Mass-Energy Equivalence
A-Level Physics Topic 23 13:08 English narration · English + 中文 subtitles burned in
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Look up at the Sun.
抬头看看太阳。
Every single second, it turns four million tonnes of itself — gone, vanished — into pure energy.
每一秒钟，它就把自身的四百万吨——消失、不见了——变成纯粹的能量。
Not burned. Converted.
不是烧掉，是转化。
That lost mass becomes the light and warmth that reaches us eight minutes later.
那些失去的质量，变成了八分钟后到达我们这里的光和热。
How can matter simply become energy?
物质怎么能就这样变成能量呢？
One equation, perhaps the most famous in all of science, tells us: mass and energy are two faces of the same thing.
有一个方程，也许是全部科学中最著名的一个，告诉我们： 质量和能量，是同一样东西的两个面。
Mass is energy, frozen.
质量就是被冻结的能量。
Today: Einstein's equation, the mass defect and binding energy that hold a nucleus together, the iron peak, and how fusion and fission unleash nuclear power.
今天：爱因斯坦的方程、把原子核束缚在一起的质量亏损和结合能、 铁峰，以及聚变和裂变如何释放核能。
Let's begin.
让我们开始吧。
Einstein's equation says it all — mass-energy equivalence: energy equals mass, times the speed of light squared.
爱因斯坦的方程道尽了一切：能量等于质量乘以光速的平方。
And the speed of light is enormous — so squaring it gives a colossal number.
而光速是极其巨大的—— 所以把它平方，就得到一个庞大得惊人的数。
This means a tiny scrap of mass holds an astonishing amount of energy.
这意味着，一丁点质量就蕴藏着惊人的能量。
In nuclear physics we use a handy conversion: one atomic mass unit is worth about nine hundred and thirty-one million electronvolts of energy.
在核物理里，我们用一个方便的换算：一个原子质量单位，大约值九亿三千一百万电子伏特的能量。
Where does that nine hundred and thirty-one come from?
这个 931 是怎么来的？
Take one atomic mass unit: one point six six one, times ten to the minus twenty-seven kilograms.
取一个原子质量单位：1.661 乘以 10 的负 27 次方千克。
Multiply by c squared — the speed of light, three times ten to the eighth, squared — and you get one point four nine times ten to the minus ten joules.
乘上 c 的平方——也就是光速 3 乘以 10 的 8 次方，再平方——就得到 1.49 乘以 10 的负 10 次方焦耳。
Divide by the charge on an electron to turn joules into electronvolts, and out comes nine hundred and thirty-one mega-electronvolts.
再除以电子的电荷，把焦耳换成电子伏特，出来的就是 931 兆电子伏特。
One u, nine hundred and thirty-one MeV.
1 u，931 MeV。
Learn that line.
把这一行记住。
Nuclear reactions are written like a chemical equation.
核反应写起来就像化学方程式。
Nitrogen-fourteen plus an alpha particle gives oxygen-seventeen plus a proton.
氮-14 加上一个 α 粒子，生成氧-17 加一个质子。
Two things must balance.
有两样东西必须配平。
The top numbers — the nucleon numbers — are conserved: fourteen plus four is seventeen plus one, eighteen on both sides.
上面的数字——核子数——是守恒的：14 加 4 等于 17 加 1，两边都是 18。
And the bottom numbers, the proton numbers, are conserved too: seven plus two is eight plus one, nine on each side.
下面的数字，质子数，也是守恒的：7 加 2 等于 8 加 1，两边都是 9。
That is conservation of charge.
这就是电荷守恒。
Use that to find a missing particle.
用它来找出缺失的粒子。
Aluminium-twenty-seven is hit by an alpha particle and gives phosphorus-thirty plus an unknown, X.
铝-27 被一个 α 粒子击中，生成磷-30 加一个未知的 X。
Balance the top row first: twenty-seven plus four is thirty-one; the products give thirty plus the unknown's nucleon number, so that number is one.
先配平上面一行：27 加 4 等于 31；产物那边是 30 加上未知粒子的核子数，所以那个数是 1。
Now the bottom row: thirteen plus two is fifteen; fifteen plus the unknown's charge, so its charge is zero.
再看下面一行：13 加 2 等于 15；15 加上未知粒子的电荷，所以它的电荷是 0。
Nucleon number one, charge zero — X is a neutron.
核子数 1、电荷 0——X 就是一个中子。
Balance both rows, and the unknown particle names itself.
把上下两行都配平，未知粒子自己就说出了身份。
Here is something strange.
这里有件奇怪的事。
Weigh a nucleus, and its mass is less than the total mass of its separate protons and neutrons added up.
称一称原子核，它比它那些分开的质子和中子加起来还要轻。
Some mass has gone missing — the mass defect.
有一部分质量不见了——这就是质量亏损。
Where did it go?
