Posted on Categories:Quantum mechanics, 物理代写, 量子力学

# 物理代写|量子力学代写Quantum mechanics代考|Scattering from the Yukawa Potential

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 物理代写|量子力学代写Quantum mechanics代考|Scattering from the Yukawa Potential

As an example consider the Yukawa potential
$$V(r)=V_0 \frac{e^{-r / r_0}}{r / r_0},$$
where $r_0$ is the distance scale that determines the range of the potential. Assuming a sufficiently high incoming energy compared to the potential strength $V_0$ we can approximate the scattering amplitude by using the first Born approximation. This amounts to the Fourier transform of the Yukawa potential as given in Eq. (14.13)
\begin{aligned} f^{(1)}\left(\mathbf{k}, \mathbf{k}^{\prime}\right) & =-\frac{2 m}{4 \pi \hbar^2} \int_0^{\infty} r^2 d r V_0 \frac{e^{-r / r_0}}{r / r_0} \int_0^{2 \pi} d \phi \int_{-1}^1 d(\cos \theta) e^{+i\left|\mathbf{k}-\mathbf{k}^{\prime}\right|(\cos \theta) r} \ & =-\frac{m}{\hbar^2} \int_0^{\infty} r^2 d r V_0 \frac{e^{-r / r_0}}{r / r_0} \frac{e^{+i\left|\mathbf{k}-\mathbf{k}^{\prime}\right| r}-e^{-i\left|\mathbf{k}-\mathbf{k}^{\prime}\right| r}}{i\left|\mathbf{k}-\mathbf{k}^{\prime}\right| r} \end{aligned}
After cancelling the $r^2$ factors in the numerator and denominator, the remaining integral is easily performed and we obtain
$$f^{(1)}\left(\mathbf{k}, \mathbf{k}^{\prime}\right)=-\frac{\left(2 m V_0 r_0^3 / \hbar^2\right)}{\left(\mathbf{k}^{\prime}-\mathbf{k}\right)^2 r_0^2+1}=-\frac{\left(2 m V_0 r_0^3 / \hbar^2\right)}{4 k^2 r_0^2 \sin ^2(\theta / 2)+1} .$$

From this expression we obtain the differential cross section due to the Yukawa potential as a function of incoming energy $E=\hbar^2 k^2 / 2 m$ and scattering angle $\theta$
$$\frac{d \sigma}{d \Omega}=\left(\frac{\left(2 m V_0 r_0^3 / \hbar^2\right)}{4 k^2 r_0^2 \sin ^2(\theta / 2)+1}\right)^2$$

## 物理代写|量子力学代写Quantum mechanics代考|Scattering from the Coulomb Potential

The Coulomb potential is obtained from the Yukawa potential in the large $r_0$ limit, provided that $r_0 V_0=q_1 q_2$ is held fixed as $r_0 \rightarrow \infty$. Then the Yukawa potential reduces to the Coulomb potential $V(r)=q_1 q_2 / r$. We can apply the same reasoning to every step of the computation of the scattering amplitude and cross section for the Yukawa potential and therefore obtain the corresponding results for the Coulomb potential by the limiting procedure. Thus, from Eq.(14.16) we obtain the first Born approximation for Coulomb scattering
$$\frac{d \sigma}{d \Omega}=\left(\frac{q_1 q_2}{4 E \sin ^2 \theta / 2}\right)^2$$
where we have replaced $E=\hbar^2 k^2 / 2 m$. Since $\hbar$ does not appear in this expression we may conclude that this must be the same as the classical result for the differential cross section. Indeed this should be expected on the basis of the correspondance principle at high energies, which is when the Born approximation is valid.

This formula was derived by Rutherford to describe the scattering of $\alpha$ particles from thin foils. From this expression that fitted his experiments he concluded that the $\alpha$ particles were scattered by a charged pointlike heavy center which resided inside the atoms that made up the thin foil. This followed from the fact that this formula was derived under the assumption of a pointlike positively charged heavy target as described by the Coulomb potential. If this assumption is replaced by another one in which the positive charge is distributed throughout the atom, then the energy and angular dependence of the differential cross section is quite different. These observations led Rutherford to propose the structure of the atom, namely that the atom must contain a small point-like heavy charged nucleus, much smaller than the size of the atom.

## 物理代写|量子力学代写Quantum mechanics代考|Scattering from the Yukawa Potential

$$V(r)=V_0 \frac{e^{-r / r_0}}{r / r_0}$$

$$f^{(1)}\left(\mathbf{k}, \mathbf{k}^{\prime}\right)=-\frac{2 m}{4 \pi \hbar^2} \int_0^{\infty} r^2 d r V_0 \frac{e^{-r / r_0}}{r / r_0} \int_0^{2 \pi} d \phi \int_{-1}^1 d(\cos \theta) e^{+i\left|\mathbf{k}-\mathbf{k}^{\prime}\right|(\cos \theta) r} \quad=-\frac{m}{\hbar^2} \int_0^{\infty} r^2 d r V_0 \frac{e^{-r / r_0}}{r / r_0} \frac{e^{+i\left|\mathbf{k}-\mathbf{k}^{\prime}\right| r}-e^{-i\left|\mathbf{k}-\mathbf{k}^{\prime}\right| r}}{i\left|\mathbf{k}-\mathbf{k}^{\prime}\right| r}$$

$$f^{(1)}\left(\mathbf{k}, \mathbf{k}^{\prime}\right)=-\frac{\left(2 m V_0 r_0^3 / \hbar^2\right)}{\left(\mathbf{k}^{\prime}-\mathbf{k}\right)^2 r_0^2+1}=-\frac{\left(2 m V_0 r_0^3 / \hbar^2\right)}{4 k^2 r_0^2 \sin ^2(\theta / 2)+1}$$

$$\frac{d \sigma}{d \Omega}=\left(\frac{\left(2 m V_0 r_0^3 / \hbar^2\right)}{4 k^2 r_0^2 \sin ^2(\theta / 2)+1}\right)^2$$

## 物理代写|量子力学代写Quantum mechanics代考|Scattering from the Coulomb Potential

$$\frac{d \sigma}{d \Omega}=\left(\frac{q_1 q_2}{4 E \sin ^2 \theta / 2}\right)^2$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。