# Difference between revisions of "Solutions 5"

From mn/safe/nukwik

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<math>E_{d}=0.0013\, MeV,\, E_{\gamma}=2.2226\, MeV</math> | <math>E_{d}=0.0013\, MeV,\, E_{\gamma}=2.2226\, MeV</math> | ||

− | <br> | + | <br> |

'''3:''' Q-Values<br> | '''3:''' Q-Values<br> | ||

− | *<sup>40</sup>Ca: M(<sup>40</sup>Ca)+M(α)-M(<sup>44</sup>Ti)=5.126 MeV<br> | + | *<sup>40</sup>Ca: M(<sup>40</sup>Ca)+M(α)-M(<sup>44</sup>Ti)=5.126 MeV<br> |

− | *<sup>22</sup>Cr: 7.61 MeV<br> | + | *<sup>22</sup>Cr: 7.61 MeV<br> |

− | *<sup>56</sup>Fe: 6.29 MeV<br> | + | *<sup>56</sup>Fe: 6.29 MeV<br> |

*<sup>58</sup>Ni: 3.367 MeV<br> | *<sup>58</sup>Ni: 3.367 MeV<br> | ||

− | <br> | + | <br> |

− | '''4:''' For ground state transitions assuming zero momentum to the neutrino<br> | + | '''4:''' For ground state transitions assuming zero momentum to the neutrino<br> |

− | #0.16 MeV<br> | + | #0.16 MeV<br> |

− | #0.78 MeV<br> | + | #0.78 MeV<br> |

− | #1.655 MeV subtracting the mass of a electron and a positron we end up at 0.633 MeV.<br> | + | #1.655 MeV subtracting the mass of a electron and a positron we end up at 0.633 MeV.<br> |

− | #β<sup>-</sup> :0.579 for 64Zn, β<sup>+</sup> : 0.653 MeV for <sup>64</sup>Ni. | + | #β<sup>-</sup> :0.579 for 64Zn, β<sup>+</sup> : 0.653 MeV for <sup>64</sup>Ni. |

#The cause of this effect is the even-even/odd-odd addition to the energy<br> | #The cause of this effect is the even-even/odd-odd addition to the energy<br> | ||

− | #<sup>228</sup>Ra is created with desintigration of <sup>232</sup>Th: <sup>232</sup>Th arrow <sup>228</sup>ra + α +Q this gives<br>M(<sup>228</sup>Ra) = M(<sup>230</sup>Th)-M(α)-Q<br>All of the naural thorium that exist is <sup>232</sup>Th. The mass that is given for thorium is therefore more or less equal to the mass <sup>232</sup>Th 232.0381 u. It is much the same for helium where the mass is 4.002602 u. To find M(<sup>228</sup>Ra) we need to find the Q-value. The excess energy (Q-value) is distributed as kinetic energy on the products <sup>228</sup>Ra and alpha. The alpha particle has very little mass compared to <sup>228</sup>Ra, and the alpha particle by conversion of momentum will receive almost all the kinetic energy: <math>E_{ | + | #<sup>228</sup>Ra is created with desintigration of <sup>232</sup>Th: <sup>232</sup>Th arrow <sup>228</sup>ra + α +Q this gives<br>M(<sup>228</sup>Ra) = M(<sup>230</sup>Th)-M(α)-Q<br>All of the naural thorium that exist is <sup>232</sup>Th. The mass that is given for thorium is therefore more or less equal to the mass <sup>232</sup>Th 232.0381 u. It is much the same for helium where the mass is 4.002602 u. To find M(<sup>228</sup>Ra) we need to find the Q-value. The excess energy (Q-value) is distributed as kinetic energy on the products <sup>228</sup>Ra and alpha. The alpha particle has very little mass compared to <sup>228</sup>Ra, and the alpha particle by conversion of momentum will receive almost all the kinetic energy:<span style="font-weight: bold;"> <math>E_{228Ra} M_{228Ra} =E_{\alpha} M_{\alpha} \rightarrow E_{^{288}Ra} \cdot 288 = E_{\alpha} \cdot 4 \rightarrow E_{^{228}Ra} = E_{\alpha} \cdot \frac{4}{228}</math></span |

+ | |||

+ | E<sub>α</sub><sub></sub> is know ( using the most powerfull α for transition to groundstate), and the Q value is found with: | ||

+ | |||

+ | |||

#<br><br> | #<br><br> | ||

## Revision as of 13:31, 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:

**3:** Q-Values

^{40}Ca: M(^{40}Ca)+M(α)-M(^{44}Ti)=5.126 MeV

^{22}Cr: 7.61 MeV

^{56}Fe: 6.29 MeV

^{58}Ni: 3.367 MeV

**4:** For ground state transitions assuming zero momentum to the neutrino

- 0.16 MeV

- 0.78 MeV

- 1.655 MeV subtracting the mass of a electron and a positron we end up at 0.633 MeV.

- β
^{-}:0.579 for 64Zn, β^{+}: 0.653 MeV for^{64}Ni. - The cause of this effect is the even-even/odd-odd addition to the energy

^{228}Ra is created with desintigration of^{232}Th:^{232}Th arrow^{228}ra + α +Q this gives

M(^{228}Ra) = M(^{230}Th)-M(α)-Q

All of the naural thorium that exist is^{232}Th. The mass that is given for thorium is therefore more or less equal to the mass^{232}Th 232.0381 u. It is much the same for helium where the mass is 4.002602 u. To find M(^{228}Ra) we need to find the Q-value. The excess energy (Q-value) is distributed as kinetic energy on the products^{228}Ra and alpha. The alpha particle has very little mass compared to^{228}Ra, and the alpha particle by conversion of momentum will receive almost all the kinetic energy: </span

E_{α}_{}is know ( using the most powerfull α for transition to groundstate), and the Q value is found with: