Quantum Physics

The Quantization of Light

§19-1 Thermal Radiation and Plank’s theory of Radiation 热辐射 普朗克的量子假设 §19-2 The Photoelectric Effect and Einstein’s Quantum Theory 光电效应 爱因斯坦的 光子理论 §19-3 The Compton Effect 康普顿效应

I. Thermal radiation §19-1 Thermal Radiation and Plank’s theory of Radiation -- At any temperature, a body emits radiation of all wavelength, but the distribution in wavelength, the spectral distribution, depends on temperature.

Some concepts about thermal radiation 1. The spectral radiancy ( 光谱辐射出射度 ) e (,T) Let de(,T): the energy emitted per unit time in radiation of wavelength in the interval +d from an unit area of the surface at absolute temperature T. e (,T) specifies the spectral distribution of an body radiation at T. Then,

2. Radiancy ( 辐射出射度 ) E(T) The total energy emitted per unit time per unit area from a body at temperature T. The integral of spectral radiancy e (,T) over all.

All bodies emit radiation to their surrounding and radiation may fall on a body. 3. Black body( 黑体 ) When radiation falls on a body, absorbed reflected Black body: can absorb all radiation falling on it. no any reflection. ----ideal model.

外壳 热电偶 保温层 加热线圈 腔体 腔芯 热屏蔽套管

II. The experiment results of black body radiation  e (,T) varies continuously with. Each e (,T) curve has a peak.  e (,T) curve increases rapidly with the increasing of T.  m decreases linearly with T increasing

Quantitatively, (1) Stefan’s law  =5.67  W/m 2 ·K 4 --Stefan constant (2) Wien’s displacement law b=2.898  m  K --Wien constant.

[Example] Stefan’s law can be used to determine the radius of a sun in astronomy. It is known that the radiation power of a sun arriving to unit area of the earth is 1.2  W/m 2. The distance between the sun and the earth is 4.3  m. The temperature of the sun’s surface is 5200K. The sun can be regarded as a black body.Find its radius. Solution Let R: sun’s radius, R : distance between sun and earth R : distance between sun and earth Sun earth

Neglecting absorption, we have  The total radiation power of the sun is

III. Classical physics encountered difficulty for explaining the radiation of black body. How can we deduce the quantitative expression of e (,T ) theoretically and make it agree with the experiment?

c 1,c 2 : constants determined by experiment. F Wien’s semi-experiment formula ： --It agrees with experiment only in short wavelength range. Wien’s line

F Rayleigh-Jeans formula It agrees with experiment only in longer wavelength range ultraviolet catastrophe ---- “ultraviolet catastrophe” Wein’s l Classical physics cannot explain the radiation of black body. R-J line

IV. Plank’s hypothesis and formula 1. “Quantum of energy ( 能量子 ) ” hypothesis F There are many oscillators in the black body. The energy of an oscillator of a given frequency cannot take arbitrary values, but can only take on the discrete values nh. where n is a positive integer or zero.  = nh is a finite amount, or quantum, of energy.  0 = h is the minimum energy of an oscillator quantum of energy ---- quantum of energy n ---- quantum number( 量子数 )

----Plank’s constant F Applying his hypothesis, Plank obtained ----Plank’s black body radiation formula Plank obtained great agreement with the experiment over the entire range of wavelength.

Furthermore, F Plank got the Nobel Prize in physics in  ---Stenfa’s law  Wien’s displacement law can be obtained.

[Example] A spring-particle oscillator system with k=15N/m, m=1kg and A=0.01m ， Calculate  the total energy of the system E=?  the quantum number of the system n= ？  if n changes from n to n+1 or n-1,  E/E=? Solution  Total energy is  According to Plank’s hypothesis,

 n changes an unit, energy changes  If n changes an unit, energy changes F No any instrument can distinguish the changing. ----Quantum effect disappear for macro-system( 宏观系统 ). And

F Electrons are ejected from metal surface when it is radiated by high frequency electromagnetic waves §19-2 The photoelectric effect photoelectron Cathode ( 阴极 ) Quartz window

I. The results of the experiments 光强较强 光强较弱 饱和 电流 截止 电压 G—sensitive ammeter I s —saturated current V 0 —stopping potential

  The saturated current is proportional with the incident light intensity. --the number of photoelectrons ejected from cathode in an unit time is proportional with --the number of photoelectrons ejected from cathode in an unit time is proportional with the incident light intensity.  The photoelectric current =0 when an inverse stopping potential -V 0 is supplied --photoelectrons have the maximum initial kinetic energy.

