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In transport models for heavy ion collision, such as the quantum molecular dynamics model and the Boltzmann-Uehling-Uhlenbeck model, the interaction between the nucleon and nucleus is treated separately by the mean field interaction and nucleon-nucleon collisions [35, 36]. The former determines the classical trajectory of the nucleon, whereas the latter causes the transition of the nucleons in the momentum space and transfers a part of the incident energy to the target nucleus. This study only focused on properties of the incident nucleons that do not undergo nucleon-nucleon collision. Therefore, classical mechanics including only the mean field was applied to describe the trajectory of the incident nucleus. The total reaction cross section of nucleon-nucleon collision was used to calculate the probability of nucleon-nucleon collision.
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It is assumed that the mean field provided by the target nucleus stays static during the interaction. The time evolution of the incident nucleon in the mean field is governed by the Newtonian equation of motion
$ \vec{F} = \mu\frac{{\rm d}^{2} \vec{r}}{{\rm d}t^{2}}, $
(1) where
$ \mu $ is the reduced mass, and$ \vec{r} $ is the distance between the incident nucleon and target nucleus. The force includes Coulomb and nuclear forces. Assuming a hard-sphere distribution of charges in the target nucleus, the Coulomb force as a function of r can be expressed as$\overrightarrow {{F_c}} (r) = \left\{ {\begin{array}{*{20}{l}} {\dfrac{{CZzr}}{{{R^3}}}\dfrac{{\vec r}}{r},\quad }&{r < R}\\ {\dfrac{{CZz}}{{{r^2}}}\dfrac{{\vec r}}{r},\quad }&{r \geqslant R} \end{array}} \right. $
(2) where C =
$ \dfrac{e^2}{4\pi\epsilon_{0}} $ = 1.44 MeV$ \cdot $ fm, Z is the charge number of target nucleus, z is the charge number of the incident nucleon (1 for proton and 0 for neutron), and R is the radius of the hard sphere, which is set as the experimental root-mean-square charge radius$ \sqrt{\langle{r}^2\rangle_{p}} $ taken from Ref. [37].The nuclear force is written as
$ \begin{aligned}[b] \vec{F_{n}}(r) =& -\frac{\partial V_{n}}{\partial r}\frac{\vec{r}}{r}, \\ V_{n} =& \alpha\frac{\rho}{\rho_{0}}+\beta\frac{\rho^\gamma}{\rho_{0}^\gamma} +\frac{C_{sp}(\gamma_{i}+1)}{2}\frac{\rho^{\gamma_{i}}}{\rho_{0}^{\gamma_{i}}} \delta ^{2} +\frac{|\tau|}{\tau} \frac{C_{sp}}{2}\frac{\rho^{\gamma_{i}+1}}{\rho_{0}^{\gamma_{i}}}2\delta, \end{aligned} $
(3) where
$ V_{n} $ is the potential of the incident nucleon,$ \delta $ = ($ \rho_{n} - \rho_{p} $ )/$ \rho $ is the isospin asymmetry, and$ \rho_{n} $ ,$ \rho_{p} $ and$ \rho $ are the neutron, proton, and nucleon densities of the target nucleus, respectively.$ \tau $ = 1/2 for neutrons and = −1/2 for protons. The parameters used in this study were$ \rho_{0} $ = 0.16 fm$ ^{-3} $ ,$ \alpha $ = −209.2 MeV,$ \beta $ = 156.4 MeV,$ \gamma $ = 1.35,$ C_{sp} $ = 38.06 MeV,$ \gamma_{i} $ = 0.75.The following form of the nucleon density was used:
$ \rho_{i}(r) = \dfrac{\rho_{i0}}{1+{\rm exp}\left(\dfrac{r-R_{i}}{d_{i}}\right)}, $
(4) where subscript i is p for proton or n for neutron. Parameters
$ \rho_{i0} $ ,$ R_{i} $ , and$ d_{i} $ are calculated by the following equations:$ \begin{aligned}[b] \rho_{p_{0}}+\rho_{n_{0}} &= \rho_{0},\\ \frac{\rho_{p_{0}}}{\rho_{n_{0}}} &= \frac{Z}{N},\\ \int\rho_{p}(r){\rm d}\vec{r} &= Z,\\ \int\rho_{n}(r){\rm d}\vec{r} &= N,\\ \int{r}^2\rho_{p}(r){\rm d}\vec{r} &= Z\langle{r}^2\rangle_{p},\\ \int{r}^2\rho_{n}(r){\rm d}\vec{r} &= N\langle{r}^2\rangle_{n}, \end{aligned} $
