Difference between revisions of "Solutions 3"

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= Amount of radioactive material (number of nuclei, number of moles, weigth) and law of radioactive decay <br> =
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= Amount of Radioactive Material (number of nuclei, number of moles, weigth) and the Law of Radioactive Decay =
  
 
====== Return to [[Problem Solving Sets]]  ======
 
====== Return to [[Problem Solving Sets]]  ======
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3002 kg <br>  
 
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#&nbsp;After 2.455 • 10<sup>5</sup> years it will be 2.299 • 10<sup>8</sup> Bq <sup>238</sup>U left, which is equal to 18.49 kg.<br><br>
 
#&nbsp;After 2.455 • 10<sup>5</sup> years it will be 2.299 • 10<sup>8</sup> Bq <sup>238</sup>U left, which is equal to 18.49 kg.<br><br>
  
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[[Category:Solved_Problem]] [[Category:Bachelor]]

Latest revision as of 10:01, 9 July 2012

Amount of Radioactive Material (number of nuclei, number of moles, weigth) and the Law of Radioactive Decay

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1:

  1. 3.2  • 108 Bq
  2. 383 Bq
  3. 2.3  • 109 Bq
  4. 8.0  • 104 Bq


2:

  1. 1.77  • 1013 atoms.
  2. 2.00  • 1011 atoms.


3:

2.40 g 60Co.


4:

  1. 2.53  • 105 Bq.
  2. 7.21 years = 7 years and 76.7 days.


5:

The mass of 99mTc is 20.5 ng. After one week 28 half-lives 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 therefore 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]


6:

  1. 1 g natural Lu contains 0.0259 g 176Lu, with a half-life equal to 3.8 • 1010 years, which gives A = 51 Bq
  2. There are two naturally occurring radioisotopes of Sm, 147Sm and 148Sm, which gives A = 127.24 Bq + 0.001 Bq = 127 Bq.


7:

3002 kg


8:

  1. 18.5 kg
  2.  After 2.455 • 105 years it will be 2.299 • 108 Bq 238U left, which is equal to 18.49 kg.