Difference between revisions of "Solutions 5"

Revision as of 13:18, 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}}\rightarrow \frac{1}{2}m_{d}^{2}v_{d}^{2}=\frac{E^{2}_{\gamma}}{2c^{2}}[/math]

         Inserting the energy of the deuterium we get:

         [math]E_{d}=\frac{E_{\gamma}^{2}}{2m_{d}c^{2}}[/math]

         we know that [math]E_{d}+E_{\gamma}=Q=2.2224\, MeV[/math] solving for this:

         [math]E_{d}=0.0013\, MeV,\, E_{\gamma}=2.2226\, MeV[/math]