Energy and momentum of a photon · 光子的能量与动量
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| photons/ˈfəʊtɒnz/ | 光子 | guāng zi |
| momentum/məʊˈmentəm/ | 动量 | dòngliàng |
| particulate nature/pəˈtɪkjʊleɪt ˈneɪtʃə/ | 粒子性 | lì zi xìng |
| quantum/ˈkwɒntəm/ | 量子 | liàng zǐ |
| Planck constant/plæŋk ˈkɒnstənt/ | 普朗克常量 | pǔ lǎng kè cháng liàng |
| wavelength/ˈweɪvleŋθ/ | 波长 | bō cháng |
| electronvolt/ɪˈlektrɒnvəʊlt/ | 电子伏特 | diàn zi fú tè |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | 动能 | dòng néng |
| electron/ɪˈlektrɒn/ | 电子 | diàn zi |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | 电势差 | diàn shì chà |
| radiation pressure/ˌreɪdɪˈeɪʃn ˈpreʃə/ | 辐射压 | fú shè yā |
Light you can count
- Turn a lamp down far enough and the light does not simply get fainter. It starts arriving in countable lumps.
- A sensitive detector clicks: one, one, one. Never half a click. The lumps have a fixed size that depends on the colour, not on the brightness.
- Those lumps are photons 光子, and this lesson is what one of them carries: an energy $hf$ and a momentum $h/\lambda$.
- This is the syllabus's evidence for the particulate nature 粒子性 of electromagnetic radiation.
可以一个一个数的光
- 把灯调得足够暗,光并不是单纯地变弱。它开始以可以数出来的一块一块到达。
- 灵敏的探测器发出咔嗒声:一下,一下,一下。从来没有半下。这些块有固定的大小,取决于颜色,而不是亮度。
- 这些块就是光子(photons),而这一课讲的是其中一个所携带的东西:能量 $hf$ 和动量 $h/\lambda$。
- 这就是考纲所说的电磁辐射的粒子性(particulate nature)的证据。
What a photon is
- A photon is a quantum 量子, a discrete packet, of energy of electromagnetic radiation. That is the two-mark definition, and both halves score.
- "A particle of light" alone is not enough. The marked words are quantum or packet or discrete amount, and of electromagnetic radiation.
- Particulate nature means energy is delivered in lumps of $hf$, never in smaller pieces. A detector receives one whole photon or none.
One packet, one arrival
光子是什么
- *一个光子是电磁辐射能量的一个量子(quantum),即一个分立的能量包。*这就是两分的定义,两半都得分。
- 只说"一个光的粒子"不够。得分的词是量子或能量包或分立的量,以及电磁辐射的。
- 粒子性意味着能量以 $hf$ 为单位成块地交付,绝不会更小。探测器要么收到整整一个光子,要么一个也收不到。
一个包,一次到达
Which phrases belong in the two-mark definition of a photon? Select all · 所有 that apply. · 哪些说法属于光子的两分定义?选出所有适用的。
"A particle of light" alone scores one at best. Both the quantum idea and the electromagnetic radiation must appear. · 只说"一个光的粒子"最多得一分。量子这个概念和电磁辐射两者都要出现。
The energy of a photon
- A photon of frequency $f$ carries
where $h$ is the Planck constant 普朗克常量.
- Using $c = f\lambda$, the same thing in terms of wavelength 波长 is
- So shorter wavelength means more energy per photon. One gamma photon carries far more than one radio photon, whatever the total power of the two sources.
光子的能量
- 频率为 $f$ 的光子携带
其中 $h$ 是普朗克常量(Planck constant)。
- 用 $c = f\lambda$,同一件事用波长(wavelength)写出来就是
- 所以波长越短,每个光子的能量越大。一个伽马光子携带的能量远大于一个射电光子,不管两个光源的总功率如何。
Energy of a photon · 光子的能量
E = h·f
Photon energy is proportional to frequency — the gradient is Planck's constant h. · 光子能量与频率 成正比——斜率是普朗克常数 h。
The energy of a photon is: · 一个光子的能量是:
$E = hf = \dfrac{hc}{\lambda}$ — proportional to frequency. · $E = hf = \dfrac{hc}{\lambda}$——与频率成正比。
Higher-frequency photons carry more energy. · 频率更高的光子携带更多能量。
$E = hf$, so energy rises with frequency — a γ-ray photon carries far more than a radio one. · $E = hf$,所以能量随频率上升——一个 γ 射线光子携带的能量远超一个无线电光子。
Worked example: a photon of green light
- Find the energy of a photon of green light of wavelength $500\ \text{nm}$.
