### 物理代写|核物理代写nuclear physics代考|PHYS729

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|核物理代写nuclear physics代考|What About Quantum Effects

One might ask whether it was correct to assume that classical physics was applicable for the description of Rutherford scattering, which probes sub-atomic scales where we might expect quantum effects to be significant. Of course, at the time of Rutherford’s calculation, Quantum Physics was unknown, but nowadays we know that the incident $\alpha$-particle has an associated de Broglie wave, and that, in general, a wave scattering from a regular configuration of gold atoms will produce a diffraction pattern. The angular scale of such diffraction patterns is of the order of the de Broglie wavelength divided by the mean separation of the gold atoms in the foil.
The de Broglie wavelength, $\lambda$, is given by
$$\lambda=\frac{h}{m_{\alpha} v}=\frac{h}{\sqrt{2 m_{\alpha} T}},$$
( $h$ is Planck’s constant), the mass of an $\alpha$-particle is $6.6 \times 10^{-27} \mathrm{~kg}$, and for $\alpha$ particles with kinetic energy $5 \mathrm{MeV}\left(8 \times 10^{-13} \mathrm{~J}\right)$ this gives a wavelength
$$\lambda \approx 6 \times 10^{-15} \mathrm{~m} .$$
In contrast, the separation of the gold atoms is around $170 \mathrm{~nm}$.
This means that the effect of diffraction from the gold atoms is negligible. On the other hand, the size of the nucleus itself is indeed of the order of the de Broglie wavelength of the incident particles, so that for projectiles with somewhat smaller wavelengths, diffraction patterns can be observed from diffraction off single nuclei and these patterns can yield useful information about the structure of nuclei. This is the subject of Chap. 2 .

## 物理代写|核物理代写nuclear physics代考|Relation Between Scattering Angle and Impact

The relation between impact parameter, $b$, and scattering angle, $\theta$, is derived using Newton’s second law of motion, Coulomb’s law for the force between the $\alpha$-particle and the nucleus, and conservation of angular momentum.

The initial and final momenta, $\boldsymbol{p}{i}, \boldsymbol{p}{f}$, have equal magnitude, $p$, (elastic scattering with no nuclear recoil is assumed). If we take $p_{i}$ to be along the $z$-axis and the scattering to be in the $x-z$ plane, then in Cartesian coordinates these two vectors are given by
$$\begin{gathered} p_{i}=p(0,0,1) \ p_{f}=p(\sin \theta, 0, \cos \theta) \end{gathered}$$
and the momentum transfer is given by
$$\boldsymbol{q} \equiv \boldsymbol{p}{f}-\boldsymbol{p}{i}=p(\sin \theta, 0,(\cos \theta-1)) .$$
Using Pythagoras’ theorem and some trigonometric manipulation, the momentum transfer, $\boldsymbol{q}$, has a magnitude
$$q=2 p \sin \left(\frac{\theta}{2}\right)$$
The direction of the vector $q$ is along the line joining the nucleus to the point of closest approach of the $\alpha$-particle. It bisects the vectors $\boldsymbol{p}{i}$ and $\boldsymbol{p}{f}$, making an angle $(\pi-\theta) / 2$ with each, as can be seen from Fig. 1.6.

The position vector, $\mathbf{r}$, from the nucleus and the $\alpha$-particle is given in terms of two-dimensional polar coordinates $(r, \phi)$ with the nucleus as the origin. The angle $\phi$ is set such that $\phi=0$ is at the point of closest approach, where $\mathbf{r}$ lies along the vector $\boldsymbol{q}$.

## 物理代写|核物理代写nuclear physics代考|What About Quantum Effects

$$\lambda=\frac{h}{m_{\alpha} v}=\frac{h}{\sqrt{2 m_{\alpha} T}},$$
( $h$ 是普朗克常数) ，质量 $\alpha$-粒子是 $6.6 \times 10^{-27} \mathrm{~kg}$ ，并且对于 $\alpha$ 具有动能的粒子 $5 \mathrm{MeV}\left(8 \times 10^{-13} \mathrm{~J}\right)$ 这给 出了一个波长
$$\lambda \approx 6 \times 10^{-15} \mathrm{~m} .$$

## 物理代写|核物理代写nuclear physics代考|Relation Between Scattering Angle and Impact

$$p_{i}=p(0,0,1) p_{f}=p(\sin \theta, 0, \cos \theta)$$

$$\boldsymbol{q} \equiv \boldsymbol{p} f-\boldsymbol{p} i=p(\sin \theta, 0,(\cos \theta-1))$$

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

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## MATLAB代写

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