U_92^238 \rightarrow Th_90^234 + \alpha_2^4
Momentum Conservation: \quad vecP_i = vec P_f
vec P_{U} = vec P_{Th} + vec P_{\alpha}
The Uranium nucleus is at rest (vec P_U = vec 0).
vec P_{Th} + vec P_{\alpha} = vec 0; \qquad vec P_{\alpha} = - vec P_{Th}; \qquad P_{\alpha} = P_{Th} = P_0
Kinetic Energies:
K_{\alpha} = P_{\alpha}^2/(2m_{\alpha}); \qquad K_{Th} = P_{Th}^2/(2m_{Th}^234);
P_{\alpha} = P_{Th} = P_0;
m_{U}^238 = 238.050788 \quad u; \qquad m_{Th}^234 = 234.04360123 \quad u;
m_{\alpha} = 4.001506 \quad u;
K_{\alpha}/K_{Th} = (P_{\alpha}/P_{Th})^2(m_{Th}/m_{\alpha}) = m_{Th}/m_{\alpha} = (234.0436\quad u)/ (4.001\quad u) = 58.49
K_{Th} = K_{\alpha}/58.49 = 0.017095.K_{\alpha}; ...... (1)
Fission Energy - Q Value:
Q = \Deltam.c^2 = [m_U - (m_{Th} + m_{\alpha})].c^2 = 0.005681 u.c^2 =
To convert the atmoc mass unit (u) to MeV use the conversion factor,
1 u = 931.5 (MeV)/c^2
Thus the energy released in the fission is, \quad Q = 5.292 MeV. This energy is distributed between the alpha particle and the Thorium-234 nucleus as kinetic energies.
K_{\alpha} + K_{Th} = Q = 5.292 MeV ...... (2)
Substituting (1) in (2) to eliminate K_{Th},
(1 + 0.017095)K_{\alpha} = 5.292 MeV
K_{\alpha} = 5.203 MeV = 8.336\times10^{-13} J
