Difference between revisions of "Solutions 3"

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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&nbsp; • 10<sup>12</sup> atomer
+
<math>N=\frac{D}{\lambda}=\frac{4\cdot 10^{7}}{3.21\cdot 10^{-5}}</math>
 
 
&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>
 
 
 
'''6:'''
 
 
 
#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.
 
 
 
<br> '''7:''' 3002 kg
 
 
 
<br>
 
 
 
'''8:'''
 
 
 
#18.5 kg
 
#&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 10:47, 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.

[math]N=\frac{D}{\lambda}=\frac{4\cdot 10^{7}}{3.21\cdot 10^{-5}}[/math]