# Difference between revisions of "Solutions 5"

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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 | + | <math>E_{1}+E_{2}=Q=764\, keV\rightarrow E_{1} = 190\, keV, E_{2}=570\, keV</math> |

− | <br> | + | <br> |

− | '''2:'''<br> | + | '''2:'''<br> |

− | #Q-Value_ 2.224MeV<br> | + | #Q-Value_ 2.224MeV<br> |

− | # 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}}\ | + | # 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 <sub></sub><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> | ||

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## Revision as of 13:18, 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 H
_{2}O, D_{2}O, 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
^{3}He has a high cross-section for the n,p-reaction. - The reaction
^{3}He+n arrow^{3}H^{+}+^{1}H(+^{2}e^{-}). 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.
^{3}H^{+}and^{1}H^{+}is created.- The two products in the reaction recives opposing recoils 180 degrees. Conversion of momentum m
_{1}v_{1}=m_{2}v_{2}gives:

**2:**

- Q-Value_ 2.224MeV

- And 3. The energy of the deuterium is not calculated relativistic but the γ-ray is completely relativistic, we then get:

Inserting the energy of the deuterium we get:

we know that _{} solving for this: