\section{Simulation of fragment evaporation.}

\hspace{1.0em} The evaporation of
 neutron, proton, deutron, thritium and alpha fragments 
 are taken into account. 

\subsection{Evaporation threshold.} 

\hspace{1.0em}One should take into account the energy condition
for fragment emission, i. e. the nucleus 
excitation energy should be higher than
the reaction threshold:
\begin{equation}
\label{SFE1}T_b^{\max }=E^{*}-Q_b-V_b > 0. 
\end{equation}
Here $T_b^{\max }$ is the maximal kinetic energy carried by emitted
fragment $b$, $Q_b = M(A,Z)-M(A_f,Z_f)-M_b$ is the fragment $b$ binding
energy. $V_b$ is the Coulomb
potential energy, i. e. the Coulomb barrier for fragment $b$.
 $M(A,Z)$ is the mass of the initial nucleus, $M(A_f,Z_f)$ is
the mass of the nucleus after emission of fragment $b$ and $M_b$ is the
fragment $b$ mass. It should be noted that expression ($\ref{SFE1}$) is 
only valid, when the recoil kinetic energy equals zero. In our code we 
apply the condition:
\begin{equation}
\label{SFE2}T_b^{\max }=E_b^{\max}-M_b-V_b > 0, 
\end{equation}  
where
\begin{equation}
\label{SFE3}E_b^{\max }=\frac{(M(A,Z)+E^{*})^2+M^2_b-M^2(A_f,Z_f)}
{2(M(A,Z)+E^{*})}. 
\end{equation}

\subsection{Coulomb barrier calculation.} 

\hspace{1.0em}The  Coulomb barrier:
\begin{equation}
  V_b = C_b\frac{Z_bZ_f}{R_f+R_b},
\end{equation}
where $C_b = 1.44$ \ MeVfm, $Z_b$ and $R_b$ are charge and
 radius of nucleus after fragment
emission, $Z_f$ and $R_f$ are charge and radius of fragment. 
The radii of nuclei are
approximated by $R = r_{C}A^{1/3}$,
 where \cite{IKP94}
\begin{equation}
\label{SFE8}r_{C} = 2.173\frac{1+0.006103Z_bZ_f}{1+0.009443Z_bZ_f} \ fm. 
\end{equation}

\subsection{The fragment evaporation probability.} 

\hspace{1.0em}Evaporation process has been  predicted by the statistical
Weisskopf-Ewing model \cite{WE40.evap}.  Probability to evaporate particle
$b$ in the energy interval $(T_b, T_b+dT_b)$ per unit of time is given
\begin{equation}
\label{SFE9}W_{b}(T_b) = \sigma_{b}(T_b)\frac{(2s_b+1)m_b}{\pi^2 h^3}
\frac{\rho(E^{*} - Q_b-T_b)}{\rho(E^{*})}T_b,
\end{equation}
where $\sigma_{b}(T_b)$ is the inverse (absorption of particle $b$)
reaction cross section, $s_b$ and $m_b$ are particle spin and mass,
  $\rho$ is level densities of  nucleus.  

\subsection{The inverse reaction cross section.}

\hspace{1.0em}To calculate inverse   reaction cross section it is
assumed to have the form \cite{Dostr59}
\begin{equation}
\label{SFE11} 
\sigma _b(T_b ) = (1+C_b)(1-k_bV_b/T_b )\pi R^2 
\end{equation}
for charged charged fragments with $A_f \leq 4$ interaction, where the
$k_b$ is the barrier penetration coefficient ( its tabulated values are
used), and
\begin{equation}
\label{SFE12} 
\sigma _b(T_b ) = \alpha (1 + \beta /T_b )\pi R^2 
\end{equation}
for neutrons. Here $R = r_{0} A^{1/3}$ denotes the absorption radius, where 
$r_0 = 1.5$ \ fm, 
$\alpha =0.76 + 2.2 A^{-1/3}$ and $\beta=(2.12 A^{-2/3} - 0.05)/(0.76 +
2.2 A^{-1/3})$.

\subsection{The level density.}

\hspace{1.0em}The level 
density is approximated by Fermi-gas approach \cite{IKP94} for the nuclear 
level density: 
\begin{equation}
\label{SFE13} \rho(E^{*}) = C\exp{(2\sqrt{aE^{*}})},
\end{equation}
where $C$ is a constant, which does not depend from nucleus properties
and excitation energy $E^{*}$ and $a$ is the level density parameter.

