Nuclear fusion and fission · 核聚变与核裂变
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| nuclear fusion/ˈnjuːklɪə ˈfjuːʒn/ | 核聚变 | hé jù biàn |
| nuclear fission/ˈnjuːklɪə ˈfɪʃn/ | 核裂变 | hé liè biàn |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | 动能 | dòng néng |
| electrostatic/ɪˌlektrəʊˈstætɪk/ | 静电 | jìng diàn |
| strong nuclear force/strɒŋ ˈnjuːklɪə fɔːs/ | 强核力 | qiáng hé lì |
| chain reaction/tʃeɪn rɪˈækʃn/ | 链式反应 | liàn shì fǎn yìng |
| critical mass/ˈkrɪtɪkl mæs/ | 临界质量 | lín jiè zhì liàng |
Two opposite processes, one rule
- Joining two tiny nuclei releases energy. Splitting one enormous nucleus releases energy. Those sound like contradictory statements.
- They are not, because both are the same move: climbing the binding-energy-per-nucleon curve towards its peak.
- A nucleus already sitting at the peak, iron-56, can release energy by neither process. That is where a star's fusion stops.
- This lesson is nuclear fusion 核聚变, nuclear fission 核裂变, and how to calculate what they give out.
Both arrows point at iron
两个相反的过程,一条规则
- 把两个极小的核并到一起会放出能量。把一个极大的核劈开也会放出能量。这听上去像是自相矛盾。
- 其实不然,因为两者是同一个动作:沿比结合能曲线朝峰值往上爬。
- 已经坐在峰上的核,也就是铁-56,靠这两个过程都放不出能量。恒星的聚变就停在那里。
- 这一课讲核聚变(nuclear fusion)、核裂变(nuclear fission),以及如何计算它们放出多少。
两支箭都指向铁
Why moving towards the peak releases energy
- On the curve, a higher position means each nucleon is more tightly bound.
- In both processes the products lie higher than the reactants, so the total binding energy increases.
- That extra binding energy is released, as the kinetic energy of the products and as photons.
- Equivalently, the total mass of the products is less than that of the reactants, by exactly $\Delta E/c^2$. Those are two descriptions of one event, and an answer wants both.
朝峰值移动为什么会放能
- 在曲线上,位置越高意味着每个核子被束缚得越紧。
- 两个过程中产物都比反应物更高,所以总结合能增大。
- 多出来的那部分结合能被放出,表现为产物的动能和光子。
- 等价地说,产物的总质量比反应物更小,恰好少 $\Delta E/c^2$。这是同一件事的两种说法,答题时两句都要有。
Nuclear fission chain reaction · 核裂变链式反应
A neutron splits a heavy nucleus, releasing energy and more neutrons — which split more nuclei. · 中子分裂重核，释放能量和更多中子——这些中子再分裂更多的核。
Energy is released when nuclei move: · 当核向哪个方向移动时释放能量:
Climbing the B/A curve toward iron means the products are more tightly bound, so energy is released. · 沿 B/A 曲线朝铁攀升意味着产物结合得更紧,所以能量被释放。
Put the explanation of why fusion and fission both release energy in order. · 把聚变和裂变都放出能量的解释按顺序排列。
The mass loss and the binding-energy gain are the same event described two ways, and a full answer says both. · 质量的减少和结合能的增加是同一件事的两种说法,完整的答案两者都要说。
An iron-56 nucleus can release energy by either fusion or fission. · 铁-56 核可以通过聚变或裂变放出能量。
It sits at the peak, so every move takes it DOWN the curve and would need energy put in. This is why fusion inside a star stops at iron. · 它正在峰上,所以任何移动都会让它沿曲线下降,需要外界输入能量。这就是恒星内部的聚变止步于铁的原因。
Fusion
- Fusion joins two light nuclei into one heavier nucleus:
- The product has greater binding energy per nucleon than the reactants, so energy comes out. This is what powers every star.
- It needs millions of kelvin, and the reason is specific: the nuclei are both positive, so they must have enough kinetic energy 动能 to overcome their electrostatic 静电 repulsion and get close enough for the strong nuclear force 强核力 to take over.
- "It needs high temperature" alone is half an answer. Name the repulsion being beaten and the force taking over.