(5) where Z and N are the proton and neutron numbers of the target nucleus, respectively, and
$ \langle{r}^2\rangle_{i} $ is the mean-square radius of the nucleus, which can be taken from Refs. [37, 38].Solving the equation of motion, i.e., Eq. (1), with initial conditions depending on impact parameter b and incident energy
$ E_{\rm in} $ , one can obtain distance$ \vec{r}(t) $ and velocity$ \vec{v}(t) $ = d$ \vec{r} $ /dt as a function of time. -
Suppose that a nucleon beam with intensity I passes through a nuclear matter with density
$ \rho $ and thickness$ {\rm d}x $ ; the probability of the nucleon-nucleon collision can then be expressed as$ \frac{{\rm d}I}{I} = -\sigma\rho {\rm d}x, $
(6) where
$ \sigma $ is the cross section of the nucleon-nucleon collision. Applying the relationship$ {\rm d}x = v(t){\rm d}t $ , one can calculate the probability for the incident nucleon passing through the target nucleus without nucleon-nucleon collision by the following time integral$ P = \frac{I^{\rm out}}{I^{\rm in}} = {\rm exp}\left [-\int_{-\infty}^{+\infty} \sigma \rho v {\rm d}t\right]. $
(7) The nucleon-nucleon collision is responsible for the transformation of the incident energy to the target nucleus. Thus, the reaction probability is
$ 1-P $ . Note that the initial conditions for Eq. (1) depend on impact parameter b. Thus, the total reaction cross section can be calculated by integrating over the impact parameter:$ \sigma_{t} = \int_{0}^{+\infty}2\pi b[1-P(b)]{\rm d}b. $
(8) We used three cross sections of the nucleon-nucleon collision to perform the calculation. The first one is the isospin independent, and constant cross section
$ \sigma $ = 40 mb. Second, we applied the cross section of the nucleon-nucleon collision in free space taken from Ref. [39].$\sigma _{pp}^{\rm free} = \sigma _{nn}^{\rm free} = \left\{ {\begin{array}{*{20}{l}} {34{{\left( {\dfrac{{{p_{{\rm{lab}}}}}}{{0.4}}} \right)}^{ - 2.104}},}&{{p_{{\rm{lab}}}}\leqslant 0.4}\\ {23.5 + 1000{{\left( {{p_{{\rm{lab}}}} - 0.7} \right)}^4},}&{0.4 < {p_{{\rm{lab}}}}\leqslant 0.8}\\ {23.5 + \dfrac{{24.6}}{{1 + \exp \left( { - \left( {{p_{{\rm{lab}}}} - 1.2} \right)/0.10} \right)}},}&{0.8 < {p_{{\rm{lab}}}}\leqslant 1.5}\\ {41 + 60\left( {{p_{{\rm{lab}}}} - 0.9} \right)\exp \left( { - 1.2{p_{{\rm{lab}}}}} \right),}&{1.5 < {p_{{\rm{lab}}}}\leqslant 5.0} \end{array}} \right.$
(9) $ \sigma _{np}^{\rm free} = \sigma _{pn}^{\rm free} = \left\{ {\begin{array}{*{20}{l}} {6.3555p_{\rm lab}^{ - 3.2481}\exp \left( { - 0.377{{\left( {\ln {p_{{\rm{lab}}}}} \right)}^2}} \right),}&{{p_{{\rm{lab}}}} \leqslant 0.4}\\ {33 + 196{{\left| {{p_{{\rm{lab}}}} - 0.95} \right|}^{2.5}},}&{0.4 < {p_{{\rm{lab}}}} \leqslant 1.0}\\ {24.2 + 8.9{p_{{\rm{lab}}}},}&{1.0 < {p_{{\rm{lab}}}} \leqslant 2.0}\\ {42,}&{2.0 < {p_{{\rm{lab}}}} \leqslant 5.0} \end{array}} \right.$
(10) $ \sigma \rho $ in Eq. (7) is expressed as$\sigma \rho = \left\{ {\begin{array}{*{20}{l}} {\sigma _{pp}^{\rm free}{\rho _p} + \sigma _{pn}^{\rm free}{\rho _n},\quad {\rm for}\; p\; {\rm reaction}}\\ {\sigma _{nn}^{\rm free}{\rho _n} + \sigma _{np}^{\rm free}{\rho _p},\quad {\rm for}\;n\; {\rm reaction}.} \end{array}} \right.$
(11) The unit of the cross sections in Eqs. (9) and (10) is
${\rm mb}$ , and$p_{\rm lab}$ (GeV/c) is the momentum in the laboratory frame.Third, the in-medium factor proposed by Cai et al. [40] was considered.
