# 物理代写|核物理代写nuclear physics代考|PHYSICS404

## 物理代写|核物理代写nuclear physics代考|Chain Reactions

The prompt neutrons that are emitted in a fission reaction can be absorbed by another parent nucleus, which then itself undergoes induced fission, producing yet more neutrons which can be absorbed to produce even more fission events, which in turn produce more neutrons. This leads to an avalanche of induced fission events, known as a “chain reaction”. The neutrons emitted in a fission reaction typically have a kinetic energy of only $1 \mathrm{MeV}$. For a non-fissile isotope, this is not usually sufficient to overcome the fission barrier and induce fission. But it is indeed possible in the case of a fissile nuclide.

The possibility of such a chain reaction was first proposed in 1939 by Herbert Anderson, Enrico Fermi and Leo Szilard [77]. In order for a chain reaction to occur, the concentration of fissile material must be above a certain value. In uranium ore, ${ }^2$ the concentration of the fissile isotope, ${ }_{92}^{235} \mathrm{U}$, is only about $0.7 \%$ – the rest is nearly all ${ }{92}^{238} \mathrm{U}$. In nuclear reactors, a concentration of $3.5-4.5 \%$ is required for a chain reaction to take place. For concentrations below this, the neutrons are simply absorbed by the non-fissile isotope and the chain reaction fizzles out. The process of increasing the concentration of the fissile isotope is called “enrichment” and usually involves isotope separation of a gaseous compound of uranium, or other fissionable material, using a powerful centrifuge. The centrifuge rotates at a very high speed $\left(50,000-70,000\right.$ r.p.m.), and the heavier isotope, ${ }{92}^{238} \mathrm{U}$, drifts towards the wall of the centrifuge where it is extracted – leaving a higher concentration of the (lighter) fissile isotope ${ }_{92}^{235} \mathrm{U}$.

We define the “neutron multiplication factor”, $k$, as the number of neutrons produced at stage $n+1$ of a chain divided by the number of neutrons produced at stage $n$. The number, $N_f(n)$, of fission events at stage $n$ is given by
$$N_f(n) \propto k^n .$$
Since the number of prompt neutrons produced in each fission reaction is $2-3, k$ is unlikely to exceed 2 . Nevertheless, this means that one fission event can trigger over 1000 fission events after 10 stages. This number, $k$, will depend on how many of the neutrons, produced at stage $n$, are absorbed by a nucleus that can undergo induced fission.

## 物理代写|核物理代写nuclear physics代考|Examples of Fusion

The simplest fusion reaction would be the formation of a helium nucleus from the fusion of two protons
$$p+p \Rightarrow{ }_2^2 \mathrm{He}+\gamma$$
The $\gamma$-ray is emitted in order for momentum to be conserved. However, there is no stable bound state of two protons – the isotope ${ }_2^2 \mathrm{He}$ does not exist. Instead, one of the protons undergoes $\beta^{+}$-decay into a neutron, emitting a positron and a neutrino. This decay is not energetically possible for a free proton, but in the fusion of two protons, energy is released, which facilitates the $\beta$-decay of the proton. The fusion process is therefore
$$p+p \rightarrow d+e^{+}+v$$
The energy released in this reaction is
$$Q=\left(2 m_p-m_d-m_e\right) c^2 .$$
The mass of the deuteron is $1875.61 \mathrm{MeV} / \mathrm{c}^2$, and inserting the masses of the proton, $938.27 \mathrm{MeV} / \mathrm{c}^2$, and electron, $0.511 \mathrm{MeV} / \mathrm{c}^2$, we find that the $Q$-value for this reaction is $0.42 \mathrm{MeV}$.
A deuteron can fuse with a proton to form ${ }_2^3 \mathrm{He}$
$$d+p \rightarrow{ }_2^3 \mathrm{He}+\gamma$$
releasing energy $Q=5.49 \mathrm{MeV}$.
A deuteron can also fuse with another deuteron. We might have expected the fusion product to be the doubly magic nuclide ${ }_2^4 \mathrm{He}$, i.e.
$$d+d \rightarrow{ }_2^4 \mathrm{He}+\gamma$$

## 物理代写|核物理代写核物理代考|链式反应

$$N_f(n) \propto k^n .$$

## 物理代写|核物理代写核物理代考|核聚变实例

The $\gamma$-射线的发射是为了动量守恒。然而，两个质子(同位素)没有稳定的束缚态 ${ }_2^2 \mathrm{He}$ 不存在。相反，其中一个质子经历 $\beta^{+}$-衰变为中子，发射一个正电子和一个中微子。这种衰变在能量上对一个自由质子来说是不可能的，但是在两个质子的聚变中，能量被释放出来，这就促进了 $\beta$-质子的衰变。因此，融合过程为
$$p+p \rightarrow d+e^{+}+v$$这个反应所释放的能量是
$$Q=\left(2 m_p-m_d-m_e\right) c^2 .$$氘核的质量是 $1875.61 \mathrm{MeV} / \mathrm{c}^2$插入质子的质量， $938.27 \mathrm{MeV} / \mathrm{c}^2$和电子， $0.511 \mathrm{MeV} / \mathrm{c}^2$，我们发现 $Q$这个反应的值是 $0.42 \mathrm{MeV}$氘核可以与质子聚变形成 ${ }_2^3 \mathrm{He}$
$$d+p \rightarrow{ }_2^3 \mathrm{He}+\gamma$$

$$d+d \rightarrow{ }_2^4 \mathrm{He}+\gamma$$

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