它去哪了？
When those nucleons snapped together, that missing mass was released as energy.
当那些核子啪地结合到一起时， 那份消失的质量以能量的形式释放了出来。
To pull the nucleus back apart, you must put exactly that energy back in — and that is the binding energy: the mass defect, times the speed of light squared.
要把原子核重新拉散开，你必须把那份能量原样补回去—— 这就是结合能：质量亏损，乘以光速的平方。
Let's use it.
来用一用。
A helium-four nucleus has a mass defect of nought point zero three zero four atomic mass units. Find its binding energy.
一个氦-4 核的质量亏损是 0.0304 个原子质量单位，求它的结合能。
Binding energy is the mass defect times c squared, and we could grind all the way through kilograms and joules — but we have the shortcut: every atomic mass unit is worth nine hundred and thirty-one MeV.
结合能等于质量亏损乘以 c 的平方，我们本可以一路换算成千克和焦耳—— 但我们有捷径：每一个原子质量单位值 931 MeV。
So multiply: nought point zero three zero four times nine hundred and thirty-one.
所以直接乘：0.0304 乘以 931。
That is about twenty-eight mega-electronvolts — twenty-eight MeV to pull a helium nucleus apart.
大约是 28 兆电子伏特——要把一个氦核拆开，需要 28 MeV。
But twenty-eight MeV on its own cannot compare nuclides — a bigger nucleus simply has more nucleons to bind.
但光看 28 MeV 没法比较不同的核素——更大的核本来就有更多核子要束缚。
So divide by the nucleon number.
所以要除以核子数。
Helium-four: twenty-eight over four is about seven point one MeV per nucleon.
氦-4：28 除以 4，大约是每个核子 7.1 MeV。
That is the binding energy per nucleon: how tightly each single nucleon is held.
这就是比结合能：每一个核子被束缚得有多紧。
A more tightly bound nucleus has a larger mass defect, a larger binding energy, and a larger value of B over A.
束缚得更紧的核，质量亏损更大、结合能更大，B 除以 A 的值也更大。
Now, how tightly is each nucleon held?
那么，每个核子被束缚得有多紧呢？
Divide the binding energy by the number of nucleons, and plot it.
把结合能除以核子的数目，再把它画出来。
The curve climbs steeply for light nuclei, reaches a peak around iron, then slowly falls for the heavy ones.
对轻核，这条曲线陡峭地上升，在铁附近到达一个顶峰，然后对重核缓缓下降。
That peak is the key: iron is the most stable nucleus of all.
那个顶峰是关键：铁是所有原子核里最稳定的。
Everything wants to move towards that peak.
一切都想朝着那个顶峰移动。
Now put numbers on that curve.
现在给这条曲线标上数字。
Below nucleon number twenty it climbs steeply, with a sharp spike at helium-four — unusually stable for such a light nucleus.
核子数小于 20 时，它陡峭上升，在氦-4 处还有一个尖峰—— 对这么轻的核来说异常稳定。
Near nucleon number fifty-six it reaches its maximum, about eight point eight MeV per nucleon; that is iron, and nothing is more stable.
在核子数 56 附近它达到最高，约为每个核子 8.8 MeV； 那就是铁，没有比它更稳定的了。
Beyond a hundred it drifts slowly down, to about seven point five MeV per nucleon for uranium.
超过 100 以后，它缓缓下降，到铀时约为每个核子 7.5 MeV。
Rise, peak, gentle fall.
上升、顶峰、缓降。
And there are two ways to climb towards it.
而要朝它爬升，有两条路。
Join two light nuclei into a heavier one — that is fusion, and it powers the stars and the Sun.
把两个轻核结合成一个更重的核——这就是聚变， 它为群星和太阳供能。
Or split one heavy nucleus into two — that is fission, the heart of nuclear reactors.
或者把一个重核分裂成两个——这就是裂变，核反应堆的心脏。
Both move their products closer to the iron peak, both raise the binding energy per nucleon, and both release the difference as energy.
两者都让它们的产物更靠近铁峰，都提高了比结合能，也都把这个差额作为能量释放出来。
Here is fusion in full.
来完整地看聚变。
Deuterium, hydrogen-two, plus tritium, hydrogen-three, gives helium-four plus a neutron, plus energy.
氘，也就是氢-2，加上氚，也就是氢-3，生成氦-4 加一个中子，再加能量。
Helium-four sits far higher on the curve than either reactant, so the binding energy per nucleon rises and the difference comes out as energy.
氦-4 在曲线上比两个反应物都高得多，所以比结合能升高，这个差额就以能量的形式放出。
But there is a price: both nuclei are positive and repel each other.
但这要付代价：两个核都带正电，会互相排斥。
You need temperatures of millions of kelvin, so their kinetic energy beats the electrostatic repulsion and they get close enough for the strong nuclear force to take over.