- $E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times10^{-34})(3.0\times10^{8})}{500\times10^{-9}} = 4.0\times10^{-19}\ \text{J}$
- In electronvolts 电子伏特 that is $4.0\times10^{-19} / 1.60\times10^{-19} = 2.5\ \text{eV}$.
- Convert the nanometres before dividing. Leaving $\lambda$ as $500$ gives an answer wrong by $10^9$, and the size of the error is the giveaway.
例题:一个绿光光子
- 求波长 $500\ \text{nm}$ 的绿光光子的能量。
- $E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times10^{-34})(3.0\times10^{8})}{500\times10^{-9}} = 4.0\times10^{-19}\ \text{J}$
- 用电子伏特(electronvolt)表示就是 $4.0\times10^{-19} / 1.60\times10^{-19} = 2.5\ \text{eV}$。
- 相除之前先把纳米换算掉。把 $\lambda$ 留成 $500$ 会让答案差 $10^9$ 倍,而误差的大小正是破绽。
The electronvolt
- $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$: the kinetic energy 动能 an electron 电子 gains moving through a potential difference 电势差 of one volt.
- eV to J, multiply by $1.60\times10^{-19}$. J to eV, divide.
- Photon energies, work functions and energy levels are all a few eV, which is why the exam quotes them that way.
- Worth memorising: $hc = 1240\ \text{eV nm}$. So a $2.0\ \text{eV}$ photon has $\lambda = 1240/2.0 = 620\ \text{nm}$, and a $400\ \text{nm}$ photon carries $3.1\ \text{eV}$, both in one step.
电子伏特
- $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$:一个电子(electron)通过一伏电势差(potential difference)所获得的动能(kinetic energy)。
- eV 换 J,乘以 $1.60\times10^{-19}$。J 换 eV,除。
- 光子能量、逸出功和能级都是几个 eV,这正是考卷用这个单位给数据的原因。
- 值得记住:$hc = 1240\ \text{eV nm}$。所以 $2.0\ \text{eV}$ 的光子 $\lambda = 1240/2.0 = 620\ \text{nm}$,而 $400\ \text{nm}$ 的光子携带 $3.1\ \text{eV}$,都是一步。
A photon has energy $3.2 \times 10^{-19}\ \text{J}$. What is this in eV? ($1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$) · 一个光子的能量是 $3.2 \times 10^{-19}\ \text{J}$。这是多少 eV?($1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$)
$\dfrac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0\ \text{eV}$. · $\dfrac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0\ \text{eV}$。
Using $hc = 1240$ eV nm, a photon of wavelength 620 nm has an energy of ____ eV. · 利用 $hc = 1240$ eV nm,波长 620 nm 的光子能量是 ____ eV。
1240/620 = 2.0 eV, in one step and with no powers of ten to lose. The same shortcut turns a 2.0 eV work function back into a 620 nm threshold wavelength. · 1240/620 = 2.0 eV,一步完成,也不会丢掉十的幂。同一个捷径也能把 2.0 eV 的逸出功换回 620 nm 的极限波长。
The momentum of a photon
- A photon has zero rest mass and yet a non-zero momentum 动量:
- Show that $p = h/\lambda$ is a standard question and the mark is for the chain, not the result: start from $E = hf$ and $p = E/c$, so $p = hf/c$; then $c = f\lambda$ gives $f/c = 1/\lambda$, so $p = h/\lambda$.
- Quote both starting equations and the wave equation. A bare final line scores nothing.
