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# 物理代写|核物理代考Nuclear Physics代写|NUC-303

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## 物理代写|核物理代考Nuclear Physics代写|Generalities

To introduce cross-sections, it is conceptually simplest to consider a thin slice of matter of area $L^2$ containing $N$ spheres of radius $R$, as shown in Fig. 3.1. A point-like particle impinging upon the slice at a random position will have a probability $\mathrm{d} P$ of hitting one of the spheres that is equal to the fraction of the surface area covered by a sphere
$$\mathrm{d} P=\frac{N \pi R^2}{L^2}=\sigma n \mathrm{~d} z \quad \sigma=\pi R^2 .$$
In the second form, we have multiplied and divided by the slice thickness $\mathrm{d} z$ and introduced the number density of spheres $n=N /\left(L^2 \mathrm{~d} z\right)$. The “crosssection” for touching a sphere, $\sigma=\pi R^2$, has dimensions of “area/sphere.”
While the cross-section was introduced here as a classical area, it can be used to define a probability $\mathrm{d} P_{\mathrm{r}}$ for any type of reaction, $\mathrm{r}$, as long as the probability is proportional to the number density of target particles and to the target thickness:
$$\mathrm{d} P_{\mathrm{r}}=\sigma_{\mathrm{r}} n \mathrm{~d} z .$$
The constant of proportionality $\sigma_{\mathrm{r}}$ clearly has the dimension of area/particle and is called the cross-section for the reaction $r$.

If the material contains different types of objects $i$ of number density and cross-section $n_i$ and $\sigma_i$, then the probability to interact is just the sum of the probabilities on each type:
$$\mathrm{d} P=\sum_i n_i \sigma_i$$

Because nuclear radii are of order of a few femtometer we can anticipate that the cross-sections for nuclear reactions involving the strong interactions will often be of order $1 \mathrm{fm}^2$. In fact, the units of cross-section most often used is the “barn,”
$$1 \mathrm{~b}=100 \mathrm{fm}^2=10^{-28} \mathrm{~m}^2 .$$

## 物理代写|核物理代考Nuclear Physics代写|Differential cross-sections

The probability for elastic scattering is determined by the elastic scattering cross-section
$$\mathrm{d} P_{\mathrm{el}}=\sigma_{\mathrm{el}} n \mathrm{~d} z .$$
Going beyond this simple probability, we can ask what is the probability that the elastic scatter results in the particle passing through a detector of area $\mathrm{d} x^2$ at a distance $r$ from the target and angle $\theta$ with respect to the initial direction. The geometry in shown in Fig. 3.2 where the detector is oriented so that it is perpendicular to the vector between it and the target. The probability is proportional to the product of the probability of a scatter and the probability that the scattered particle goes through the detector. If the scattering angle is completely random, the second is just the ratio of $\mathrm{d} x^2$ and the area of the sphere surrounding the target
$$\mathrm{d} P_{\mathrm{el}, \theta}=\sigma_{\mathrm{el}} n \mathrm{~d} z \frac{\mathrm{d} x^2}{4 \pi r^2} \quad \text { isotropic scattering. }$$
The solid angle covered by the detector is $\mathrm{d} \Omega=\mathrm{d} x^2 / r^2$ so
$$\mathrm{d} P_{\mathrm{el}, \theta}=\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} n \mathrm{~d} z \mathrm{~d} \Omega,$$
where the differential scattering cross section is $\mathrm{d} \sigma / \mathrm{d} \Omega=\sigma_{\mathrm{el}} / 4 \pi$ for isotropic scattering. In general, the scattering is not isotropic so $\mathrm{d} \sigma / \mathrm{d} \Omega$ is a function of $\theta$. If the target or beam particles are polarized, it can be a function of the azimuthal angle $\phi$.

The total elastic cross-section determines the total probability for elastic scattering so
$$\sigma_{\mathrm{el}}=\int \mathrm{d} \Omega \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=\int_0^{2 \pi} \mathrm{d} \phi \int_0^\pi \sin \theta \mathrm{d} \theta \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}(\theta, \phi) .$$

# 核物理代考

## 物理代写|核物理代考Nuclear Physics代写|Generalities

$$\mathrm{d} P=\frac{N \pi R^2}{L^2}=\sigma n \mathrm{~d} z \quad \sigma=\pi R^2 .$$

$$\mathrm{d} P_{\mathrm{r}}=\sigma_{\mathrm{r}} n \mathrm{~d} z .$$

$$\mathrm{d} P=\sum_i n_i \sigma_i$$

$$1 \mathrm{~b}=100 \mathrm{fm}^2=10^{-28} \mathrm{~m}^2 .$$

## 物理代写|核物理代考Nuclear Physics代写|Differential cross-sections

$$\mathrm{d} P_{\mathrm{el}}=\sigma_{\mathrm{el}} n \mathrm{~d} z .$$

$$\mathrm{d} P_{\mathrm{el}, \theta}=\sigma_{\mathrm{el}} n \mathrm{~d} z \frac{\mathrm{d} x^2}{4 \pi r^2} \quad \text { isotropic scattering. }$$

$$\mathrm{d} P_{\mathrm{el}, \theta}=\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} n \mathrm{~d} z \mathrm{~d} \Omega,$$

$$\sigma_{\mathrm{el}}=\int \mathrm{d} \Omega \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=\int_0^{2 \pi} \mathrm{d} \phi \int_0^\pi \sin \theta \mathrm{d} \theta \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}(\theta, \phi) .$$

## Matlab代写

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