# # The Beginning Of The Quantum Story

K = 273.15 + C, 是是否會奇怪為什麼K和C之間的轉換不像m和inch之間的轉換? 為什麼出現了這個數273.15, 這個數僅僅是一種保持自洽的約定嗎?

### # Character of the Spectrum of a Blackbody

$I_{T} = \int_0^\infty I(\lambda) \mathrm{d}\lambda = \sigma T^4$

where $I(\lambda)$ is the energy per wavelength emitted per unit time per unit area of the blackbody, $I_T$ is the energy per unit time per are (from experiment), $\sigma$ is $5.67\times 10^{-8} \text{W}/(\text{m}^2\text{K}^4)$ is a fitting parameter.

$\lambda_\mathrm{max} \propto \frac{1}{T}$

Because of this equation, $c = \lambda \nu$, we can deduce the relation between the wave frequency and temprature, that is,

$\nu_\mathrm{max} \propto T$

### # Plank's Theory

1. Treat blackbody as large number of atomic oscillators (simple harmonic oscillator), each of which emits and absorbs EM waves.

2. Each atomic oscillator can have only discrete values of energy,

$E = n h \nu, n=0, 1, 2, ...$

where $h = 6.63\times 10^{-34}$ is called Planck's constant.

1. The energy of the EM wave emitted by the atomic oscillators must be in multiples of $h\nu$. He called $h \nu$ as quanta. 這樣做的動機就是湊答案，做fitting. 而普朗克常數就是為了得到好的fitting參數而產生的擬和參數。如果和答案對得到，那麼他的假設就有可能是對的。

## # Photoelectric Effect

Plank只是把Atomic Ocilator的能量量子化了，但是還沒有粒子的觀念。用頻率為$\nu$的單色光monochromatic light透過真空管打到陰極金屬板Cathode metal plate上, 光被吸收就是被金屬板上的電子吸收, 那麼電子就有可能掙脫原子核的束縛以一個初始的動能跑出來, 從而形成photoelectron光電子，這就是光電效應中的光電啊。 另外一側裝了一個陽極Anode Collector, 構建一個迴路，添加一個負電壓到陽極，使陽極充斥這電子，其作用是阻止光電子撞到collector上面。阻止光電子過來的負電壓Retarding voltage越大，能夠過來的光電子就越少。另外在迴路裡面安裝一個Ammeter to measure photoelectron current。In this experiment arrangement, there are four variables.

1. $I$ monochromatic light intensity, which is proportional to the square of the amplitude of the wave(electric field amplitude) in classical physics, i.e. $I \propto \varepsilon_0^2$.
2. $V$ retarding voltage
3. $\nu$ light frenquency
4. $i$ photocurrent

There are four main reults of this experiment.

1. When $V$ and $\nu$ are constants. The light intensity $I$ is propotional to the photocurrent $i$.
2. Stoping Potential 居然和Light intensity的強度沒有關係。憑想像應該要變大如果入射光的強度變大。
3. Frequency和Stoping potential有關係，在入射頻率大於某個值之後，兩者成正比。感覺到光的強度和frenquency有關係。不同的材料的Stoping potential的起始點不一樣，但是斜率是一樣的。
4. When emission occurs, it occurs almost instantaneously, independent of $I$.

In classical physics, the light intensity is only relate to the amplitude of this light and has nothing to do with the frenquency. Results 2, 3, and 4 can not be explained by classical physics.

### # Einstein Theory

According to Einstein's hypothesis,

• light of frequency $\nu$ can be regarded as a collection of discrete packets of energy, each energy packet ( called photon) containing energy

$E_\mathrm{photon} = h \nu$

• a beam of electromagnetic radiation is a beam of particles, or photons.
• Intensity $I$ of electromagnetic radiation $propot$ density of photons
• photoelectric effect can be visualized as a particle-paticle (photon-electron) collision.

