[1222] | 1 | \section{Fission process simulation.} |
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| 2 | |
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| 3 | \subsection{Atomic number distribution of fission products.} |
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| 4 | |
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| 5 | \hspace{1.0em}As follows from experimental data \cite{VH73} mass |
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| 6 | distribution of fission products consists of the symmetric and the |
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| 7 | asymmetric components: |
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| 8 | \begin{equation} |
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| 9 | \label{FPS1} F(A_f) = F_{sym}(A_f) + \omega F_{asym}(A_f), |
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| 10 | \end{equation} |
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| 11 | where $\omega(U,A,Z)$ defines relative contribution of each component |
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| 12 | and it depends from excitation energy $U$ and $A,Z$ of fissioning |
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| 13 | nucleus. It was found in \cite{ABIM93} that experimental data can be |
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| 14 | approximated with a good accuracy, if one take |
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| 15 | \begin{equation} |
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| 16 | \label{FPS2} F_{sym}(A_f) = \exp{[-\frac{(A_f - A_{sym})^2}{2\sigma_{sym}^2}]} |
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| 17 | \end{equation} |
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| 18 | and |
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| 19 | \begin{equation} |
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| 20 | \begin{array}{c} |
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| 21 | \label{FPS3} F_{asym}(A_f) = \exp{[-\frac{(A_f - A_{2})^2}{2\sigma_{2}^2}]} + |
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| 22 | \exp{[-\frac{{A_f - (A - A_{2})}^2}{2\sigma_{2}^2}]} + \\ |
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| 23 | + C_{asym}\{\exp{[-\frac{(A_f - A_{1})^2}{2\sigma_{1}^2}]} + |
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| 24 | \exp{[-\frac{{A_f - (A - A_{1})}^2}{2\sigma_{2}^2}]}\}, |
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| 25 | \end{array} |
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| 26 | \end{equation} |
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| 27 | where $A_{sym} = A/2$, $A_1$ and $A_2$ are the mean values and |
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| 28 | $\sigma^2_{sim}$, $\sigma^2_1$ and $\sigma^2_2$ are dispertions of the |
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| 29 | Gaussians respectively. From an analysis of experimental data |
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| 30 | \cite{ABIM93} the parameter $C_{asym} \approx 0.5$ was defined and the |
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| 31 | next values for dispersions: |
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| 32 | \begin{equation} |
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| 33 | \label{FPS4} \sigma^2_{sym} = \exp{(0.00553U + 2.1386)}, |
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| 34 | \end{equation} |
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| 35 | where $U$ in MeV, |
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| 36 | \begin{equation} |
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| 37 | \label{FPS5} 2\sigma_1 = \sigma_2 = 5.6 \ MeV |
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| 38 | \end{equation} |
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| 39 | for $A \leq 235$ and |
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| 40 | \begin{equation} |
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| 41 | \label{FPS6} 2\sigma_1 = \sigma_2 = 5.6 + 0.096 (A - 235) \ MeV |
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| 42 | \end{equation} |
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| 43 | for $A > 235$ were found. |
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| 44 | |
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| 45 | The weight $\omega(U,A,Z)$ was approximated as follows |
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| 46 | \begin{equation} |
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| 47 | \label{FPS7} \omega = \frac{\omega_{a} - F_{asym}(A_{sym})} |
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| 48 | {1 - \omega_a F_{sym}((A_1 + A_2)/2)}. |
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| 49 | \end{equation} |
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| 50 | The values of $\omega_a$ for nuclei with $96 \geq Z \geq 90$ were |
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| 51 | approximated by |
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| 52 | \begin{equation} |
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| 53 | \label{FPS8} \omega_a(U) = \exp{(0.538U - 9.9564)} |
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| 54 | \end{equation} |
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| 55 | for $U \leq 16.25$ MeV, |
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| 56 | \begin{equation} |
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| 57 | \label{FPS9} \omega_a(U) = \exp{(0.09197U - 2.7003)} |
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| 58 | \end{equation} |
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| 59 | for $U > 16.25$ MeV and |
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| 60 | \begin{equation} |
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| 61 | \label{FPS10} \omega_a(U) = \exp{(0.09197U - 1.08808)} |
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| 62 | \end{equation} |
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| 63 | for $z = 89$. |
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| 64 | For nuclei with $Z \leq 88$ the authors of \cite{ABIM93} constracted |
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| 65 | the following approximation: |
