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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

Nuclei, whose charge distributions are not spherically symmetric, possess an electric quadrupole moment $\mathcal{Q}$ (which has the dimension of area) defined in (2.17). This electric quadrupole interacts with the electric field gradient (EFG) given by $$V_{i j} \equiv \frac{\partial}{\partial x_i} E_j(\boldsymbol{x})$$
which is generated by the electrons in the atom in which the nucleus is embedded and, in the case of a crystal, also by the surrounding ions. The electric field gradient couples to the spin of the nuclear state, denoted by quantum number $I$, and also to its $z$-component, $m_I \hbar$.

For nuclear states with spin exceeding $I=\frac{1}{2}$, the contribution to the energy of the nuclear state is 4
$$\Delta E_{\text {quad }}=\frac{V_{z z} \mathcal{Q}}{4 I(2 I-1)}\left[3 m_I^2-I(I+1)\right] .$$
Unlikê the Zeeman effect, this depends on the modulus, $\left|m_l\right|$, of the quantum number $m_I$ and not its sign. Nevertheless, it separates the energies of states with different values of $\left|m_I\right|$. So, for example, in a transition from a state with $I=\frac{1}{2}$ to a state with $I=\frac{3}{2}$, the initial state has $m_I=\pm \frac{1}{2}$, but the final state can have either $m_I=\pm \frac{1}{2}$ or $m_I=\pm \frac{3}{2}$. These final states have slightly different energies, and so the $\gamma$-ray emitted or absorbed in these transitions will have slightly different energies as shown in Fig. 8.1. These energy differences are only of the order of $10^{-6} \mathrm{eV}$, but they can nevertheless be detected [65].

These various effects all depend on the atomic electron configuration. Different atomic structures lead to different atomic radii, different values for wavefunctions at the origin and different electric field gradients at the nucleus. This means that there will be a very small difference in the $\gamma$-ray energy spectrum of a given nuclide in different chemical environments – for example, in ions with different electric charges or in different crystal structures.

## 物理代写|核物理代写nuclear physics代考|The Mössbauer Effect

In Atomic Physics, it is possible to promote atoms into their excited states by bombarding them with photons with the resonant frequencies, i.e. with energies equal to the energies between the ground state and the excited states.

In nuclei this is not usually possible. The reason for this is to do with the nuclear recoil. The energy of the emitted photon, $E_\gamma$, is not exactly equal to the energy difference, $E_{i f}=E_i-E_f$, between an initial nuclear state of energy, $E_i$, and a final state oo énergy, $E_f$. The phooton carries mómentum $E_\gamma / c$, and sô the récoioiling nucleus must have equal and opposite momentum. Consequently, it acquires a recoil kinetic energy of
$$T=\frac{E_\gamma^2}{2 m_N c^2},$$
where $m_N$ is the nuclear mass. The excitation energy, $E_{i f}$, is the sum of the photon energy plus this recoil kinetic energy
$$E_{i f}=E_\gamma\left(1+\frac{E_\gamma}{2 m_N c^2}\right) .$$
The photon energy is slightly less than the energy difference between the initial and final nuclear states. On the other hand, in order to absorb a photon and promote the nucleus into a higher energy state, the incident photon needs to have an energy, $E_\gamma^{\prime}$, which exceeds the energy difference, $E_{i f}$, in order to provide the energy of the nuclear recoil
$$E_\gamma^{\prime}=E_{i f}+\frac{E_\gamma^{\prime 2}}{2 m_N c^2} \approx E_\gamma\left(1+\frac{E_\gamma}{m_N c^2}\right) .$$
For a photon of energy $100 \mathrm{keV}$ and a nucleus with $A=100$, the energy of the photon required for absorption exceeds that of the emitted photon by about $0.1 \mathrm{cV}$. This may not seem like a very large energy difference compared with a photon energy of $100 \mathrm{keV}$. However, the problem is that the $\gamma$-ray line-widths are extremely narrow. Even for fast decaying excited states with lifetimes, $\tau$, of about $10^{-12} \mathrm{~s}$, the line-width is given by
$$\Gamma=\frac{\hbar}{\tau} \approx 10^{-3} \mathrm{eV},$$
so the difference between the absorbing and emitted photons is much larger than the photon width, thereby making the reabsorption of an emitted photon impossible.

# 核物理代写

## 物理代写|核物理代写核物理代考|电四极移

$$\Delta E_{\text {quad }}=\frac{V_{z z} \mathcal{Q}}{4 I(2 I-1)}\left[3 m_I^2-I(I+1)\right] .$$
Unlikê塞曼效应，这取决于量子数$m_I$的模量$\left|m_l\right|$，而不是它的符号。然而，它分离了具有不同值$\left|m_I\right|$的状态的能量。因此，例如，在从带有$I=\frac{1}{2}$的状态到带有$I=\frac{3}{2}$的状态的转换中，初始状态有$m_I=\pm \frac{1}{2}$，但最终状态可以有$m_I=\pm \frac{1}{2}$或$m_I=\pm \frac{3}{2}$。这些最终态的能量略有不同，因此在这些跃迁中发射或吸收的$\gamma$射线的能量略有不同，如图8.1所示。这些能量差异仅为$10^{-6} \mathrm{eV}$量级，但仍然可以被检测到[65]

## 物理代写|核物理代写核物理代考|Mössbauer效应

$$T=\frac{E_\gamma^2}{2 m_N c^2},$$
where $m_N$ 是核质量。激发能， $E_{i f}$，为光子能量加上这个后坐力动能
$$E_{i f}=E_\gamma\left(1+\frac{E_\gamma}{2 m_N c^2}\right) .$$光子能量略小于初始态和最终态之间的能量差。另一方面，为了吸收光子并将原子核提升到更高的能量状态，入射光子需要有能量， $E_\gamma^{\prime}$，超过了能量差， $E_{i f}$，以提供核后坐力

，所以吸收光子和发射光子之间的差异远大于光子宽度，因此使发射光子的重吸收不可能

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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。

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