[1208] | 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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