聚变
- 聚变把两个轻核并成一个较重的核:
- 产物的比结合能大于反应物,所以能量放了出来。每一颗恒星都靠这个供能。
- 它需要数百万开尔文,原因很具体:两个核都带正电,所以必须有足够的动能(kinetic energy)克服它们之间的静电(electrostatic)排斥,靠得足够近,好让强核力(strong nuclear force)接管。
- 只说"需要高温"是半个答案。要说出被克服的是什么排斥、接管的是什么力。
Fusion needs very high temperatures to overcome the electrostatic repulsion between nuclei. · 聚变需要非常高的温度来克服核之间的静电排斥。
The nuclei must get close enough for the strong force to act, despite repelling — hence the millions of kelvin in stars. · 核必须足够接近,强力才能起作用,尽管它们互相排斥——所以恒星里有几百万开尔文。
Why does fusion need millions of kelvin? Select all · 所有 that apply. · 聚变为什么需要数百万开尔文?选出所有适用的。
"It needs high temperature" is half an answer. Name the electrostatic repulsion being beaten and the strong nuclear force taking over. · 只说"需要高温"是半个答案。要说出被克服的静电排斥和接管的强核力。
Fission
- Fission splits a heavy nucleus into two lighter ones, usually after it absorbs a neutron:
- The products sit higher on the curve than uranium-235, so energy is released.
- The extra neutrons can trigger further fissions: a chain reaction 链式反应, which sustains itself only in a large enough mass of fuel, the critical mass 临界质量.
- Below the critical mass too many neutrons escape from the surface before finding a nucleus, and the reaction dies out.
One neutron in, three out
裂变
- 裂变把一个重核劈成两个较轻的核,通常是在它吸收一个中子之后:
- 产物在曲线上比铀-235 更高,所以能量被放出。
- 多出来的中子能引发进一步的裂变:链式反应(chain reaction),只有燃料的质量足够大,即达到临界质量(critical mass),它才能维持下去。
- 低于临界质量时,太多中子还没碰到核就从表面跑掉了,反应因此熄灭。
进去一个中子,出来三个
The extra neutrons released in fission can trigger a ____ reaction. · 裂变中释放的额外中子能引发 ____ 反应。
Each fission releases neutrons that cause more fissions — a chain reaction in a critical mass of fuel. · 每次裂变释放的中子引起更多裂变——在临界质量的燃料中形成链式反应。
A sustained chain reaction needs a critical mass of fuel. · 持续的链式反应需要临界质量的燃料。
Below the critical mass too many neutrons escape, and the reaction dies out. · 低于临界质量时太多中子逃逸,反应就熄灭了。
The marked contrast
- Fission: a heavy nucleus, large $A$, splits into two lighter nuclei of roughly similar mass, usually after absorbing a neutron, and releases further neutrons.
- Fusion: two light nuclei, small $A$, join into one heavier nucleus, and it needs very high temperature and pressure to overcome the electrostatic repulsion.
- Both release energy, but per kilogram of fuel, fusion releases more.
- Write the contrast in matched pairs: heavy against light, splits against joins, absorbs a neutron against needs high temperature. A list of facts about only one of them scores half.
评分要的那组对比
- 裂变:一个重核($A$ 大)劈成两个较轻、质量大致相近的核,通常先吸收一个中子,并放出更多中子。
- 聚变:两个轻核($A$ 小)并成一个较重的核,需要很高的温度和压力来克服静电排斥。
- 两者都放出能量,但按每千克燃料算,聚变放得更多。
- 要成对地写这组对比:重对轻、劈开对并合、吸收中子对需要高温。只罗列其中一方的事实,只能得一半。
Match each process to what it does. · 把每个过程与它的作用配对。
Both move toward the iron peak, so both can release energy. · 两者都朝铁峰移动,所以都能释放能量。
Match each feature to the process it describes. · 把每个特征与它描述的过程配对。
Write the contrast in matched pairs. A list of facts about only one process collects half the marks. · 要成对地写这组对比。只罗列其中一个过程的事实,只能拿一半的分。
Worked example: energy from a mass change
- In a nuclear reaction the total mass decreases by $0.020\ \text{u}$. Find the energy released.
- $\Delta E = \Delta m\,(\text{u}) \times 931 = 0.020 \times 931 = 19\ \text{MeV}$.
- The recipe is always the same four moves: total mass of reactants, total mass of products, $\Delta m = m_{\text{reactants}} - m_{\text{products}}$, then $\Delta E = c^2\Delta m$.
- A positive $\Delta m$ means energy is released. If you get a negative answer you have subtracted the wrong way round, not discovered a reaction that absorbs energy.
例题:由质量变化求能量
- 某核反应中总质量减少了 $0.020\ \text{u}$。求放出的能量。
- $\Delta E = \Delta m\,(\text{u}) \times 931 = 0.020 \times 931 = 19\ \text{MeV}$。
- 做法永远是同样四步:反应物总质量、产物总质量、$\Delta m = m_{\text{reactants}} - m_{\text{products}}$,然后 $\Delta E = c^2\Delta m$。
- $\Delta m$ 为正表示放出能量。若算出负值,那是你减反了方向,而不是发现了一个吸收能量的反应。
A reaction has a mass change of $0.020\ \text{u}$. How much energy is released, in MeV? · 一个反应的质量变化是 $0.020\ \text{u}$。释放多少能量(用 MeV)?
