Difference between revisions of "Solutions 3"
From mn/safe/nukwik
(→Amount of radioactive material (number of nuclei <-> number of moles <-> weigth) and law of radioactive decay) |
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| − | = Amount of radioactive material (number of nuclei, number of moles, | + | = Amount of radioactive material (number of nuclei, number of moles, weigth) and law of radioactive decay <br> = |
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====== Return to [[Problem Solving Sets]] ====== | ====== Return to [[Problem Solving Sets]] ====== | ||
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#2.00 • 10<sup>11</sup> atoms. | #2.00 • 10<sup>11</sup> atoms. | ||
| − | <br>'''3:''' 2.40 g <sup>60</sup>Co. | + | <br>'''3:''' |
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| + | 2.40 g <sup>60</sup>Co. | ||
<br>'''4:'''<br> | <br>'''4:'''<br> | ||
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#7.21 years = 7 years and 76.7 days. | #7.21 years = 7 years and 76.7 days. | ||
| − | <br>'''5:''' The mass of <sup>99m</sup>Tc is 20.5 ng. After one week 28 half-lives will have past, and the activity will be the following:<br> | + | <br>'''5:''' |
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| + | The mass of <sup>99m</sup>Tc is 20.5 ng. After one week 28 half-lives will have past, and the activity will be the following:<br> | ||
D = D<sub>0</sub> • (½)<sup>28</sup> = 4 • 10<sup>7</sup> Bq • (½)<sup>28</sup> = 0.15 Bq. | D = D<sub>0</sub> • (½)<sup>28</sup> = 4 • 10<sup>7</sup> Bq • (½)<sup>28</sup> = 0.15 Bq. | ||
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<br> | <br> | ||
| − | '''6:''' | + | '''6:'''<br> |
#1 g natural Lu contains 0.0259 g <sup>176</sup>Lu, with a half-life equal to 3.8 • 10<sup>10</sup> years, which gives A = 51 Bq <br> | #1 g natural Lu contains 0.0259 g <sup>176</sup>Lu, with a half-life equal to 3.8 • 10<sup>10</sup> years, which gives A = 51 Bq <br> | ||
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<br> | <br> | ||
| − | '''7: '''3002 kg <br> | + | '''7: ''' |
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| + | 3002 kg <br> | ||
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'''8:''' <br> | '''8:''' <br> | ||
Revision as of 16:45, 25 June 2012
Amount of radioactive material (number of nuclei, number of moles, weigth) and law of radioactive decay
Return to Problem Solving Sets
1:
- 3.2 • 108 Bq
- 383 Bq
- 2.3 • 109 Bq
- 8.0 • 104 Bq
2:
- 1.77 • 1013 atoms.
- 2.00 • 1011 atoms.
3:
2.40 g 60Co.
4:
- 2.53 • 105 Bq.
- 7.21 years = 7 years and 76.7 days.
5:
The mass of 99mTc is 20.5 ng. After one week 28 half-lives 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 therefore be the same as the original number of 99mTc atoms.
6:
- 1 g natural Lu contains 0.0259 g 176Lu, with a half-life equal to 3.8 • 1010 years, which gives A = 51 Bq
- There are two naturally occurring radioisotopes of Sm, 147Sm and 148Sm, which gives A = 127.24 Bq + 0.001 Bq = 127 Bq.
7:
3002 kg
8:
- 18.5 kg
- After 2.455 • 105 years it will be 2.299 • 108 Bq 238U left, which is equal to 18.49 kg.
