Difference between revisions of "Solutions 3"

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'''1:''' <br>  
 
'''1:''' <br>  
  
#3.2 <span style="background-color: navy; color: white;" /> • 10<sup>8</sup> Bq  
+
#3.2&nbsp; • 10<sup>8</sup> Bq  
 
#383 Bq  
 
#383 Bq  
#2.3 <span style="background-color: navy; color: white;" /> • 10<sup>9</sup> Bq  
+
#2.3&nbsp; • 10<sup>9</sup> Bq  
#8 <span style="background-color: navy; color: white;" /> • 10<sup>4</sup> Bq
+
#8&nbsp; • 10<sup>4</sup> Bq
  
 
<br>'''2:''' <br>  
 
<br>'''2:''' <br>  
  
#1.78 <span style="background-color: navy; color: white;" /> • 10<sup>13</sup> atoms.  
+
#1.78&nbsp; • 10<sup>13</sup> atoms.  
#2 <span style="background-color: navy; color: white;" /> • 10<sup>11</sup> atoms.
+
#2&nbsp; • 10<sup>11</sup> atoms.
  
 
<br>'''3:''' 2.40 g <sup>60</sup>Co.  
 
<br>'''3:''' 2.40 g <sup>60</sup>Co.  
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<br>'''4:'''<br>  
 
<br>'''4:'''<br>  
  
#2.53 <span style="background-color: navy; color: white;" /> • 10<sup>5</sup> Bq.  
+
#2.53&nbsp; • 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> <span style="background-color: navy; color: white;" /> • (½)<sup>28</sup> = 4 <span style="background-color: navy; color: white;" /> • 10<sup>7</sup> Bq <span style="background-color: navy; color: white;" /> • (½)<sup>28</sup> = 0.15 Bq.  
+
D = D<sub>0</sub>&nbsp; • (½)<sup>28</sup> = 4&nbsp; • 10<sup>7</sup> Bq&nbsp; • (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 <span style="background-color: navy; color: white;" /> • 10<sup>12</sup> atomer  
+
N = = = 1,25&nbsp; • 10<sup>12</sup> atomer  
  
&nbsp;The activity will then be: D = λN = 1.05 <span style="background-color: navy; color: white;" /> • 10<sup>-13</sup> s<sup>-1</sup> <span style="background-color: navy; color: white;" /> • 1.25 <span style="background-color: navy; color: white;" /> • 10<sup>12</sup> = 0.13 Bq.  
+
&nbsp;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<span style="background-color: navy; color: white;" /> •10<sup>10</sup> years. This gives D = 51 Bq  
+
#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.
  
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#18.5 kg  
 
#18.5 kg  
#&nbsp;After 2.455<span style="background-color: navy; color: white;" /> •10<sup>5</sup> years it remains 2.299<span style="background-color: navy; color: white;" /> •10<sup>8</sup> Bq <sup>238</sup>U,which is equal to18.49 kg. <br><br>
+
#&nbsp;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 09:34, 19 June 2012

Amount of radioactive material (number of nuclei <-> number of moles <-> weigth) and law of radioactive decay


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.