\subsection{The total evaporation probability.} 

\hspace{1.0em}The total probability $W_b$ or total partial width
$\Gamma_b=\hbar W_b$ to evaporate particle $b$ can be obtained from 
Eq. ($\ref{SFE9}$) by integration over $T_b$:
\begin{equation}
\label{SFE17} 
W_{b}=\int_{V_b}^{E^{*}-Q_b} W_b(T_b)dT_b.
\end{equation}
Here the summation is carried out over all excited states of the
fragment.
 
Integration in Eq. ($\ref{SFE17}$) for probability to emit fragment $b$
can be performed analiticaly, if we will use Eq. ($\ref{SFE11}$) for level
density and the Eqs. ($\ref{SFE11}$)-($\ref{SFE12}$) for inverse cross
section.
The probability
to emit a charged particle:
\begin{equation}
\label{SFE18} 
\begin{array}{c}
W_{b}=\gamma _bA_b^{2/3}B\exp [-2 
\sqrt{aE^{*}}]\frac{(1+C_b)}{a_b^2}\{a_bT_b^{\max }[2\exp (2\sqrt{%
a_bT_b^{\max }})+1]- \\
- 3\sqrt{a_bT_b^{\max }}\exp (2\sqrt{a_bT_b^{\max }}%
)-3[1-\exp (2\sqrt{a_bT_b^{\max }})]/2\}, 
\end{array}
\end{equation}
where $T_b^{max}$ is defined by the equation ($\ref{SFE1}$).
The following notations were introduced: $A_b=A-\Delta A_b$, $%
B=m_Nr_0^2/(2\pi h^2)$, $\gamma _b=(2s_b+1)m_b/m_N$. $\Delta A_b$ is the
number of nucleons in $b$ particle. $m_b$, $m_N$ and $s_b$ are mass of
particle $b$, mass of nucleon and spin of particle $b$ respectively. 
  The $a_b$ is level density parameter for
nucleus after emission of fragment $b$.
Similarly for the neutron evaporation probability we obtain the
following equation:
\begin{equation}
\begin{array}{c}
\label{SFE19}W_{n}
=\gamma _nA_n^{2/3}B\frac \alpha {2a_n^2} \\
\exp [-2\sqrt{aE^{*}}+2\sqrt{a_nT_n^{\max }}]
[4a_nT_n^{\max }+(2a_n\beta -3)(2\exp (2\sqrt{a_nT_n^{\max }}-1)].
\end{array} 
\end{equation}
Using probabilities Eq. ($\ref{SFE18}$) and Eq. ($\ref{SFE19}$) we are 
able to
choose randomly the type of emitted fragment.

\subsection{Kinetic energy of emitted fragment.}

\hspace{1.0em}The equation ($\ref{SFE9}$) can be used to sample kinetic
energies of evaporated fragments. 
 For example, keeping terms in Eq. ($\ref{SFE9}$),
which depend from $T_b$ and using the approximations for inverse cross
section is given by Eq. ($\ref{SFE11}$) and for level densities are given
by Eq. ($\ref{SFE13}$), we obtain for charged fragments
\begin{equation}
\label{SFE20}W(x)=C_1 x\exp (2\sqrt{a(T_b^{\max }-x)}= C_2 
T_b\exp (2\sqrt{aE^{*}}), 
\end{equation}
where $C_1$ and $C_2$ do not depend from $T_b$, $x=T_b-V_b$. To generate
values of $x$ we can use the next procedure, changing the expression for
$W(x)$ to have $W(x^{\max })=1$ ($x^{\max }=[(a_b+T_b^{\max
}+1/4)^{1/2}-1/2]/a_b$). Choose two random numbers $\xi _1$ and $\xi _2$
(distributed with equal probabilities between 0 and 1) and find kinetic
energy of particle $b$ as $T_b=T_b^{\max }\xi _1+V_b$ at condition $\xi
_2\leq W(\xi _1T_b^{\max })$. If this condition is not fulfilled we
should choose another pair of random numbers.


\subsection{Angular distribution of evaporated fragments.}

\hspace{1.0em}We consider the angular distribution for emitted 
fragments as
isotropical since we have no information about spin and polarization of 
nuclei.


\subsection{Parameters of residual nucleus.}

\hspace{1.0em}After fragment emission we  update parameter
of decaying nucleus:
\begin{equation}
\label{SFE21} 
\begin{array}{c}
A_f=A-A_b; Z_f=Z-Z_b; P_f = P_0 - p_b; \\ 
E_f^{*}=\sqrt{E_f^2-\vec{P}^2_f} - M(A_f,Z_f). 
\end{array}
\end{equation}
Here $p_b$ is the evaporated fragment four momentum.