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| 66 | \begin{equation} |
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| 67 | \label{FPS11}\omega_a(U) = |
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| 68 | \exp{[0.3(227 - a)]} \exp{ \{0.09197[U - (B_{fis} - 7.5)] |
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| 69 | - 1.08808 \}}, |
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| 70 | \end{equation} |
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| 71 | where for $A > 227$ and $U < B_{fis} - 7.5$ the corresponding factors occuring |
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| 72 | in exponential functions vanish. |
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| 73 | |
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| 74 | \subsection{Charge distribution of fission products.} |
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| 75 | |
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| 76 | \hspace{1.0em}At given mass of fragment $A_f$ the |
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| 77 | experimental data \cite{VH73} on the charge $Z_f$ distribution of |
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| 78 | fragments are well approximated by Gaussian with dispertion |
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| 79 | $\sigma^2_{z} = 0.36$ and the average $<Z_f>$ is described by |
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| 80 | expression: |
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| 81 | \begin{equation} |
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| 82 | \label{FPS12} <Z_f> = \frac{A_f}{A}Z + \Delta Z, |
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| 83 | \end{equation} |
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| 84 | when parameter $\Delta Z = -0.45$ for $A_f \geq 134$, $\Delta Z = - |
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| 85 | 0.45(A_f -A/2)/(134 - A/2)$ for $ A - 134 < A_f < 134$ and $\Delta Z = |
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| 86 | 0.45$ for $A \leq A - 134$. |
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| 87 | |
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| 88 | After sampling of fragment atomic masses numbers and fragment charges, |
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| 89 | we have to check that fragment ground state masses do not exceed initial |
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| 90 | energy and calculate the maximal fragment kinetic energy |
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| 91 | \begin{equation} |
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| 92 | \label{FPS13a}T^{max} < U + M(A,Z) - M_1(A_{f1}, Z_{f1}) - M_2(A_{f2}, Z_{f2}), |
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| 93 | \end{equation} |
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| 94 | where $U$ and $M(A,Z)$ are the excitation energy and mass of initial |
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| 95 | nucleus, $M_1(A_{f1}, |
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| 96 | Z_{f1})$, and $M_2(A_{f2}, Z_{f2})$ are masses |
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| 97 | of the first and second fragment, respectively. |
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| 98 | |
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| 99 | |
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| 100 | \subsection{Kinetic energy distribution of fission products.} |
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| 101 | |
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| 102 | \hspace{1.0em}We use the empiricaly defined \cite{VKW85} dependence of |
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| 103 | the average kinetic energy $<T_{kin}>$ (in MeV) of fission fragments on |
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| 104 | the mass and the charge of a fissioning nucleus: |
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| 105 | \begin{equation} |
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| 106 | \label{FPS13}<T_{kin}> = 0.1178 Z^2/A^{1/3} + 5.8. |
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| 107 | \end{equation} |
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| 108 | This energy is distributed differently in cases of symmetric and |
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| 109 | asymmetric modes of fission. It follows from the analysis of data |
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| 110 | \cite{ABIM93} that in the asymmetric mode, the average kinetic energy of |
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| 111 | fragments is higher than that in the symmetric one by approximately |
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| 112 | $12.5$ MeV. To approximate the average numbers of kinetic energies |
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| 113 | $<T_{kin}^{sym}$ and $<T_{kin}^{asym}>$ for the symmetric and asymmetric |
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| 114 | modes of fission the authors of \cite{ABIM93} suggested empirical |
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| 115 | expressions: |
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| 116 | \begin{equation} |
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| 117 | \label{FPS14} <T_{kin}^{sym}> = <T_{kin}> - 12.5 W_{asim}, |
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| 118 | \end{equation} |
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| 119 | \begin{equation} |
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| 120 | \label{FPS15} <T_{kin}^{asym}> = <T_{kin}> + 12.5 W_{sim}, |
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| 121 | \end{equation} |
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| 122 | where |
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| 123 | \begin{equation} |
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| 124 | \label{FPS16} W_{sim} = \omega \int F_{sim}(A)dA/\int F(A)dA |
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| 125 | \end{equation} |
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| 126 | and |
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| 127 | \begin{equation} |
