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.  
  
<math>N=\frac{D}{\lambda}=\frac{4\cdot 10^{7}}{3.21\cdot 10^{-5}}</math>
+
<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>

Revision as of 10:49, 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} \rightarrow \frac{4\cdot 10^{7}Bq}{3.21\cdot 10^{-5}s^{-1}}=1.25\cdot 10^{12}atoms[/math]