Suggested+Solutions+to+Particles+and+Nuclear+Reactions

  Exercises with Particles and Nuclear Reactions
 * 1: **
 * 1) Thermal neutrons have a kinetic energy of about 0.025 eV
 * 2) Fast neutrons are slowed by elastic collisions with particles with the similar mass as the neutrons. A moderator also needs to have a low cross-section for absorbing neutrons.
 * 3) A moderator is a material that slows down the neutrons, for instance H 2 O, D 2 O, C, Be. A neutron absorbent catches the neutrons and reacts with the neutron. Good absorbents are e.g. borium and gadolinium.
 * 4) The reaction 3 He+n→ 3 H + + 1 H(+2e - ). The charged particles from the reaction creates more ionizations when they travel towards the anode and the cathode, this gives a electric signal.
 * 5) The Q-value from the reaction is 763 keV, the reaction is exothermic.
 * 6) 3 H + and 1 H + is created.
 * 7) From the nuclide carte we can see that 3 He has a high cross-section for the n,p-reaction.
 * 8) The two products in the reaction receive opposing recoils 180 degrees. Conversion of momentum m 1 v 1 =m 2 v 2 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_dv_d=\frac{E_{\gamma}}{c}\rightarrow m_d^2v^2=\frac{1}{2}m_d^2v^2=\frac{E_{\gamma}^2}{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

Q-Values
 * 3: **
 * 40 Ca: M( 40 Ca)+M(α)-M( 44 Ti)=5.13 MeV
 * 52 Cr: 7.61 MeV
 * 56 Fe: 6.29 MeV
 * 58 Ni: 3.37 MeV

For ground state transitions assuming zero momentum to the neutrino > M( 228 Ra) = M( 232 Th)-M(α)-Q > All of the natural 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 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:
 * 4: **
 * 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 64 Zn, β + <span style="font-family: 'Times New Roman','serif'; font-size: 16px;"> : 0.653 MeV for 64 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Ni.
 * 5) <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">The cause of this effect is the even-even/odd-odd addition to the energy
 * 6) 228 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Ra is created with disintegration of 232 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Th: 232 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Th→ 228 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">ra + α +Q this gives

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

<span style="font-family: 'Times New Roman','serif'; font-size: 16px;">E α <span style="font-family: 'Times New Roman','serif'; font-size: 16px;"> is known (using the most powerful α for transition to ground state), and the Q value is found

math Q=E_{228Ra}+E_{\alpha}=4.013 \, MeV \cdot \left(1+\frac{4}{228}\right)=4.083 \, MeV

math <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">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:**

(^{232}Th(n,\gamma )^{233}Th \rightarrow ^{233}Pa \rightarrow ^{233}U) math
 * 1) math

(^{238}U(n ,\gamma ) ^{239}U\rightarrow ^{239}Np \rightarrow ^{239}Pu) math
 * 1) math

(^{10}B+n \rightarrow ^{7}Li + \alpha) math
 * 6:**
 * 1) math
 * 1) Q-value is 2.789 MeV

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
 * 1) math

Q=M(^{89}Y)+M(^{1}H)-M(^{89}Zr) - M(n) = 13.887 \, MeV math >
 * 7:**
 * 1) math
 * 1) 1 GBq 89 Zr = 4.1•10 14 atoms, there must be created 1.9•10 10 per second
 * 2) Yttrium is a mono isotope, you will only find 99Y in nature. Therefore there is no need to purify the sample for other isotopes.

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