光子的动量
- 光子的静止质量为零,却有不为零的动量(momentum):
- 证明 $p = h/\lambda$ 是常考题,而分给的是那条推理链,不是结果:从 $E = hf$ 与 $p = E/c$ 出发,得 $p = hf/c$;再由 $c = f\lambda$ 得 $f/c = 1/\lambda$,所以 $p = h/\lambda$。
- 两个出发方程和波速公式都要写出来。只写最后一行得不到分。
A photon's momentum is: · 一个光子的动量是:
$p = \dfrac{E}{c} = \dfrac{h}{\lambda}$. · $p = \dfrac{E}{c} = \dfrac{h}{\lambda}$。
A photon has zero rest mass but still carries momentum. · 光子的静止质量为零,但仍携带动量。
Its momentum is $\dfrac{E}{c}$ — non-zero even though its rest mass is zero. This gives radiation pressure. · 它的动量是 $\dfrac{E}{c}$——尽管静止质量为零,动量却非零。这产生辐射压。
Put the derivation of $p = h/\lambda$ in order. · 把 $p = h/\lambda$ 的推导按顺序排列。
The marks are on the chain. A bare final line scores nothing, even though the result is right. · 分给的是这条链。只写最后一行得不到分,尽管结果是对的。
Worked example: identifying a photon from its momentum
- A photon in free space has momentum $9.5\times10^{-28}\ \text{N s}$. Show that it is a photon of red light.
- $\lambda = \dfrac{h}{p} = \dfrac{6.63\times10^{-34}}{9.5\times10^{-28}} = 7.0\times10^{-7}\ \text{m} = 700\ \text{nm}$.
- The visible range is $400$ to $700\ \text{nm}$, and $700\ \text{nm}$ is its red end. Saying which end, and quoting the range, is what "show that" wants.
- Its energy is $pc = 2.9\times10^{-19}\ \text{J} = 1.8\ \text{eV}$.
例题:从动量认出光子
- 自由空间中一个光子的动量是 $9.5\times10^{-28}\ \text{N s}$。证明它是红光光子。
- $\lambda = \dfrac{h}{p} = \dfrac{6.63\times10^{-34}}{9.5\times10^{-28}} = 7.0\times10^{-7}\ \text{m} = 700\ \text{nm}$。
- 可见光范围是 $400$ 到 $700\ \text{nm}$,而 $700\ \text{nm}$ 正是它的红端。说出是哪一端并给出范围,才是"证明"要的。
- 它的能量是 $pc = 2.9\times10^{-19}\ \text{J} = 1.8\ \text{eV}$。
Radiation pressure
- Force is the rate of change of momentum, so a stream of photons pushes on whatever it hits. This is radiation pressure 辐射压.
- On a mirror each photon bounces back, so its momentum changes by $2p$: pressure $= 2I/c$.
- On a black surface each photon is absorbed, the change is only $p$: pressure $= I/c$.
- The pressure depends on the intensity, not the colour. Blue light of the same intensity delivers fewer photons per second, but each carries proportionally more momentum, and the two effects cancel exactly.
Bounce and you push twice as hard
辐射压
- 力是动量的变化率,所以一束光子会推它打到的任何东西。这就是辐射压(radiation pressure)。
- 打在镜面上,每个光子被弹回,动量改变 $2p$:压强 $= 2I/c$。
- 打在黑色表面上,每个光子被吸收,改变只有 $p$:压强 $= I/c$。
- 压强取决于强度,而不是颜色。同样强度的蓝光每秒送来的光子更少,但每个携带的动量按比例更大,两个效应恰好抵消。
被弹回来,推力就翻倍
Match each surface to the radiation pressure a beam of intensity $I$ exerts on it. · 把每种表面与强度为 $I$ 的光束对它施加的辐射压配对。
Force is the rate of change of momentum, and a reversal changes it by 2p rather than p. · 力是动量的变化率,而反向使动量改变 2p 而不是 p。
Replacing a red beam with a blue beam of the same intensity increases the radiation pressure. · 把红光束换成同样强度的蓝光束会增大辐射压。
The blue beam delivers fewer photons per second, but each carries proportionally more momentum. The two effects cancel exactly, so the pressure depends on intensity alone. · 蓝光束每秒送来的光子更少,但每个携带的动量按比例更大。两个效应恰好抵消,所以压强只取决于强度。
Worked example: counting photons from a laser
- A laser emits $2.0\ \text{mW}$ at $650\ \text{nm}$. Find the photons emitted per second, and the force on a surface that absorbs the beam completely.
