Difference between revisions of "Solutions 3"

From mn/safe/nukwik
Jump to: navigation, search
m (moved Solutions 2 to Solutions 3: It is where it belongs)
 
(15 intermediate revisions by 2 users not shown)
Line 1: Line 1:
<br>
+
= Amount of Radioactive Material (number of nuclei, number of moles, weigth) and the Law of Radioactive Decay  =
  
<br>
+
====== Return to [[Problem Solving Sets]]  ======
  
'''1:''' <br>
+
<br>  
  
#3.2 * 10<sup>8</sup> Bq
+
'''1:''' <br>  
#383 Bq
 
#2.3 * 10<sup>9</sup> Bq
 
#8 * 10<sup>4</sup> Bq
 
  
<br>'''2:''' <br>
+
#3.2&nbsp; • 10<sup>8</sup> Bq
 +
#383 Bq
 +
#2.3&nbsp; • 10<sup>9</sup> Bq
 +
#8.0&nbsp; • 10<sup>4</sup> Bq
  
#1.78 * 10<sup>13</sup> atoms.
+
<br>'''2:''' <br>  
#2 * 10<sup>11</sup> atoms.
 
  
<br>'''3:''' 2.40 g <sup>60</sup>Co.
+
#1.77&nbsp; • 10<sup>13</sup> atoms.
 +
#2.00&nbsp; • 10<sup>11</sup> atoms.
  
<br>'''4:'''<br>
+
<br>'''3:'''  
  
#2.53 * 10<sup>5</sup> Bq.
+
2.40 g <sup>60</sup>Co.  
#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>'''4:'''<br>  
  
D = D<sub>0</sub> * (½)<sup>28</sup> = 4 * 10<sup>7</sup> Bq * (½)<sup>28</sup> = 0.15 Bq.
+
#2.53&nbsp; • 10<sup>5</sup> Bq.
 +
#7.21 years = 7 years and 76.7 days.
  
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.
+
<br>'''5:'''
  
N = = = 1,25 * 10<sup>12</sup> atomer
+
The mass of <sup>99m</sup>Tc is 20.5 ng. After one week 28 half-lives will have past, and the activity will be the following:<br>
  
&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.
+
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.  
  
<br>
+
All <sup>99m</sup>Tc has disintegrated and turned into <sup>99</sup>Tc. The number of <sup>99</sup>Tc atoms will therefore be the same as the original number of <sup>99m</sup>Tc atoms.
  
'''6:'''
+
<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>
  
#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
+
<br>  
#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.
 
  
 +
'''6:'''<br>
  
'''7:''' 3002 kg
+
#1 g natural Lu contains 0.0259 g <sup>176</sup>Lu, with a half-life equal to 3.8 • 10<sup>10</sup> years, which gives A = 51 Bq <br>
 +
#There are two naturally occurring radioisotopes of Sm, <sup>147</sup>Sm and <sup>148</sup>Sm, which gives A = 127.24 Bq + 0.001 Bq = 127 Bq.
  
 +
<br>
  
 +
'''7: '''
  
'''8:'''
+
3002 kg <br>
  
#18.5 kg
+
<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>
+
 
 +
'''8:''' <br>
 +
 
 +
#18.5 kg  
 +
#&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>
 +
 
 +
<br>
 +
 
 +
[[Category:Solved_Problem]] [[Category:Bachelor]]

Latest revision as of 09:01, 9 July 2012

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

Return to Problem Solving Sets


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.