 V 0 depends linearly on the frequency of the incident light and is independent of its intensity. i.e. cutoff frequency 0 Below 0, no photoelectric effect occurs.

 Electron emission takes place immediately as the light is incident on the surface with no detectable time delay. II. The classical wave theory of light cannot explain the results. According to the classical theory,  the initial kinetic energy would be decided by the intensity of light instead of the frequency of light.

 photoelectric effect would take place for any frequency of light as long as its intensity is large enough instead of existing a cutoff frequency 0.  the photoelectron would not escape from the metal immediately if the intensity of light is very small.

III. Einstein’s quantum theory of the PE-effect l The electromagnetic field itself is quantized and that light consists of corpuscles, called light quanta or photons. Each photon travels with the speed of light c and carries a quantum of energy of magnitude E=h l The electromagnetic field itself is quantized and that light consists of corpuscles, called light quanta or photons. Each photon travels with the speed of light c and carries a quantum of energy of magnitude E=h l The energy flow density ( or intensity ) of light is S =N h l The energy flow density ( or intensity ) of light is S =N h Einstein offered his “photon postulate” for explanation the results of PE-effect in 1905.

the energy of photon the work function of electron --Einstein’s photoelectric effect equation If v m =0, l When a photon falls on a metallic surface, then,

IV.Einstein’s explanation for photoelectric effect  larger intensity of light i.e. larger number of photons larger number of photoelectrons i.e. larger photoelectric current.  as i.e.

 there is a cutoff frequency  there is a cutoff frequency ， no photoelectric effect for  W/h ，  A photon is absorbed by an electron immediately if  0 and the electron will eject immediately. F Conclusion ： light is the flow of particles. Einstein got Noble prize on physics in 1921

[Example] A beam of ultraviolet light with =2500Å, intensity S=2W/m 2 irradiates on a potassium foil. The work function of potassium is W=2.21eV. Find  the maximum kinetic energy of the photoelectrons,  the maximum number of the photoelectrons per unit time per area from the surface of the potassium foil. Solution  Using Einstein’s equation,

Because one photon can knock out one electron only, the maximum number of the photoelectrons is  The energy of a photon is

I. The Compton effect §19-3 The Compton Effect When a beam of x-rays with sharply defined wavelength 0 falls on some target (such as graphite), it will be scattered and the scattered radiation have two components of wavelength:the origin wavelength 0 and the larger wavelength.

探测器 石墨 光阑 入射光 散射光 x 射 线管 Device Results

Results of experiment   = - 0 increase with the increment of scattering angle .  has nothing to do with 0 and scatter (target).  the intensity of 0 decreases and the intensity of increase with the increment of .

14 S i 16 S 19 K 20 C a 24 C r 26 F e 28 N i 29 C u  the intensity of is larger for the target with lighter atoms than the target with heavier atoms.

Classical theory: x-ray is electromagnetic wave. It acts on the free electrons of the target. There should not be the component of in scattered rays. II.The explanations for Compton effect. Forcing the electrons oscillate at same frequency. radiate electromagnetic wave with same same frequency.

Photon collides with electron in elastic. Photon theory ：  Photons collide with the outer bound electrons in the atoms of the target,

outer bound electron looked as free electron Ze          h 0 Ze          h mc 2 Part of energy is transferred to electron

 Photons collide with the inner bound electrons in the atoms of the target, inner bound electron looked as free electron Ze          h 0 equivalent to collide with the whole atoms of target. As m atom >> m photon,  photon does not lose energy. So we have = 0 or = 0

 Ze          h 0 Compton effect is evident for smaller Z than bigger Z.

III. The Compton equation 1.Photons collide with the outer bound electrons Before collision photon ： energy momentum electron ： After collision photon ： electron ： E,PE,P Ee,PeEe,Pe

Conservation of momentum Conservation of energy    

Considering relativistic energy and momentum electron ： For photon ：  and  can be re-written substituting We get ： and ：

or ----Compton wavelength of the electron Here ----depend on  only The magnitude of c is closer to the magnitude of the wavelength 0 of x-ray (0.1~100Å) =2.43  Å So  is evident.

2.Photon collides with the inner bound electron The mass of atom >> m 0 So Compton effect is not evident.

[Example] x-ray with 0 =1.00  m is scattered by electron. Find  The Compton shift at the scattering angle  =90 0 and  The energy that each electron gets from x-ray. Solution ： 

 The energy that each electron gets from x-ray So the energy that each electron gets = the energy that each photon lose