Difference between revisions of "Solutions 3"
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(Created page with "<br> <br> '''1:''' <br> #3.2 * 10<sup>8</sup> Bq #383 Bq #2.3 * 10<sup>9</sup> Bq #8 * 10<sup>4</sup> Bq <br>'''2:''' <br> #1.78 * 10<sup>13</sup> atoms. #2 * 10<sup>11</sup...") |
m (moved Solutions 2 to Solutions 3: It is where it belongs) |
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Revision as of 09:44, 19 June 2012
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.