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| 128 | \label{FPS17} W_{asim} = \int F_{asim}(A)dA/\int F(A)dA, |
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| 129 | \end{equation} |
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| 130 | respectively. In the symmetric fission the experimental data for the |
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| 131 | ratio of the average kinetic energy of fission fragments |
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| 132 | $<T_{kin}(A_f)>$ to this maximum energy $<T^{max}_{kin}>$ as a function |
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| 133 | of the mass of a larger fragment $A_{max}$ can be approximated by |
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| 134 | expressions |
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| 135 | \begin{equation} |
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| 136 | \label{FPS18} <T_{kin}(A_f)>/<T^{max}_{kin}> = |
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| 137 | 1 - k [(A_f - A_{max})/A]^2 |
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| 138 | \end{equation} |
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| 139 | for $A_{sim} \leq A_f \leq A_{max} + 10$ and |
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| 140 | \begin{equation} |
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| 141 | \label{FPS19} <T_{kin}(A_f)>/<T^{max}_{kin}> = |
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| 142 | 1 - k(10/A)^2 - 2 (10/A)k(A_f - A_{max} - 10)/A |
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| 143 | \end{equation} |
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| 144 | for $A_f > A_{max} + 10$, where $A_{max} = A_{sim}$ and $k = 5.32$ and |
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| 145 | $A_{max} = 134$ and $k = 23.5$ for symmetric and asymmetric fission |
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| 146 | respectively. For both modes of fission the distribution over the |
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| 147 | kinetic energy of fragments $T_{kin}$ is choosen Gaussian with their own |
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| 148 | average values $<T_{kin}(A_f)>= <T_{kin}^{sym}(A_f)>$ or |
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| 149 | $<T_{kin}(A_f)>=<T_{kin}^{asym}(A_f)>$ and dispersions $\sigma^2_{kin}$ |
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| 150 | equal $8^2$ MeV or $10^2$ MeV$^2$ for symmetrical and asymmetrical |
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| 151 | modes, respectively. |
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| 152 | |
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| 153 | \subsection{Calculation of the excitation energy of fission products.} |
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| 154 | |
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| 155 | \hspace{1.0em}The total excitation energy of fragments $U_{frag}$ |
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| 156 | can be defined according to equation: |
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| 157 | \begin{equation} |
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| 158 | \label{FPS21} U_{frag} = U + M(A,Z) - M_1(A_{f1}, Z_{f1}) - M_2(A_{f2}, Z_{f2}) - |
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| 159 | T_{kin}, |
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| 160 | \end{equation} |
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| 161 | where $U$ and $M(A,Z)$ are the excitation energy and mass of initial |
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| 162 | nucleus, $T_{kin}$ is the fragments kinetic energy, $M_1(A_{f1}, |
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| 163 | Z_{f1})$, and $M_2(A_{f2}, Z_{f2})$ are masses |
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| 164 | of the first and second fragment, respectively. |
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| 165 | |
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| 166 | The value of excitation energy of fragment $U_f$ determines the fragment |
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| 167 | temperature ($T = \sqrt{U_f/a_f}$, where $a_f \sim A_f$ is the parameter |
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| 168 | of fragment level density). Assuming that after disintegration |
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| 169 | fragments have the same temperature as initial nucleus than the total |
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| 170 | excitation energy will be distributed between fragments in proportion to |
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| 171 | their mass numbers one obtains |
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| 172 | \begin{equation} |
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| 173 | \label{FPS22} U_f = U_{frag} \frac{A_f}{A}. |
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| 174 | \end{equation} |
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| 175 | |
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| 176 | \subsection{Excited fragment momenta.} |
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| 177 | |
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| 178 | \hspace{1.0em}Assuming that fragment kinetic energy $T_f= |
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| 179 | P^2_f/(2(M(A_{f},Z_{f}+U_f)$ we are |
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| 180 | able to calculate the absolute value of fragment c.m. momentum |
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| 181 | \begin{equation} |
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| 182 | \label{FPS23} |
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| 183 | P_f=\frac{(M_1(A_{f1},Z_{f1}+U_{f1})(M_2(A_{f2},Z_{f2}+U_{f2})}{ |
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| 184 | M_1(A_{f1},Z_{f1})+U_{f1} + M_2(A_{f2},Z_{f2})+U_{f2}}T_{kin}. |
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| 185 | \end{equation} |
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| 186 | and its components, assuming fragment isotropical distribution. |
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