- Energy per photon: $E = hc/\lambda = (6.63\times10^{-34})(3.00\times10^{8})/(650\times10^{-9}) = 3.06\times10^{-19}\ \text{J}$.
- Rate: $\dfrac{P}{E} = \dfrac{2.0\times10^{-3}}{3.06\times10^{-19}} = 6.5\times10^{15}$ per second.
- Force: momentum delivered per second $= P/c = \dfrac{2.0\times10^{-3}}{3.00\times10^{8}} = 6.7\times10^{-12}\ \text{N}$.
- A mirror would feel twice that. Even full sunlight, at about $1\ \text{kW/m}^2$, exerts only a few micropascals.
例题:数一数激光的光子
- 一台激光器以 $650\ \text{nm}$ 发出 $2.0\ \text{mW}$。求每秒发出的光子数,以及作用在完全吸收该光束的表面上的力。
- 每个光子的能量:$E = hc/\lambda = (6.63\times10^{-34})(3.00\times10^{8})/(650\times10^{-9}) = 3.06\times10^{-19}\ \text{J}$。
- 速率:$\dfrac{P}{E} = \dfrac{2.0\times10^{-3}}{3.06\times10^{-19}} = 6.5\times10^{15}$ 每秒。
- 力:每秒交付的动量 $= P/c = \dfrac{2.0\times10^{-3}}{3.00\times10^{8}} = 6.7\times10^{-12}\ \text{N}$。
- 镜面受到的力是这个的两倍。即便是约 $1\ \text{kW/m}^2$ 的直射阳光,也只产生几微帕的压强。
A 2.0 mW laser beam is completely absorbed by a surface. What is the force on the surface, in piconewtons? (1 pN = 1e-12 N) · 一束 2.0 mW 的激光被表面完全吸收。表面受到的力是多少皮牛?(1 pN = 1e-12 N)
F = P/c = 2.0e-3 / 3.00e8 = 6.7e-12 N. A mirror would feel twice as much. You never need the photon count for the force, only for "how many photons per second". · F = P/c = 2.0e-3 / 3.00e8 = 6.7e-12 N。镜面受到的力是这个的两倍。求力从不需要光子数,只有问"每秒多少个光子"时才需要。
Marks that slip away
- A photon is a quantum of energy of electromagnetic radiation. "A particle of light" is half an answer.
- Convert nm to m before dividing, and eV to J before using $\tfrac{1}{2}mv^2$.
- In "show that $p = h/\lambda$" the marks are on $E = hf$, $p = E/c$ and $c = f\lambda$. Write the chain out.
- A photon has zero rest mass but non-zero momentum. It is not "massless therefore momentumless".
- Radiation pressure on a mirror is double that on a black surface of the same area.
容易丢掉的分
- 光子是电磁辐射能量的量子。"一个光的粒子"只是半个答案。
- 相除之前把 nm 换成 m,用 $\tfrac{1}{2}mv^2$ 之前把 eV 换成 J。
- "证明 $p = h/\lambda$"的分在 $E = hf$、$p = E/c$ 和 $c = f\lambda$ 上。把链条写出来。
- 光子静止质量为零但动量不为零。不能推成"没有质量所以没有动量"。
- 同样面积上,镜面受到的辐射压是黑面的两倍。
You've got it
- a photon is a quantum of energy of electromagnetic radiation, which is what the particulate nature of radiation means
- $E = hf = hc/\lambda$, and $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$, with $hc = 1240\ \text{eV nm}$ as the one-step shortcut
- a photon has momentum $p = E/c = h/\lambda$, derived from $E = hf$, $p = E/c$ and $c = f\lambda$
- radiation pressure is $I/c$ on an absorber and $2I/c$ on a mirror, and depends on intensity, not colour
你掌握了
- 光子是电磁辐射能量的量子,这正是辐射的粒子性的含义
- $E = hf = hc/\lambda$,且 $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$,而 $hc = 1240\ \text{eV nm}$ 是一步到位的捷径
- 光子的动量 $p = E/c = h/\lambda$,由 $E = hf$、$p = E/c$ 和 $c = f\lambda$ 推出
- 辐射压在吸收面上是 $I/c$、在镜面上是 $2I/c$,取决于强度而非颜色