Difference between revisions of "Solutions 5"
From mn/safe/nukwik
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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}+ | + | <math>E_{1}+E_{2}=Q=764 keV\rightarrow E_{1} = 190 Kev, E_{2}=570 keV</math> |
Revision as of 12:54, 18 June 2012
1:
- Thermal neutrons have a kinetic energy of about 0.025eV
- 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.
- 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.
- From the nuclide carte we can see that 3He has a high cross-section for the n,p-reaction.
- 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.
- The Q-value from the reaction is 764 keV, the reaction is exothermic.
- 3H+ and 1H+ is created.
- The two products in the reaction recives opposing recoils 180 degrees. Conversion of momentum m1v1=m2v2 gives: