Difference between revisions of "Solutions 4"

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'''2: '''<br>  
 
'''2: '''<br>  
  
binding energy B per nucleon:<br>  
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binding energy B per nucleon:<br> <span style="font-weight: bold;"><math>\frac{B}{A}=\frac{Z\cdot M(^{1}H)+N\cdot M(^{1}n)-M(A,Z)}{A}=</math></span>
<span style="font-weight: bold;"><math>\frac{B}{A}=\frac{Z\cdot M(^{1}H)+N\cdot M(^{1}n)-M(A,Z)}{A}=</math></span><br>
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<br><br>
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<math>=\frac{12 \cdot 1.00782503207 \, u + 12 \cdot 1.0086649156 \, cdot - 23.985041 \, u}{24}=0.00868 \, u</math><br> <br><br>

Revision as of 15:16, 18 June 2012

1:
Mass excess is given by deltaM(Z,A) = M(Z,A)-A. The masses can therefore be found with M(Z,A) = delta M (Z,A ) + A by using 1 u = 931.5 MeV we get:

  • M(n) = 1.008665 u
  • M(1H) = 1.007825 u
  • M(4He) = 4.002603 u
  • M(56Fe) = 55.934938 u
  • M(142Ce) = 141.909244 u
  • M(238U) = 238.050788 u
  1. The most stable nuclei is 56Fe. It has the highest binding energy compared to the number of nucleons.
  2. 1.00 kg 2H is 496.5 mol by fusion 248,2 mol 4He is created which is 993.6 g. The difference in mass is equal to the energy 1000 g-993.6 g = 6.36 g. The energy is then 5.7 [math]\cdot[/math]1014 J, 3.56 [math]\cdot[/math] 1027 MeV or 1.59 [math]\cdot[/math]108 kWh.
  3. 1.00 kg 233U fission to 92Rb, 138Cs and three neutrons. The mass difference is 0.785 g. Which gives 7,1[math]\cdot[/math]1013 J, 4.40 [math]\cdot[/math]1026 MeV or 1.96[math]\cdot[/math]107 kWh.
  4. The energy from the fission is distributed in different ways: most of it goes to the kinetic energy of for the fission products. Other parts goes to the kinetic energy of the neutrinos and the neutrons, and some of it goes to “prompt” gamma-rays and beta/gamma rays from the fission products.


2:

binding energy B per nucleon:
[math]\frac{B}{A}=\frac{Z\cdot M(^{1}H)+N\cdot M(^{1}n)-M(A,Z)}{A}=[/math]

[math]=\frac{12 \cdot 1.00782503207 \, u + 12 \cdot 1.0086649156 \, cdot - 23.985041 \, u}{24}=0.00868 \, u[/math]