# Difference between revisions of "Solutions 5"

From mn/safe/nukwik

Line 45: | Line 45: | ||

#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 | + | #β<sup>-</sup> :0.579 for <sup>64</sup>Zn, 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:<span style="font-weight: bold;"> | #<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;"> | ||

Line 51: | Line 51: | ||

+ | |||

+ | <br> | ||

<math>E_{228Ra} M_{228Ra} =E_{\alpha} M_{\alpha} \rightarrow E_{288Ra} \cdot 288 = E_{\alpha} \cdot 4 \rightarrow E_{228Ra} = E_{\alpha} \cdot \frac{4}{228}</math><br> | <math>E_{228Ra} M_{228Ra} =E_{\alpha} M_{\alpha} \rightarrow E_{288Ra} \cdot 288 = E_{\alpha} \cdot 4 \rightarrow E_{228Ra} = E_{\alpha} \cdot \frac{4}{228}</math><br> | ||

Line 72: | Line 74: | ||

#<math>Q=M(^{89}Y)+M(^{1}H)-M(^{89}Zr) - M(n) = -3.6 \, MeV</math> | #<math>Q=M(^{89}Y)+M(^{1}H)-M(^{89}Zr) - M(n) = -3.6 \, MeV</math> | ||

− | #1 GBq <sup>89</sup>Zr = 4.1<math>\cdot</math>10<sup>14</sup> atoms, there must be created 1.9<math>\cdot</math>10<sup>10</sup> per second | + | #1 GBq <sup>89</sup>Zr = 4.1 <math>\cdot</math>10<sup>14</sup> atoms, there must be created 1.9 <math>\cdot</math>10<sup>10</sup> per second |

#Yttrium is a mono isotope, it is only <sup>99</sup>Y is the only natural isotope. Therefore there is no need to purify the sample for other isotopes.<br> | #Yttrium is a mono isotope, it is only <sup>99</sup>Y is the only natural isotope. Therefore there is no need to purify the sample for other isotopes.<br> | ||

## Revision as of 10:16, 19 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^{64}Zn, 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:

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

This gives

**5: **

** 6:**

- Q-value is 2.788 MeV

**7:**

- 1 GBq
^{89}Zr = 4.1 10^{14}atoms, there must be created 1.9 10^{10}per second - Yttrium is a mono isotope, it is only
^{99}Y is the only natural isotope. Therefore there is no need to purify the sample for other isotopes.