Difference between revisions of "Solutions 3"
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| − | = Amount of | + | = Amount of Radioactive Material (number of nuclei, number of moles, weigth) and the Law of Radioactive Decay = |
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| + | ====== Return to [[Problem Solving Sets]] ====== | ||
<br> | <br> | ||
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#383 Bq | #383 Bq | ||
#2.3 • 10<sup>9</sup> Bq | #2.3 • 10<sup>9</sup> Bq | ||
| − | #8 • 10<sup>4</sup> Bq | + | #8.0 • 10<sup>4</sup> Bq |
<br>'''2:''' <br> | <br>'''2:''' <br> | ||
| − | #1. | + | #1.77 • 10<sup>13</sup> atoms. |
| − | #2 • 10<sup>11</sup> atoms. | + | #2.00 • 10<sup>11</sup> atoms. |
| − | <br>'''3:''' 2.40 g <sup>60</sup>Co. | + | <br>'''3:''' |
| + | |||
| + | 2.40 g <sup>60</sup>Co. | ||
<br>'''4:'''<br> | <br>'''4:'''<br> | ||
#2.53 • 10<sup>5</sup> Bq. | #2.53 • 10<sup>5</sup> Bq. | ||
| − | #7. | + | #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- | + | <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. | ||
| − | All <sup>99m</sup>Tc has disintegrated and turned into <sup>99</sup>Tc. The number of <sup>99</sup>Tc atoms will | + | All <sup>99m</sup>Tc has disintegrated and turned into <sup>99</sup>Tc. The number of <sup>99</sup>Tc atoms will therefore be the same as the original number of <sup>99m</sup>Tc atoms. |
| + | |||
| + | <math>N=\frac{D}{\lambda} \rightarrow \frac{4\cdot 10^{7}Bq}{3.21\cdot 10^{-5}s^{-1}}=1.25\cdot 10^{12}atoms</math> | ||
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| + | <br> | ||
| − | + | '''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> | |
| + | #There are two naturally occurring radioisotopes of Sm, <sup>147</sup>Sm and <sup>148</sup>Sm, which gives A = 127.24 Bq + 0.001 Bq = 127 Bq. | ||
<br> | <br> | ||
| − | ''' | + | '''7: ''' |
| − | + | 3002 kg <br> | |
| − | |||
| − | |||
| − | <br> | ||
<br> | <br> | ||
| − | '''8:''' | + | '''8:''' <br> |
#18.5 kg | #18.5 kg | ||
| − | # After 2.455 • 10<sup>5</sup> years it | + | # After 2.455 • 10<sup>5</sup> years it will be 2.299 • 10<sup>8</sup> Bq <sup>238</sup>U left, which is equal to 18.49 kg.<br><br> |
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| + | <br> | ||
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| + | [[Category:Solved_Problem]] [[Category:Bachelor]] | ||
Latest revision as of 10:01, 9 July 2012
Amount of Radioactive Material (number of nuclei, number of moles, weigth) and the 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.