$ \sigma \rho $ in Eq. (7) is expressed as$ \sigma \rho = \left\{{\begin{array}{*{20}{l}} {\sigma _{pp}^{\rm med}{\rho _p} + \sigma _{pn}^{\rm med}{\rho _n},\quad {\rm for}\; p\; {\rm reaction}}\\ {\sigma _{nn}^{\rm med}{\rho _n} + \sigma _{np}^{\rm med}{\rho _p},\quad {\rm for}\; n\; {\rm reaction},} \end{array}} \right.$
(12) $ \begin{aligned}[b] &\sigma_{p p}^{\rm med} = \sigma_{n n}^{\rm med } = \frac{1.0+0.1667 E_{\rm lab}^{1.05} \rho^{3}}{1.0+9.704 \rho^{1.2}} \sigma_{p p}^{\rm free}, \\ &\sigma_{n p}^{\rm med} = \sigma_{pn}^{\rm med} = \frac{1.0+0.0034 E_{\rm lab}^{1.51} \rho^{2}}{1.0+21.55 \rho^{1.34}} \sigma_{n p}^{\rm free}, \end{aligned} $
(13) where
$ E_{\rm lab} $ (MeV) is the kinetic energy in the laboratory frame. -
Let us assume that the positions of proton and neutron in the deuteron satisfy a Gaussian distribution. Let the center of the deuteron be at the origin of the coordinates; then, the probability density function that the neutron or proton is at point
$ (x_{d}, y_{d}, z_{d}) $ is$\begin{aligned}[b] f =& A\exp\left(-\frac{x_{d}^2+y_{d}^2+z_{d}^2}{L}\right)\\ =& A\exp\left(-\frac{r_{d}^2}{L}\right),\end{aligned} $
(14) where
$ r_{d} $ is the distance between the nucleon and the center of the deuteron. The root-mean-square radius of deuteron is$ \sqrt{\langle{r}^2\rangle} $ = 2.1421 fm. Based on the normalization conditions and the root-mean-square radius of deuteron, this function satisfies the following two equations:$ \int f(r_{d}){\rm d}\vec{r_{d}} = 1, $
(15) $ \int {r_{d}}^2 f(r_{d}){\rm d}\vec{r_{d}} = \langle{r}^2\rangle = (2.1421)^2{\rm fm}^{2}. $
(16) Using these two equations, the parameters A and L of the Gaussian distribution can be found.
The deuteron-induced reaction is equivalent to one neutron and one proton reacting with the target nucleus. Let deuteron be incident with impact parameter b in the z-axis direction, and let us establish coordinates with deuteron center as origin. Then, according to the determined Gaussian distribution function, the probability of the neutron being located at point
$ (x_{d}, y_{d}, z_{d}) $ can be found to be$ f{\rm d}x{\rm d}y{\rm d}z $ . In this condition, the proton is at point$ (-x_{d}, -y_{d}, -z_{d}) $ , and it can be seen from the geometric relationship that the impact parameter of the proton and neutron are functions of b, x, and y:$ b_{p} = \sqrt{(b-x_{d})^2+y_{d}^2},\quad b_{n} = \sqrt{(b+x_{d})^2+y_{d}^2}. $
(17) The probabilities of no collision
$ P_{p} $ and$ P_{n} $ depend on$ b_{p} $ and$ b_{n} $ , respectively. Similar to the calculation method of cross section in nucleon-induced reactions in subsection II B, the three components of the deuteron-nucleus total reaction cross section can be obtained by integrating over$ x_{d} $ ,$ y_{d} $ ,$ z_{d} $ and b. The cross section for the deuteron absorption is$\begin{aligned}[b] \sigma_{d-ABS} =& \int 2\pi bdb\:\:[1-P_{n}(b_{n})][1\\&-P_{p}(b_{p})] \:\:f {\rm d}x_{d}{\rm d}y_{d}{\rm d}z_{d}, \end{aligned}$
(18) the cross section for the nonelastic breakup where the proton is absorbed is
$ \sigma_{p-NEB} = \int 2\pi bdb\:\: P_{n}(b_{n})[1-P_{p}(b_{p})] \:\: f {\rm d}x_{d}{\rm d}y_{d}{\rm d}z_{d}, $
(19) and the cross section for the neutron-stripping nonelastic breakup where the neutron is absorbed is
$ \sigma_{n-NEB} = \int 2\pi bdb\:\:[1-P_{n}(b_{n})]P_{p}(b_{p}) \:\: f {\rm d}x_{d}{\rm d}y_{d}{\rm d}z_{d}. $
(20) It is evident that the three reaction cross sections depend only on the incident energy of deuteron and the nuclear density distribution of target nucleus. The total reaction cross section of deuteron-nucleus reactions is
$ \sigma_{t} = \sigma_{d-ABS}+\sigma_{p-NEB}+\sigma_{n-NEB}. $
(21)
Nucleon stripping in deuteron-induced spallation reactions at hundreds MeV/nucleon
- Received Date: 2020-10-22
- Available Online: 2021-04-15
Abstract: A nucleon-nucleus dynamics model was developed to investigate the proton-, neutron-, and deuteron-induced reactions at hundreds of MeV/nucleon. In this model, the trajectory of incident nucleon is described by classical mechanics, and the probability of reaction between the nucleon and nucleus is calculated by exponential damping. It is shown that the total reaction cross sections calculated by the model agree in general with the predictions by the CDCC and the experimental data. The model was applied to investigate the nucleon stripping in deuteron-induced reactions and its symmetry energy dependence.