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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|Nuclear Size and Shape

When we talk of the size of a macroscopic object, we all understand what is meant. With sub-microscopic ohjects such as atoms or nuclei, we need to he a hit more careful since the classical picture of electrons moving in fixed orbits around the nucleus is incompatible with Quantum Physics.

Due to Heisenberg’s uncertainty principle, we cannot know exactly where an electron is in an atom. We can, however, determine its wavefunction $\Psi(\boldsymbol{r})$ (provided we can solve the Schroedinger equation for this wavefunction). The interpretation of the wavefunction is that the probability density, $P(\boldsymbol{r})$, (probability per unit volume) at position $r$ is given by
$$P(\boldsymbol{r})-|\Psi(\boldsymbol{r})|^2$$
and the size of the atom is the distance from the nucleus beyond which there is only ā small probábility of finding thê eelectron.1

More precisely, the radius, $R$, of an atom is defined as the expectation value of the radial coordinate, $r$, of the argument, $r$ of the wavefunction, $\Psi(r)$,
$$R \equiv \int r|\Psi(\boldsymbol{r})|^2 d^3 \boldsymbol{r}$$
Unfortunately, we can only solve the Schroedinger equation exactly for the hydrogen atom. Nevertheless, this serves as a good example. In its ground state, the wavefunction of a hydrogen atom is given by

$$\Psi_{000}(\boldsymbol{r})=\frac{1}{\sqrt{\pi a_0^3}} e^{-r / a_0},$$
where $a_0$ is the Bohr radius and takes the value $5.292 \times 10^{-11} \mathrm{~m}(52920 \mathrm{fm})$.
The expectation value of the radial coordinate, $R$, can be calculated from (2.2) and takes the value
$$R=\frac{3}{2} a_0 .$$

## 物理代写|粒子物理代写particle physics代考|The Limitations of Rutherford Scattering

As explained in the previous chapter, one of the axioms that was postulated in order to derive the Rutherford scattering formula was that the nucleus could be considered to be a point particle. The finite size of a nucleus will lead to deviations from this formula when the incident projectile particle has sufficient energy to be able to penetrate the repulsive electric field down to a distance which is of the order of the nuclear size.

It is difficult to produce $\alpha$-particles with sufficient energy to probe the charge distribution of the nucleus, so we use high energy electrons instead.

For electrons the projectile charge number, $z$, is set to unity in the Rutherford scattering formula. There is one further change which is due to the fact that these electrons are moving relativistically with a velocity $v$ close to the speed of light, $c$. This correction depends on the spin of the incident particle. In the case of a singly charged, spin- $\frac{1}{2}$ projectile of mass $m$ (such as an electron) the correction factor (first calculated by Nevill Mott [20]) is
$$\left(1+\frac{p^2}{m^2 c^2} \cos ^2\left(\frac{\theta}{2}\right)\right),$$
so that the (Mott) differential cross section is
$$\frac{d \sigma}{d \Omega}{ }_{\mid \text {Mott } \mid}=\left(\frac{Z \alpha \hbar c}{p^2}\right)^2 \frac{1}{4 \sin ^4(\theta / 2)}\left[m^2+\frac{p^2}{c^2} \cos ^2\left(\frac{\theta}{2}\right)\right] .$$
In the non-relativistic limit, it is the first term in the square bracket that dominates and we recover the Rutherford scattering formula, (1.10) (with $z=1$ ), whereas in the ultra-relativistic case it is the second term that dominates and the differential cross section becomes insensitive to the mass of the incident particle.

# 粒子物理代考

## 物理代写|粒子物理代写particle physics代考|Nuclear Size and Shape

$$P(\boldsymbol{r})-|\Psi(\boldsymbol{r})|^2$$

$$R \equiv \int r|\Psi(\boldsymbol{r})|^2 d^3 \boldsymbol{r}$$

$$\Psi_{000}(\boldsymbol{r})=\frac{1}{\sqrt{\pi a_0^3}} e^{-r / a_0},$$

$$R=\frac{3}{2} a_0$$

## 物理代写|粒子物理代写particle physics代考|The Limitations of Rutherford Scattering

$$\left(1+\frac{p^2}{m^2 c^2} \cos ^2\left(\frac{\theta}{2}\right)\right),$$

$$\frac{d \sigma}{d \Omega} \mid \text { Mott } \mid=\left(\frac{Z \alpha \hbar c}{p^2}\right)^2 \frac{1}{4 \sin ^4(\theta / 2)}\left[m^2+\frac{p^2}{c^2} \cos ^2\left(\frac{\theta}{2}\right)\right] .$$

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