1 | \section{Simulation of fragment evaporation.} |
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2 | |
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3 | \hspace{1.0em} The evaporation of |
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4 | neutron, proton, deutron, thritium and alpha fragments |
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5 | are taken into account. |
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6 | |
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7 | \subsection{Evaporation threshold.} |
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8 | |
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9 | \hspace{1.0em}One should take into account the energy condition |
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10 | for fragment emission, i. e. the nucleus |
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11 | excitation energy should be higher than |
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12 | the reaction threshold: |
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13 | \begin{equation} |
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14 | \label{SFE1}T_b^{\max }=E^{*}-Q_b-V_b > 0. |
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15 | \end{equation} |
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16 | Here $T_b^{\max }$ is the maximal kinetic energy carried by emitted |
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17 | fragment $b$, $Q_b = M(A,Z)-M(A_f,Z_f)-M_b$ is the fragment $b$ binding |
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18 | energy. $V_b$ is the Coulomb |
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19 | potential energy, i. e. the Coulomb barrier for fragment $b$. |
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20 | $M(A,Z)$ is the mass of the initial nucleus, $M(A_f,Z_f)$ is |
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21 | the mass of the nucleus after emission of fragment $b$ and $M_b$ is the |
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22 | fragment $b$ mass. It should be noted that expression ($\ref{SFE1}$) is |
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23 | only valid, when the recoil kinetic energy equals zero. In our code we |
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24 | apply the condition: |
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25 | \begin{equation} |
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26 | \label{SFE2}T_b^{\max }=E_b^{\max}-M_b-V_b > 0, |
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27 | \end{equation} |
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28 | where |
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29 | \begin{equation} |
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30 | \label{SFE3}E_b^{\max }=\frac{(M(A,Z)+E^{*})^2+M^2_b-M^2(A_f,Z_f)} |
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31 | {2(M(A,Z)+E^{*})}. |
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32 | \end{equation} |
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33 | |
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34 | \subsection{Coulomb barrier calculation.} |
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35 | |
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36 | \hspace{1.0em}The Coulomb barrier: |
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37 | \begin{equation} |
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38 | V_b = C_b\frac{Z_bZ_f}{R_f+R_b}, |
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39 | \end{equation} |
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40 | where $C_b = 1.44$ \ MeVfm, $Z_b$ and $R_b$ are charge and |
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41 | radius of nucleus after fragment |
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42 | emission, $Z_f$ and $R_f$ are charge and radius of fragment. |
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43 | The radii of nuclei are |
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44 | approximated by $R = r_{C}A^{1/3}$, |
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45 | where \cite{IKP94} |
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46 | \begin{equation} |
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47 | \label{SFE8}r_{C} = 2.173\frac{1+0.006103Z_bZ_f}{1+0.009443Z_bZ_f} \ fm. |
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48 | \end{equation} |
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49 | |
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50 | \subsection{The fragment evaporation probability.} |
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51 | |
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52 | \hspace{1.0em}Evaporation process has been predicted by the statistical |
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53 | Weisskopf-Ewing model \cite{WE40.evap}. Probability to evaporate particle |
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54 | $b$ in the energy interval $(T_b, T_b+dT_b)$ per unit of time is given |
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55 | \begin{equation} |
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56 | \label{SFE9}W_{b}(T_b) = \sigma_{b}(T_b)\frac{(2s_b+1)m_b}{\pi^2 h^3} |
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57 | \frac{\rho(E^{*} - Q_b-T_b)}{\rho(E^{*})}T_b, |
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58 | \end{equation} |
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59 | where $\sigma_{b}(T_b)$ is the inverse (absorption of particle $b$) |
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60 | reaction cross section, $s_b$ and $m_b$ are particle spin and mass, |
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61 | $\rho$ is level densities of nucleus. |
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62 | |
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63 | \subsection{The inverse reaction cross section.} |
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64 | |
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65 | \hspace{1.0em}To calculate inverse reaction cross section it is |
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66 | assumed to have the form \cite{Dostr59} |
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67 | \begin{equation} |
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68 | \label{SFE11} |
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69 | \sigma _b(T_b ) = (1+C_b)(1-k_bV_b/T_b )\pi R^2 |
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70 | \end{equation} |
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71 | for charged charged fragments with $A_f \leq 4$ interaction, where the |
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72 | $k_b$ is the barrier penetration coefficient ( its tabulated values are |
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73 | used), and |
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74 | \begin{equation} |
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75 | \label{SFE12} |
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76 | \sigma _b(T_b ) = \alpha (1 + \beta /T_b )\pi R^2 |
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77 | \end{equation} |
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78 | for neutrons. Here $R = r_{0} A^{1/3}$ denotes the absorption radius, where |
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79 | $r_0 = 1.5$ \ fm, |
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80 | $\alpha =0.76 + 2.2 A^{-1/3}$ and $\beta=(2.12 A^{-2/3} - 0.05)/(0.76 + |
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81 | 2.2 A^{-1/3})$. |
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82 | |
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83 | \subsection{The level density.} |
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84 | |
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85 | \hspace{1.0em}The level |
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86 | density is approximated by Fermi-gas approach \cite{IKP94} for the nuclear |
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87 | level density: |
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88 | \begin{equation} |
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89 | \label{SFE13} \rho(E^{*}) = C\exp{(2\sqrt{aE^{*}})}, |
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90 | \end{equation} |
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91 | where $C$ is a constant, which does not depend from nucleus properties |
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92 | and excitation energy $E^{*}$ and $a$ is the level density parameter. |
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93 | |
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94 | \subsection{The total evaporation probability.} |
