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)
(Amount of Radioactive Material (Number of Nuclei, Number of Moles, Weigth) and Law of Radioactive Decay)
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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]]  ======

Revision as of 16:08, 25 June 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.