Solutions 5

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Particles and Nuclear Reactions

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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[math] \rightarrow[/math] 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 763 keV, the reaction is exothermic.
  7. 3H+ and 1H+ is created.
  8. The two products in the reaction receives 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=763\, keV\rightarrow E_{1} = 190.75\, keV, E_{2}=572.25\, keV[/math]


2:

  1. Q-Value 2.224 MeV
  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}v^{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}=2.2211\, MeV[/math]


3:

Q-Values

  • 40Ca: M(40Ca)+M(α)-M(44Ti)=5.13 MeV
  • 52Cr: 7.61 MeV
  • 56Fe: 6.29 MeV
  • 58Ni: 3.37 MeV


4:

For ground state transitions assuming zero momentum to the neutrino

  1. 0.16 MeV
  2. 0.78 MeV
  3. 1.655 MeV subtracting the mass of a electron and a positron we end up at 0.633 MeV.
  4. β- :0.579 for 64Zn,   β+ : 0.653 MeV for 64Ni.
  5. The cause of this effect is the even-even/odd-odd addition to the energy
  6. 228Ra is created with disintegration of 232Th: 232Th [math]\rightarrow[/math]228ra + α +Q this gives
    M(228Ra) = M(232Th)-M(α)-Q
    All of the natural thorium that exist is 232Th. The mass that is given for thorium is therefore more or less equal to the mass 232Th 232.0381 u. It is much the same for helium where the mass is 4.002602 u. To find M(228Ra) we need to find the Q-value. The excess energy (Q-value) is distributed as kinetic energy on the products 228Ra and alpha. The alpha particle has very little mass compared to 228Ra, and the alpha particle by conversion of momentum will receive almost all the kinetic energy:

[math]E_{228Ra} M_{228Ra} =E_{\alpha} M_{\alpha} \rightarrow E_{228Ra} \cdot 228 = E_{\alpha} \cdot 4 \rightarrow E_{228Ra} = E_{\alpha} \cdot \frac{4}{228}[/math]Eα is know ( using the most powerful α for transition to ground state), and the Q value is found with:[math]Q=E_{228Ra}+E_{\alpha}=4.013 \, MeV \cdot \left(1+\frac{4}{228}\right)=4.083 \, MeV[/math]  This gives [math]M(^{228}Ra)=232.0381 \, u-4.002602 \, u - \frac{4.083\, MeV}{931.5 \, MeV/u}=228.03\, u[/math]


5:

  1. [math](^{232}Th(n,\gamma )^{233}Th \rightarrow ^{233}Pa \rightarrow ^{233}U)[/math]  
  2. [math](^{238}U(n ,\gamma ) ^{239}U\rightarrow ^{239}Np \rightarrow ^{239}Pu)[/math]


 6:

  1. [math](^{10}B+n \rightarrow ^{7}Li + \alpha)[/math] 
  2. Q-value is 2.789 MeV
  3. [math]2.789 \, MeV \cdot 10^{14}\, s^{-1} cm ^{-2} \cdot 100\, cm^{2} \cdot 1.602 \cdot 10^{-13} \, W/MeV = 4468 \, W[/math]


7:

  1. [math]Q=M(^{89}Y)+M(^{1}H)-M(^{89}Zr) - M(n) = 13.887 \, MeV[/math]
  2. 1 GBq 89Zr = 4.1 [math]\cdot[/math]1014 atoms, there must be created 1.9 [math]\cdot[/math]1010 per second
  3. Yttrium is a mono isotope, it is only 99Y is the only natural isotope. Therefore there is no need to purify the sample for other isotopes.