Statistics-lab™可以为您提供charlotte.eduPHYS4140 Nuclear Physics核物理课程的代写代考和辅导服务！

PHYS4140 Nuclear Physics课程简介

The nucleus is the central part of an atom that contains protons and neutrons. Radioactivity refers to the spontaneous emission of particles or radiation from a nucleus, which can occur through several different processes, including alpha decay, beta decay, and gamma decay.

Nuclear reactions involve changes in the nucleus of an atom, such as the fusion of two atomic nuclei to form a heavier nucleus, or the fission of a heavy nucleus into two or more lighter nuclei. These processes can release large amounts of energy and are the basis for nuclear power and weapons.

Fission is the process of splitting a heavy nucleus into two or more lighter nuclei, releasing energy in the form of kinetic energy of the fission products and also releasing several neutrons. These neutrons can then induce fission in other nuclei, leading to a chain reaction. Fission is used in nuclear power plants to generate electricity.

PREREQUISITES 

Credit Hours: (3)
Prerequisite(s): PHYS 3141; and MATH 2171 or MATH 2241; both with grades of $\mathrm{C}$ or above.
Most Recently Offered (Day): Fall 2020, Fall 2018
Most Recently Offered (Evening): Course has not been offered at this time in the past 3 years

PHYS4140 Nuclear Physics HELP（EXAM HELP， ONLINE TUTOR）

We can convert from SI units to $[\mathrm{MeV} \cdot \mathrm{fm}]$ by using the following conversions: \begin{align*} 1 \mathrm{\ eV} &= 1.602 \times 10^{-19} \mathrm{\ J} \ 1 \mathrm{\ fm} &= 10^{-15} \mathrm{\ m} \ 1 \mathrm{\ J} &= 6.24 \times 10^{18} \mathrm{\ MeV} \end{align*}

1. We have $h c$ in SI units, which is given by $h c = 6.626 \times 10^{-34} \mathrm{\ J} \cdot \mathrm{s} \cdot 2.998 \times 10^{8} \mathrm{\ m/s}$. Converting to $[\mathrm{MeV} \cdot \mathrm{fm}]$ gives: \begin{align*} h c &= (6.626 \times 10^{-34} \mathrm{\ J} \cdot \mathrm{s}) (2.998 \times 10^{8} \mathrm{\ m/s}) \ &= (6.626 \times 10^{-34} \mathrm{\ J} \cdot \mathrm{s}) (2.998 \times 10^{17} \mathrm{\ fm/s}) \ &= 19.849 \mathrm{\ MeV} \cdot \mathrm{fm} \end{align*}
2. We have $\hbar c$ in SI units, which is given by $\hbar c = (1.055 \times 10^{-34} \mathrm{\ J} \cdot \mathrm{s}) (2.998 \times 10^{8} \mathrm{\ m/s})$. Converting to $[\mathrm{MeV} \cdot \mathrm{fm}]$ gives: \begin{align*} \hbar c &= (1.055 \times 10^{-34} \mathrm{\ J} \cdot \mathrm{s}) (2.998 \times 10^{17} \mathrm{\ fm/s}) \ &= 0.197 \mathrm{\ MeV} \cdot \mathrm{fm} \end{align*}
3. We have $\frac{e^2}{4 \pi \epsilon_0}$ in SI units, where $e$ is the elementary charge and $\epsilon_0$ is the vacuum permittivity. The SI unit of $\frac{e^2}{4 \pi \epsilon_0}$ is $\mathrm{\ J \cdot m}$. Converting to $[\mathrm{MeV} \cdot \mathrm{fm}]$ gives: \begin{align*} \frac{e^2}{4 \pi \epsilon_0} &= \frac{(1.602 \times 10^{-19} \mathrm{\ C})^2}{4 \pi \epsilon_0 (8.854 \times 10^{-12} \mathrm{\ F/m})} \ &= 8.987 \times 10^{9} \mathrm{\ N \cdot m^2/C^2} \ &= 8.987 \times 10^{9} \mathrm{\ J \cdot m} \cdot \frac{1 \mathrm{\ eV}}{1.602 \times 10^{-19} \mathrm{\ J}} \cdot \frac{1 \mathrm{\ fm}}{10^{-15} \mathrm{\ m}} \ &= 0.439 \mathrm{\ MeV} \cdot \mathrm{fm} \end{align*}

1. We can use the values we calculated earlier to determine $\alpha$:

$\alpha=\frac{e^2}{\hbar c \cdot 4 \pi \varepsilon_0}=\frac{(1.44 \mathrm{MeV-fm})^2}{(197.3 \mathrm{MeV-fm})(1.0546 \times 10^{-34} \mathrm{J \cdot s}) \cdot 4 \pi \cdot 8.854 \times 10^{-12} \mathrm{F/m}}\approx 0.0073$

2. The electromagnetic interaction (or the electromagnetic force) has its strength determined by the fine structure constant $\alpha$.

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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