Difference between revisions of "Solutions 5"

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#Thermal neutrons have a kinetic energy of about 0.025eV  
 
#Thermal neutrons have a kinetic energy of about 0.025eV  
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<math>m_{1}^{2}v_{1}^{2}=m_{2}^{2}v_{2}^{2}\rightarrow m_{1}E_{1}=m_{2}E_{2}\rightarrow E_{1}=\frac{m_{2}}{m{1}}E_{2}=\frac{1}{3}E_{2}</math>  
 
<math>m_{1}^{2}v_{1}^{2}=m_{2}^{2}v_{2}^{2}\rightarrow m_{1}E_{1}=m_{2}E_{2}\rightarrow E_{1}=\frac{m_{2}}{m{1}}E_{2}=\frac{1}{3}E_{2}</math>  
  
<math>E_{1}+E_{2}=Q=764 keV\rightarrow E_{1} = 190 Kev, E_{2}=570 keV</math>
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<math>E_{1}+E_{2}=Q=764 keV\rightarrow E_{1} = 190 Kev, E_{2}=570 keV</math>
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#Q-Value_ 2.224MeV<br>
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#&nbsp;And 3. The energy of the deuterium is not calculated relativistic but the γ-ray is completely relativistic, we then get: <math>m_{d}v_{d}=\frac{E_{\gamma}}{c}\rightarrow m_{d}^{2}=\frac{E_{\gamma}^{2}}{c^{2}}\leftarrow \frac{1}{2}m_{d}^{2}v_{d}^{2}=\frac{E^{2}_{\gamma}}{2c^{2}}</math><br>
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Revision as of 12:09, 18 June 2012

1:

  1. Thermal neutrons have a kinetic energy of about 0.025eV
  2. Fast neutrons are “braked” with elastic collisions with particles with the same size as the neutrons. A moderator also needs to have a low cross-section for absorbing neutrons.
  3. A moderator is a material that brakes the neutrons, for instance H2O, D2O, C, Be. A neutron absorbent catches the neutrons and reacts with the neutron. Good absorbants are borium and gadolinium.
  4. From the nuclide carte we can see that 3He has a high cross-section for the n,p-reaction.
  5. The reaction 3He+n arrow 3H++1H(+2e-). The charged particles from the reaction creates more ionisations when they travel towards the anode and the cathode, this gives a electric signal.
  6. The Q-value from the reaction is 764 keV, the reaction is exothermic.
  7. 3H+ and 1H+ is created.
  8. The two products in the reaction recives opposing recoils 180 degrees. Conversion of momentum m1v1=m2v2 gives:

[math]m_{1}^{2}v_{1}^{2}=m_{2}^{2}v_{2}^{2}\rightarrow m_{1}E_{1}=m_{2}E_{2}\rightarrow E_{1}=\frac{m_{2}}{m{1}}E_{2}=\frac{1}{3}E_{2}[/math]

[math]E_{1}+E_{2}=Q=764 keV\rightarrow E_{1} = 190 Kev, E_{2}=570 keV[/math]


2:

  1. Q-Value_ 2.224MeV
  2.  And 3. The energy of the deuterium is not calculated relativistic but the γ-ray is completely relativistic, we then get: [math]m_{d}v_{d}=\frac{E_{\gamma}}{c}\rightarrow m_{d}^{2}=\frac{E_{\gamma}^{2}}{c^{2}}\leftarrow \frac{1}{2}m_{d}^{2}v_{d}^{2}=\frac{E^{2}_{\gamma}}{2c^{2}}[/math]