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)
(No difference)

Revision as of 08:44, 19 June 2012



1:

  1. 3.2 * 108 Bq
  2. 383 Bq
  3. 2.3 * 109 Bq
  4. 8 * 104 Bq


2:

  1. 1.78 * 1013 atoms.
  2. 2 * 1011 atoms.


3: 2.40 g 60Co.


4:

  1. 2.53 * 105 Bq.
  2. 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. 1 g nat-Lu contains 0.0259 g 176Lu, which has a half-life of 3.8*1010 years. This gives D = 51 Bq
  2. 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:

  1. 18.5 kg
  2.  After 2.455*105 years it remains 2.299*108 Bq 238U,which is equal to18.49 kg.