Difference between revisions of "Solutions 3"
From mn/safe/nukwik
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'''1:''' <br> | '''1:''' <br> | ||
− | #3.2 | + | #3.2 • 10<sup>8</sup> Bq |
#383 Bq | #383 Bq | ||
− | #2.3 | + | #2.3 • 10<sup>9</sup> Bq |
− | #8 | + | #8 • 10<sup>4</sup> Bq |
<br>'''2:''' <br> | <br>'''2:''' <br> | ||
− | #1.78 | + | #1.78 • 10<sup>13</sup> atoms. |
− | #2 | + | #2 • 10<sup>11</sup> atoms. |
<br>'''3:''' 2.40 g <sup>60</sup>Co. | <br>'''3:''' 2.40 g <sup>60</sup>Co. | ||
Line 19: | Line 19: | ||
<br>'''4:'''<br> | <br>'''4:'''<br> | ||
− | #2.53 | + | #2.53 • 10<sup>5</sup> Bq. |
#7.19 years = 7 years and 68 days. | #7.19 years = 7 years and 68 days. | ||
<br>'''5:''' The mass of <sup>99m</sup>Tc is 20.5 ng. After one week 28 half-times will have past, and the activity will be the following:<br> | <br>'''5:''' The mass of <sup>99m</sup>Tc is 20.5 ng. After one week 28 half-times will have past, and the activity will be the following:<br> | ||
− | D = D<sub>0</sub> | + | D = D<sub>0</sub> • (½)<sup>28</sup> = 4 • 10<sup>7</sup> Bq • (½)<sup>28</sup> = 0.15 Bq. |
All <sup>99m</sup>Tc has disintegrated and turned into <sup>99</sup>Tc. The number of <sup>99</sup>Tc atoms will therfore be the same as the original number of <sup>99m</sup>Tc atoms. | All <sup>99m</sup>Tc has disintegrated and turned into <sup>99</sup>Tc. The number of <sup>99</sup>Tc atoms will therfore be the same as the original number of <sup>99m</sup>Tc atoms. | ||
− | N = = = 1,25 | + | N = = = 1,25 • 10<sup>12</sup> atomer |
− | The activity will then be: D = λN = 1.05 | + | The activity will then be: D = λN = 1.05 • 10<sup>-13</sup> s<sup>-1</sup> • 1.25 • 10<sup>12</sup> = 0.13 Bq. |
<br> | <br> | ||
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'''6:''' | '''6:''' | ||
− | #1 g nat-Lu contains 0.0259 g <sup>176</sup>Lu, which has a half-life of 3.8 | + | #1 g nat-Lu contains 0.0259 g <sup>176</sup>Lu, which has a half-life of 3.8 • 10<sup>10</sup> years. This gives D = 51 Bq |
#There are two naturally occuring Sm radioisotopes, <sup>147</sup>Sm and <sup>148</sup>Sm, which gives D = 127.24 Bq + 0.001 Bq = 127 Bq. | #There are two naturally occuring Sm radioisotopes, <sup>147</sup>Sm and <sup>148</sup>Sm, which gives D = 127.24 Bq + 0.001 Bq = 127 Bq. | ||
Line 46: | Line 46: | ||
#18.5 kg | #18.5 kg | ||
− | # After 2.455 | + | # After 2.455 • 10<sup>5</sup> years it remains 2.299 • 10<sup>8</sup> Bq <sup>238</sup>U,which is equal to18.49 kg. <br><br> |
Revision as of 10:34, 19 June 2012
Amount of radioactive material (number of nuclei <-> number of moles <-> weigth) and law of radioactive decay
1:
- 3.2 • 108 Bq
- 383 Bq
- 2.3 • 109 Bq
- 8 • 104 Bq
2:
- 1.78 • 1013 atoms.
- 2 • 1011 atoms.
3: 2.40 g 60Co.
4:
- 2.53 • 105 Bq.
- 7.19 years = 7 years and 68 days.
5: The mass of 99mTc is 20.5 ng. After one week 28 half-times will have past, and the activity will be the following:
D = D0 • (½)28 = 4 • 107 Bq • (½)28 = 0.15 Bq.
All 99mTc has disintegrated and turned into 99Tc. The number of 99Tc atoms will therfore be the same as the original number of 99mTc atoms.
N = = = 1,25 • 1012 atomer
The activity will then be: D = λN = 1.05 • 10-13 s-1 • 1.25 • 1012 = 0.13 Bq.
6:
- 1 g nat-Lu contains 0.0259 g 176Lu, which has a half-life of 3.8 • 1010 years. This gives D = 51 Bq
- There are two naturally occuring Sm radioisotopes, 147Sm and 148Sm, which gives D = 127.24 Bq + 0.001 Bq = 127 Bq.
7: 3002 kg
8:
- 18.5 kg
- After 2.455 • 105 years it remains 2.299 • 108 Bq 238U,which is equal to18.49 kg.