你需要几百万开尔文的高温， 让它们的动能战胜静电斥力，靠得足够近，强核力才能接手。
That is why fusion needs the heart of a star to get going.
这就是为什么聚变需要恒星内部那样的条件才能启动。
And fission in full.
再完整地看裂变。
A slow neutron strikes uranium-two-three-five, which splits into barium-one-four-one and krypton-ninety-two, plus three more neutrons, plus energy.
一个慢中子撞上铀-235，它裂成钡-141 和氪-92，再加三个中子和能量。
Both fragments sit higher on the curve than uranium, so again energy is released.
两个碎片在曲线上都比铀更高，所以同样放出能量。
Now notice those three neutrons: each one can split another uranium nucleus.
现在注意那三个中子：每一个都能再裂开一个铀核。
Two, four, eight — a chain reaction, growing generation by generation, as long as the lump of fuel is big enough.
二、四、八——只要燃料块足够大，就形成链式反应，一代一代地增长。
That minimum amount is the critical mass.
这个最小的量就是临界质量。
So how much energy comes out?
那么放出多少能量呢？
Calculating the energy released takes four steps, every time.
每次都是四步。
One: add up the masses of the reactants.
第一步：把反应物的质量加起来。
Two: add up the masses of the products.
第二步：把产物的质量加起来。
Three: subtract — the mass change is reactants minus products, and it is positive when energy is released.
第三步：相减——质量变化等于反应物减产物，放出能量时它是正的。
Four: multiply by c squared.
第四步：乘以 c 的平方。
Work in kilograms and the answer is in joules; work in atomic mass units and simply multiply by nine hundred and thirty-one to get MeV.
用千克算，答案是焦耳；用原子质量单位算，直接乘 931 就得到 MeV。
Try it.
试一试。
In a nuclear reaction the total mass falls by nought point zero two zero atomic mass units.
某个核反应中，总质量减少了 0.020 个原子质量单位，求放出的能量。
Find the energy released. The masses are already in u, so we can skip the kilograms entirely: the energy released in MeV equals the mass change in u, times nine hundred and thirty-one.
质量已经是以 u 为单位，所以我们完全不用碰千克：以 MeV 为单位的放出能量， 等于以 u 为单位的质量变化乘以 931。
Nought point zero two zero times nine hundred and thirty-one is about nineteen MeV.
0.020 乘以 931，大约是 19 MeV。
Nineteen mega-electronvolts, from a mass change you could never weigh.
19 兆电子伏特，来自一个你根本称不出来的质量变化。
That is the power of c squared.
这就是光速平方的威力。
Unstable nuclei also shed energy, by radioactive decay — spontaneous and random.
不稳定的原子核也会放出能量，靠的是放射性衰变——它是自发的、随机的。
You cannot say when any one nucleus will decay, only the chance that it does.
你说不出任何一个核什么时候会衰变，只能说出它衰变的概率。
Yet across billions, a clear pattern emerges: the number left falls exponentially.
然而在亿万个核之中， 一个清晰的规律浮现出来：剩下的数目按指数下降。
The time to halve is the half-life — fixed for each nuclide.
减半所需的时间，就是半衰期—— 每种核素各自固定。
And the activity, the decays each second, is the decay constant times the number remaining.
而活度，也就是每秒的衰变数，等于衰变常数乘以剩下的数目。
Two words describe every decay, and examiners want both.
每一次衰变都有两个关键词，考官两个都要。
Spontaneous: nothing triggers it.
自发：没有任何东西触发它。
Heat the source, squeeze it, bond it into a compound — the rate does not change.
加热放射源、加压、把它变成化合物——衰变率都不变。
And random: you cannot predict which nucleus goes next, only the probability that one decays in a given time.
还有随机： 你无法预测下一个衰变的是哪个核，只能说出一个核在给定时间内衰变的概率。
The evidence is audible.
证据是能听见的。
The count rate fluctuates: a Geiger counter beside a source clicks at uneven intervals, never a steady beat, even though the long-run mean rate is perfectly well defined.
盖革计数器放在放射源旁边，滴答声间隔不均匀，从不是稳定的节拍， 尽管长期的平均速率是完全确定的。
Now let's count those decays.
现在来数这些衰变。
Activity is the number of decays per unit time, measured in becquerel — one becquerel is one decay per second.
活度是单位时间内的衰变次数，单位是贝克勒尔——1 贝克勒尔就是每秒衰变 1 次。
For N undecayed nuclei of a radionuclide it equals the decay constant times that number: A equals lambda N.
它等于衰变常数乘以未衰变的核数：A 等于 λN。
Lambda, the decay constant, is the probability per unit time that any one nucleus decays, and its unit is per second.
λ，也就是衰变常数， 是任何一个核在单位时间内衰变的概率，单位是每秒。
Lambda is fixed for a nuclide, so double the sample and you double the activity.