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95 | |
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96 | \hspace{1.0em}The total probability $W_b$ or total partial width |
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97 | $\Gamma_b=\hbar W_b$ to evaporate particle $b$ can be obtained from |
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98 | Eq. ($\ref{SFE9}$) by integration over $T_b$: |
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99 | \begin{equation} |
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100 | \label{SFE17} |
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101 | W_{b}=\int_{V_b}^{E^{*}-Q_b} W_b(T_b)dT_b. |
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102 | \end{equation} |
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103 | Here the summation is carried out over all excited states of the |
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104 | fragment. |
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105 | |
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106 | Integration in Eq. ($\ref{SFE17}$) for probability to emit fragment $b$ |
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107 | can be performed analiticaly, if we will use Eq. ($\ref{SFE11}$) for level |
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108 | density and the Eqs. ($\ref{SFE11}$)-($\ref{SFE12}$) for inverse cross |
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109 | section. |
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110 | The probability |
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111 | to emit a charged particle: |
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112 | \begin{equation} |
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113 | \label{SFE18} |
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114 | \begin{array}{c} |
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115 | W_{b}=\gamma _bA_b^{2/3}B\exp [-2 |
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116 | \sqrt{aE^{*}}]\frac{(1+C_b)}{a_b^2}\{a_bT_b^{\max }[2\exp (2\sqrt{% |
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117 | a_bT_b^{\max }})+1]- \\ |
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118 | - 3\sqrt{a_bT_b^{\max }}\exp (2\sqrt{a_bT_b^{\max }}% |
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119 | )-3[1-\exp (2\sqrt{a_bT_b^{\max }})]/2\}, |
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120 | \end{array} |
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121 | \end{equation} |
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122 | where $T_b^{max}$ is defined by the equation ($\ref{SFE1}$). |
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123 | The following notations were introduced: $A_b=A-\Delta A_b$, $% |
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124 | B=m_Nr_0^2/(2\pi h^2)$, $\gamma _b=(2s_b+1)m_b/m_N$. $\Delta A_b$ is the |
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125 | number of nucleons in $b$ particle. $m_b$, $m_N$ and $s_b$ are mass of |
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126 | particle $b$, mass of nucleon and spin of particle $b$ respectively. |
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127 | The $a_b$ is level density parameter for |
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128 | nucleus after emission of fragment $b$. |
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129 | Similarly for the neutron evaporation probability we obtain the |
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130 | following equation: |
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131 | \begin{equation} |
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132 | \begin{array}{c} |
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133 | \label{SFE19}W_{n} |
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134 | =\gamma _nA_n^{2/3}B\frac \alpha {2a_n^2} \\ |
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135 | \exp [-2\sqrt{aE^{*}}+2\sqrt{a_nT_n^{\max }}] |
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136 | [4a_nT_n^{\max }+(2a_n\beta -3)(2\exp (2\sqrt{a_nT_n^{\max }}-1)]. |
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137 | \end{array} |
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138 | \end{equation} |
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139 | Using probabilities Eq. ($\ref{SFE18}$) and Eq. ($\ref{SFE19}$) we are |
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140 | able to |
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141 | choose randomly the type of emitted fragment. |
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142 | |
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143 | \subsection{Kinetic energy of emitted fragment.} |
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144 | |
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145 | \hspace{1.0em}The equation ($\ref{SFE9}$) can be used to sample kinetic |
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146 | energies of evaporated fragments. |
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147 | For example, keeping terms in Eq. ($\ref{SFE9}$), |
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148 | which depend from $T_b$ and using the approximations for inverse cross |
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149 | section is given by Eq. ($\ref{SFE11}$) and for level densities are given |
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150 | by Eq. ($\ref{SFE13}$), we obtain for charged fragments |
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151 | \begin{equation} |
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152 | \label{SFE20}W(x)=C_1 x\exp (2\sqrt{a(T_b^{\max }-x)}= C_2 |
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153 | T_b\exp (2\sqrt{aE^{*}}), |
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154 | \end{equation} |
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155 | where $C_1$ and $C_2$ do not depend from $T_b$, $x=T_b-V_b$. To generate |
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156 | values of $x$ we can use the next procedure, changing the expression for |
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157 | $W(x)$ to have $W(x^{\max })=1$ ($x^{\max }=[(a_b+T_b^{\max |
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158 | }+1/4)^{1/2}-1/2]/a_b$). Choose two random numbers $\xi _1$ and $\xi _2$ |
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159 | (distributed with equal probabilities between 0 and 1) and find kinetic |
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160 | energy of particle $b$ as $T_b=T_b^{\max }\xi _1+V_b$ at condition $\xi |
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161 | _2\leq W(\xi _1T_b^{\max })$. If this condition is not fulfilled we |
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162 | should choose another pair of random numbers. |
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163 | |
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164 | |
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165 | \subsection{Angular distribution of evaporated fragments.} |
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166 | |
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167 | \hspace{1.0em}We consider the angular distribution for emitted |
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168 | fragments as |
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169 | isotropical since we have no information about spin and polarization of |
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170 | nuclei. |
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171 | |
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172 | |
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173 | \subsection{Parameters of residual nucleus.} |
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174 | |
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175 | \hspace{1.0em}After fragment emission we update parameter |
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176 | of decaying nucleus: |
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177 | \begin{equation} |
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178 | \label{SFE21} |
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179 | \begin{array}{c} |
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180 | A_f=A-A_b; Z_f=Z-Z_b; P_f = P_0 - p_b; \\ |
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181 | E_f^{*}=\sqrt{E_f^2-\vec{P}^2_f} - M(A_f,Z_f). |
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182 | \end{array} |
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183 | \end{equation} |
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184 | Here $p_b$ is the evaporated fragment four momentum. |
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