A photon of energy $h\nu$ hits a electron, and its energy can be converted to two parts,

$h\nu = E_\mathrm{k} + E_\mathrm{b}$

where $E_\mathrm{k}$ is K.E. of the ejected electron, and $E_\mathrm{b}$ is the binding energy of the electron. With fixed frequency $\nu$, if $E_\mathrm{b}$ decreases, then $E_\mathrm{k}$ increases. When define a parameter $\phi$ is the minimum binding energy and is called the work function of the metal. Hence,

$E_\mathrm{k,max} = h\nu - \phi$

and

$V_0 = \frac{h}{e} \nu - \frac{\phi}{e}$

$h/e = 4.1\times10^{-15} \text{J}\cdot \text{sec/C}$

WARNING

Why $E_\mathrm{k,max}$ is measured by determining the stoping potential $V_0$ ? Why $E_\mathrm{k,max} = eV_0$

## # X-Ray

X-Ray 就是用高壓形成電子流，衝擊金屬靶材，打出光子，形成X-Ray。X-Ray的頻譜有幾個尖峰，我們用的X-ray的頻率就是用的尖峰出的頻率，因為能量大是嗎?

Incident electrons interact with

1. atomic electrons, leads to the characterist X-ray, which depend on the target meterial, independent of $V$
2. nuclei, leads to the continuous spectrum (Bremsstrahlung spectrum) , $\lambda_\mathrm{min} \propto \frac{1}{V}$, i.e., $\nu_\mathrm{max} \propto V$

If now wo consider the electron approaching the nucleus with energy $E_k$, emitting several photons of energy $h\nu_1$, $h\nu_2$, and so on, and finally leaving the nucleus with energy $E_k^\prime$, we can, from energy consevation considerrations, write

$E_k - E_k^\prime = h\nu_1 + h\nu_2 + ...$

The most energetic photon that can be produced by the interaction of the electron with the nucleus is the one that is produced when the electron loses all its energy in the emission of a single photon. In such a case $E_k^\prime$ is zero and therefore

$E_k = h\nu_\text{max}$

Question?

X一般是指未知的東西，當年(1895)倫琴不知道這個光叫做什麼，所以取了這樣一個名字。 X-rays in Computer-Assisted Tomography (CAT, tomos means slice in Greek)

X 光和一般的單色可見光的差別是什麼?

Gamma 和 X-ray和其他三種射線有本質上的差異是不是?

• Alpha 射線，He去掉電子形成的射線，連紙張都穿不過
• Beta 電子流, 可以穿過紙張，但是穿不過塑料
• Gamma and X-ray 可以穿過塑料, 但是穿不過lead

Why? 這個要等理解了原子結構之後就可以回答了。

### # Mementum of the Photon

From Einstein's theory of special relativity, the total relativistic energy of a particle is given by

$E = E_k + E_0$

where $E_k$ is the kinetic energy of the particle, $E_0 = m_0c^2$ is the rest energy($m_0$) is the rest mass of the particle, that is, the mass of the particle when its velocity is 0), $m$ is the relativistic maxx ( the mass when the particle is moving), and $c$ is the velocity of light. For the photon, the rest energy and therfore the rest mass is zero; a photon at rest does not exist.

We can get an expression for the momentum of the photon.

$E_\text{photon} = h\nu = mc^2$

Dividing by $c$, we obtain

$mc = \frac{h\nu}{c} = \frac{h}{\lambda} = p_\text{photon}$

Why the rest energy of photon is zero?

Photon is always moving, so it's impossible its velocity is zero. Or photon don't have mass, because it virtual. ???

What is the mechanism of crookes radiometer?

The torch lights imit to the target. Different colors correspond to different meterials. The black target receives more light energy. So the temperature of the balck side is much higher. The molecules on the black side is moving much faster. So the black sides push the white sides.

If you think the spin is caused by the mometum transfer of the photon. Muth of energy is absord by the black side of the photon. Much of energy is absord by the black sides. 應該是白色的那一面推黑色的那一面才對。還有一點是光子的能量是真的很小

[1] Convert kelvin to celsius (opens new window)
[2] Visible spectrum (opens new window)
[3] By Nicoguaro - Own work, CC BY 4.0, https://commons.wikimedia.org/w/index.php?curid=49734309
[4] Wikipedia: Black-body radiation (opens new window)
[5] Blackbody Radiation Cannot Be Explained Classically (opens new window)
[6] Wikipedia:Michelson Morley experiment (opens new window)
[7] Absolute Zero (opens new window)
[8] Youtobe: How a Crookes radiometer works (opens new window)