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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|粒子物理代写Particle Physics代考|Diffraction

We account for the charge distribution of the nucleus by writing the differential cross section as
$$\frac{d \sigma}{d \Omega}=\left.\frac{d \sigma}{d \Omega}\right|_{\text {Mott }}\left|F\left(q^2\right)\right|^2 .$$
The correction factor $F\left(q^2\right)$ is called the “electric form factor” and $\mathbf{q}$ is the momentum transferred by the electron in the scattering, i.e. the difference between the final electron momentum, $\boldsymbol{p}_f$, and the initial momentum, $\boldsymbol{p}_i$, both of which have the same magnitude, $p$, but differ in direction by the scattering angle, $\theta$. This electric form factor, $F\left(q^2\right)$, is in general a complex quantity, and it is the square modulus of this complex quantity which enters into the expression for the differential cross section.

From Fig. 2.1, and a little trigonometry, the magnitude of the momentum transferred (also shown in Appendix 1) is given by
$$q=2 p \sin \left(\frac{\theta}{2}\right) .$$
To understand the structure of the electric form factor, we need to consider quantum effects. Recall that in Quantum Physics the electron behaves as a wave with de Broglie wavelength
$$\lambda=h / p .$$
More precisely, we can relate the “wave-vector”, $\mathbf{k}$, to the momentum $\mathbf{p}$,
$$p=\hbar \boldsymbol{k},$$
where the magnitude, $k$, of the wave-vector, $k$, is $2 \pi / \lambda$ and its direction is the direction of the wave-motion.

## 物理代写|粒子物理代写Particle Physics代考|The Saxon–Woods Distribution

A more realistic model for the charge distribution is the Saxon-Woods distribution [22] for which
$$\rho_p(r)=\rho_0 f_{R, \delta}(r),$$
where $R$ is the nuclear radius and the function $f_{a, b}(r)$ is called a “Saxon-Woods potential”, with parameters $(a, b)$. Such a potential is given by
$$f_{a, b}(r)=\frac{1}{1+e^{(r-a) / b}},$$
The overall normalization, $\rho_0$ is chosen such that total charge is $Z e$, i.e.
$$Z e=4 \pi \rho_0 \int r^2 d r \frac{1}{1+\exp ((r-R) / \delta)} .$$
The Saxon-Woods distribution is shown in Fig. 2.6. We take $R$ to be the nuclear charge radius and $\delta$ is the “surface depth” – it measures the range in $r$ over which the charge distribution is substantially reduced from its value at $r=R$. It is a parameter, which needs to be fit to data for every nucleus.

This leads to a differential cross section which is shown in Fig. 2.7, where we have taken the values $R=3.4 \mathrm{fm}$ and $\delta=0.58 \mathrm{fm}$. We see that this predicted differential cross section has dips but no zeros and is much more similar in shape to the experimental results.

In fact, the Saxon-Woods model fits data from most nuclei rather well for nuclei with atomic mass number $A>40$, with the charge radius given by (2.4) for atomic number, $Z$, and atomic mass number, $A$, and the parameter $\delta$ in the range $0.4-0.5 \mathrm{fm}$.

# 粒子物理代考

## 物理代写|粒子物理代写粒子物理代考|衍射

$$q=2 p \sin \left(\frac{\theta}{2}\right) .$$

$$\lambda=h / p .$$

$$p=\hbar \boldsymbol{k},$$

## 物理代写|粒子物理代写粒子物理代考|撒克逊-伍兹分布

$$\rho_p(r)=\rho_0 f_{R, \delta}(r),$$
，其中$R$为核半径，函数$f_{a, b}(r)$称为“撒克逊-伍兹势”，参数为$(a, b)$。这样的电势由
$$f_{a, b}(r)=\frac{1}{1+e^{(r-a) / b}},$$

$$Z e=4 \pi \rho_0 \int r^2 d r \frac{1}{1+\exp ((r-R) / \delta)} .$$

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