$\Delta E = \Delta m\,(\text{u}) \times 931 = 0.020 \times 931 \approx 18.6\ \text{MeV}$. · $\Delta E = \Delta m\,(\text{u}) \times 931 = 0.020 \times 931 \approx 18.6\ \text{MeV}$。
Worked example: fusing two deuterium nuclei
- The mass defect of $^{2}_{1}\text{H}$ is $0.002388\ \text{u}$ and that of $^{4}_{2}\text{He}$ is $0.030377\ \text{u}$. Find the energy released when two deuterium nuclei fuse into one helium-4 nucleus.
- Work with binding energies, since mass defects are given. Before: $2 \times 0.002388 = 0.004776\ \text{u}$ of defect. After: $0.030377\ \text{u}$.
- The defect has increased by $0.030377 - 0.004776 = 0.025601\ \text{u}$, and an increase in mass defect is energy given out.
- $\Delta E = 0.025601 \times 931 = 23.8\ \text{MeV}$.
- Note the shape of it: you may work either from the masses of the nuclei or from their mass defects, but never mix the two in one sum.
例题:两个氘核聚变
- $^{2}_{1}\text{H}$ 的质量亏损是 $0.002388\ \text{u}$,$^{4}_{2}\text{He}$ 的是 $0.030377\ \text{u}$。求两个氘核聚成一个氦-4 核时放出的能量。
- 既然给的是质量亏损,就用结合能来做。反应前:亏损共 $2 \times 0.002388 = 0.004776\ \text{u}$。反应后:$0.030377\ \text{u}$。
- 亏损增加了 $0.030377 - 0.004776 = 0.025601\ \text{u}$,而质量亏损的增加就是放出的能量。
- $\Delta E = 0.025601 \times 931 = 23.8\ \text{MeV}$。
- 注意这道题的形状:你可以从核的质量出发,也可以从它们的质量亏损出发,但绝不能在同一个算式里把两者混用。
Two deuterium nuclei (mass defect 0.002388 u each) fuse into helium-4 (mass defect 0.030377 u). How much energy is released, in MeV? (931 MeV per u.) · 两个氘核(各自质量亏损 0.002388 u)聚成氦-4(质量亏损 0.030377 u)。放出多少能量(MeV)?(每 u 折合 931 MeV。)
The mass defect increases by 0.030377 - 2(0.002388) = 0.025601 u, and an increase in defect is energy released: 23.8 MeV. Work from masses OR from defects, never a mixture. · 质量亏损增加了 0.030377 - 2(0.002388) = 0.025601 u,而亏损的增加就是放出的能量:23.8 MeV。要么全用质量、要么全用亏损,绝不混用。
Marks that slip away
- Explaining the energy release needs the curve: the products are higher, so the total binding energy increases and mass is lost.
- For fusion, name what the temperature is for: overcoming electrostatic repulsion so the strong nuclear force can act.
- A chain reaction needs a critical mass, otherwise too many neutrons escape.
- Fission usually absorbs a neutron first and releases more. Say how many if the equation shows them.
- Iron-56 releases energy by neither process. That is a fact worth a mark, and it is the reason stellar fusion stops there.
容易丢掉的分
- 解释放能要用到曲线:产物位置更高,所以总结合能增大,质量减少。
- 讲聚变时要说清高温是为了什么:克服静电排斥,好让强核力发挥作用。
- 链式反应需要临界质量,否则太多中子跑掉。
- 裂变通常先吸收一个中子,再放出更多。方程里给了个数就要说出来。
- 铁-56 靠这两个过程都放不出能量。这一点值一分,也是恒星聚变止步于此的原因。
You've got it
- energy is released whenever nuclei move towards the iron peak, because the products are more tightly bound and the total mass falls by $\Delta E/c^2$
- fusion joins two light nuclei and needs millions of kelvin so the nuclei can beat their electrostatic repulsion and let the strong nuclear force act
- fission splits a heavy nucleus after absorbing a neutron, releasing more neutrons that can sustain a chain reaction above the critical mass
- to find the energy: total reactant mass, total product mass, $\Delta m$, then $\Delta E = c^2\Delta m$, or $\Delta m\,(\text{u}) \times 931$ in MeV
你掌握了
- 只要核朝铁峰移动就会放出能量,因为产物被束缚得更紧,总质量减少 $\Delta E/c^2$
- 聚变把两个轻核并起来,需要数百万开尔文,好让核克服静电排斥、让强核力发挥作用
- 裂变在吸收一个中子后把重核劈开,放出的中子能在临界质量以上维持链式反应
- 求能量:反应物总质量、产物总质量、$\Delta m$,再 $\Delta E = c^2\Delta m$,或以 MeV 计时用 $\Delta m\,(\text{u}) \times 931$