λ 对一种核素是固定的，所以样品加倍，活度也加倍。
So why exponential?
那为什么是指数式呢？
Because lambda is a fixed fractional rate: the rate of loss, d N by d t, equals minus lambda N.
因为 λ 是固定的比例速率：减少的速率，dN 比 dt，等于负 λN。
Solve that and you get N equals N-nought e to the minus lambda t.
解这个方程，就得到 N 等于 N 零乘以 e 的负 λt 次方。
Since activity is proportional to N, activity follows the same law, and so does any count rate you measure.
因为活度与 N 成正比， 活度遵循同样的规律，你测到的任何计数率也一样。
The deep reason is this: each nucleus has the same fixed chance per second, whatever its age, so the same fraction disappears in every equal interval.
更深的原因是：每个核每秒都有同样固定的概率，与它的年龄无关， 所以在每一段相等的时间里，消失的都是同样的比例。
A fixed fraction per unit time is exactly what an exponential means.
单位时间内消失固定的比例，这正是指数式的含义。
Half-life is the time for the number of undecayed nuclei — or the activity, or the count rate — to fall to half.
半衰期是未衰变核数——或者活度、或者计数率——降到一半所需的时间。
Put N equals N-nought over two into the exponential and the N-noughts cancel; take natural logs and you are left with a beautifully simple link: lambda times the half-life equals the natural log of two, about nought point six nine three.
把 N 等于 N 零除以 2 代入指数式，N 零就消掉了；再取自然对数， 就剩下一个漂亮而简单的关系：λ 乘以半衰期等于 2 的自然对数，约等于 0.693。
A big decay constant therefore means a short half-life.
所以衰变常数大，就意味着半衰期短。
After n half-lives a fraction one-half to the n survives — after five, only about three percent is left.
经过 n 个半衰期，剩下的比例是二分之一的 n 次方—— 经过 5 个，只剩大约百分之三。
A worked example.
一道例题。
An isotope has a half-life of six point zero hours. Find its decay constant.
某同位素的半衰期是 6.0 小时，求它的衰变常数。
First the units — lambda comes out per second, so the half-life must be in seconds: six hours is twenty-one thousand six hundred seconds.
先看单位——λ 的单位是每秒，所以半衰期必须换成秒：6 小时等于 21600 秒。
Now divide: nought point six nine three over twenty-one thousand six hundred.
现在做除法：0.693 除以 21600。
That gives three point two times ten to the minus five per second.
得到 3.2 乘以 10 的负 5 次方每秒。
Miss the conversion and you lose the mark.
漏了这个换算，这一分就没了。
With real data, take logs.
面对真实数据，就取对数。
From activity equals A-nought e to the minus lambda t, the log of A equals the log of A-nought minus lambda t.
由活度等于 A 零乘以 e 的负 λt 次方， 得到 A 的对数等于 A 零的对数减去 λt。
So plot the natural log of activity against time: you get a straight line, with gradient minus lambda.
所以把活度的自然对数对时间作图： 你会得到一条直线，斜率为负 λ。
That uses every data point, not just two, and the straightness itself proves the decay really is exponential.
这用上了每一个数据点，而不只是两个， 而且直线本身就证明了衰变真的是指数式的。
From a single pair of readings, lambda is one over t, times the log of A-nought over A.
如果只有一对读数，λ 等于 1 除以 t，再乘以 A 零比 A 的对数。
Three marks to secure.
三个要拿稳的分。
First, mass and energy are linked by energy equals mass times the speed of light squared.
第一，质量和能量由“能量等于质量乘以光速的平方”联系起来。
Second, the mass defect, times light-speed squared, is the binding energy — and binding energy per nucleon peaks at iron.
第二，质量亏损乘以光速的平方，就是结合能——而比结合能在铁处达到最高。
Third, decay is exponential, with half-life equal to the natural log of two over the decay constant.
第三，衰变是指数式的，半衰期等于二的自然对数除以衰变常数。
Master these, and mass-energy is yours.
掌握这些，质能等价就是你的了。
And four more marks people throw away.
还有四个常被丢掉的分。
Balance every nuclear equation by conserving nucleon number and proton number — the top row and the bottom row.
配平每一个核反应方程式，要让核子数和质子数都守恒——上面一行和下面一行。
Per nucleon means divide by A; never quote B when the question asks for B over A.
比结合能就是除以 A；题目问 B 除以 A 时，千万不要只写 B。
Convert half-lives to seconds before you find lambda.
求 λ 之前，先把半衰期换成秒。
And remember why fusion and fission both give out energy: their products sit closer to the iron peak.
还要记住聚变和裂变为什么都放出能量：它们的产物离铁峰更近。
Now go and earn them.
现在去把这些分拿到手。