ln(2)/lambda = ln(2)/5.4 x 10-3 = 128.4 sec, or 2.1 minutes.

2.1 min

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38. Question

1.6 x 109 decays/s

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39. Question

0.0625

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40. Question

5.60 x 1019 nuclei

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41. Question

(a) 0.0625
(b) 0.0442

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42. Question

Solution goes here

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43. Question

(a) 2.44 X 1012 decays/s
(b) 2.43 X 1012 decays/s
(c) 4.48 X 105 decays/s

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44. 55-Cs-124 has a half life of 30.8 s. (a) if we have 7.8 micrograms initially, how many nuclei are present? (b) How many are present 2.0 min later? (c) What is the activity after this time? (d) After how much time will the activity drop to less than 1 per second?

This process is to solve for part (d) as represented by the upcoming test. To do so, we will need to take the steps to be able to use the time formula, \lambdaN = \lambdaNoe^(-\lambdat). The decay constant can be found with

\lambda = 0.693 / 30.8 = .0225 s-1

To find the initial number of nuclei (No):

No = ((7.8E-6 g) / (124 g/mol))*(6.02E23 atoms/mol) = 3.8E16 nuclei

Now, to find the time:

\lambdaN = \lambdaNoe^(-\lambdat)
1 decay/sec = (.0225 s-1)*(3.8E16)*(e^(.0225 s-1*t))
t = 1.53E3 s = 25.5 minutes

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45. Question

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46. Question

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47. Question

(a) 1.38 x 10-13 s-1
(b) 5.387 x 108 decays/min

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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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