## 物理代写|核物理代写nuclear physics代考|Kinematics

In order for fusion to take place, the (centres of the) nuclei must be within the range of the strong inter-nuclear forces (a few fm). For this to happen, the (positively charged) nuclei need to overcome a Coulomb barrier, which tends to repel them. They therefore need to have some kinetic energy in order to be able to do this.

In order to be able to conserve momentum, there must be at least two final-state particles (one of which could be a photon). For two final-state particles, we have a reaction
$$1+2 \rightarrow 3+4$$
It is most convenient to consider the kinematics in the centre-of-mass $(\mathrm{CM})$ frame in which the incident nuclei have equal and opposite momentum, $\pm \boldsymbol{p}$. In the nonrelativistic limit, the magnitude of this momentum is related to the total kinetic energy, $T_{\mathrm{CM}}\left(\equiv T_1+T_2\right)$, in the CM frame, by $$p=\sqrt{2 \tilde{m}{12} T{\mathrm{CM}}},$$
where
$$\tilde{m}{12}=\frac{m_1 m_2}{m_1+m_2}$$ is the reduced mass of the incident particles whose masses are $m_1$ and $m_2$, respectively. The kinetic energies, $T_i(i=1,2)$, of the two incident particles are $$T_i=\frac{p^2}{2 m_i}=\frac{\tilde{m}{12}}{m_i} T_{\mathrm{CM}},(i=1,2) .$$
The fusion process generates supplementary energy $Q$ equal to the difference between the total rest energy of the incident nuclei and that of the fusion product or products. This means that in the $\mathrm{CM}$ frame, the total kinetic energy of the fusion products is $T_{\mathrm{CM}}+Q$. In most cases, $Q$ is much larger than $T_{\mathrm{CM}}$, but this initial energy nevertheless has to be taken into account.

## 物理代写|核物理代写nuclear physics代考|Fusion Rates

The range of the strong nuclear force is a few fermi, and at this distance, the Coulomb potential is around $1 \mathrm{MeV}$. Therefore, the height of the potential barrier which the incident nucleus needs to overcome in order for fusion to occur is of the order of $1 \mathrm{MeV}$.

However, nuclei with energies less than this can tunnel through the barrier. This is the reverse process of the quantum tunnelling that occurs in $\alpha$-decay. The probability for a nucleus with kinetic energy $T$ to tunnel through the barrier is the Gamow factor discussed in Chap. 6 (see 6.18)).
$$\exp \left{-\sqrt{\frac{E_G}{T}}\right}$$
The Gamow energy, $E_G$, is given by
$$E_G=2 \pi^2 \alpha^2 Z_1^2 Z_2^2 \tilde{m}_{12} c^2$$

where $Z_1$ and $Z_2$ are the atomic numbers of the two fusion nuclei and $\tilde{m}_{12}$ is their reduced mass. ${ }^3$

To obtain the fusion cross section, we multiply this probability by the collision cross section. If we were to consider the nucleus as a hard sphere, we would estimate its collision cross section to be of the order of the area of a disc of radius equal to the nuclear radius. However, for microscopic objects, the “effective size” of a nucleus with momentum $p$ is of the order of the area of a disk whose radius is equal to the its reduced de Broglie wavelength
$$\lambda=\frac{\hbar}{p} .$$
The collision cross section, $\sigma_c$, can therefore be written as
$$\sigma_c \approx \eta_{12}^2 \pi \lambda^2=\eta_{12}^2 \pi \frac{\hbar^2}{2 \widetilde{m}{12} T{\mathrm{CM}}} .$$
where $\eta_{12}$ is an overall factor of order unity, which depends on the fusion reaction, and must be fit to experimental data.

## 物理代写|核物理代写核物理代考|Kinematics

$$1+2 \rightarrow 3+4$$
。最方便的考虑运动学是在质心$(\mathrm{CM})$坐标系中，其中入射的原子核具有相等且相反的动量$\pm \boldsymbol{p}$。在非相对论极限中，这个动量的大小与总动能$T_{\mathrm{CM}}\left(\equiv T_1+T_2\right)$有关，在CM坐标系中，$$p=\sqrt{2 \tilde{m}{12} T{\mathrm{CM}}},$$
，其中
$$\tilde{m}{12}=\frac{m_1 m_2}{m_1+m_2}$$是入射粒子的减少质量，它们的质量分别为$m_1$和$m_2$。两个入射粒子的动能$T_i(i=1,2)$为$$T_i=\frac{p^2}{2 m_i}=\frac{\tilde{m}{12}}{m_i} T_{\mathrm{CM}},(i=1,2) .$$
。聚变过程产生的补充能量$Q$等于入射核的剩余总能量与聚变产物的剩余总能量之差。这意味着在$\mathrm{CM}$坐标系中，聚变产物的总动能为$T_{\mathrm{CM}}+Q$。在大多数情况下，$Q$比$T_{\mathrm{CM}}$大得多，但这个初始能量仍然必须考虑在内

## 物理代写|核物理代写核物理代考|熔合率

.强核力的范围是几个费米，在这个距离上，库仑势在$1 \mathrm{MeV}$左右。因此，为了发生核聚变，入射核需要克服的势垒高度为$1 \mathrm{MeV}$ .然而，能量低于这个值的原子核可以穿过势垒。这是发生在$\alpha$ -衰变中的量子隧道的相反过程。动能为$T$的原子核穿过势垒的概率是第六章讨论的伽莫夫因子(见6.18)。
$$\exp \left{-\sqrt{\frac{E_G}{T}}\right}$$

$$E_G=2 \pi^2 \alpha^2 Z_1^2 Z_2^2 \tilde{m}_{12} c^2$$ 给出

$$\lambda=\frac{\hbar}{p} .$$

$$\sigma_c \approx \eta_{12}^2 \pi \lambda^2=\eta_{12}^2 \pi \frac{\hbar^2}{2 \widetilde{m}{12} T{\mathrm{CM}}} .$$
，其中$\eta_{12}$是一个整体的阶单位因子，它取决于聚变反应，必须与实验数据相拟合

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