湖北大学：《大学物理》第十九章（19-1） 量子物理

量子物理 Quantum Physics) 经典物理的困惑 1)1887年, AA Michelson- EW.Morley实验否 定了绝对参考系的存在; )1990年,“紫外灾难”; 3)1896年,首次发现放射性现象。 热辐射 光是电磁波还是微粒 1.采体辐射、普朗克量子假说 ( Black-body Radiation, Planck's Quantum Supposition

经典物理的困惑： 量子物理 （ Quantum Physics） 2）1990年，“紫外灾难”； 1）1887年，A.A.Michelson-E.W.Morley实验否 定了绝对参考系的存在； 光是电磁波还是微粒 ？ 1. 黑体辐射、普朗克量子假说 （Black-body Radiation、Planck’s Quantum Supposition） 热辐射 实 验 3）1896年，首次发现放射性现象

1) Heat radiation(热辐射) Experiment shows: 19 Radiation relates to 火炉 temperature

1）Heat Radiation （ 热辐射） Experiment shows: 火 炉 1000 600 4度度 Radiation relates to temperature

2) black body(黑体) black-body radiation(黑体辐射) 抛光的铜镜表面吸收率:=0.02 般金属表面吸收率:a=06-08 煤烟吸收率: =0.95-0.98 Let us construct cavity in a metal block, through the wall of which a small hole is drilled. this small hole is called a black body.(as=1) 如远处不点灯的窗翻 若室内点灯 (是自身辐射)

2）black body (黑体) black-body radiation (黑体辐射) 抛光的铜镜表面吸收率： 一般金属表面吸收率： 煤烟吸收率： a 总 = 0.02 a 总 = 0.6−0.8 a 总 = 0.95−0.98 Let us construct cavity in a metal block, through the wall of which a small hole is drilled. This small hole is called a black body. (a总=1) 如远处不点灯的窗 若室内点灯 （是自身辐射）

a) spectral radiancy(单色辐出度) The quantity Map is the rate at which energy is radiated per unit time, per unit wavelength and per unit area of surface for wavelengths lying in the intervalλto+d入 at the temperature T 总 dM,(wm) (元.T) 黑体 显然,它是波长和温度的函数 b) radiancy (辐出度) The appropriate quantity Mon is the rate at which energy is radiated per unit time and per unit area of surface at the temperature T

a）spectral radiancy (单色辐出度) The quantity M(T) is the rate at which energy is radiated per unit time, per unit wavelength and per unit area of surface for wavelengths lying in the interval  to +d at the temperature T. 黑体    d dM M( .T ) = ( . ) −3 a W m 总=1 显然，它是波长和温度的函数 b) radiancy (辐出度) The appropriate quantity M(T) is the rate at which energy is radiated per unit time and per unit area of surface at the temperature T. dM d

b) radiancy (辐出度) 单位时间内从物体单位面积上所辐射的各种 波长的总的辐射能,用M(T)表示。 Mo-J Manda(m) 黑体 显然,它是温度的函数

黑体 ( . ) −2 W m a总=1 显然，它是温度的函数. b) radiancy (辐出度) 单位时间内从物体单位面积上所辐射的各种 波长的总的辐射能，用M（T）表示。 MT   = 0 M(T ) M(.T ) d

3)Studing black-body radiation experimentally equipment实验装置 P B A 平行光管三棱镜 测量系统 黑体

3) Studing black-body radiation experimentally equipment 实验装置 A L1 B P L2 C 黑体 平行光管 三棱镜 测量系统

TThe law about black- body radiation黑体辐射规律 a)J. Stefan--Ludwig 入0, (wcm um) Boltzmann law. 2200K 50 M 0(T) 2000K =5.67×108w/mK 30 800K J. Stefan--Ludwig 20 1000k Boltzmann constant 10 a(nm) 1.02.03.04.05.0 Moot) depend only on the temperature and are quite independent of the material and of the shape and size of the cavity

The law about black-body radiation黑体辐射规律 1.0 2.0 3.0 4.0 5.0 20 30 10 40 50 60 (nm) ( . ) 1 1 0( ) − − M T wcm m 2200K 2000K 1800K 1600K a) J.Stefan—Ludwig Boltzmarn law: 4 M0(T ) =T 5.67 10 w/ mK −8  =  J.Stefan—Ludwig Boltzmarn constant M0(T) depend only on the temperature and are quite independent of the material and of the shape and size of the cavity

The law a bout black-body radiation 炉火纯青 b)Wien displacement law Mour (wcm um) (维恩位移定律) 2200K 1: absolute temperatur卓早2 TA=b 290度 b=2.898×10-3mK 200K 1800K 炉 m:wavelength for peak value 60K ^(nm) 1.02.03.04.05.0 注意:以上两规律只适用于黑体,对非黑体只 近似成立

The law about black-body radiation b)Wien displacement law (维恩位移定律)： Tm = b b 2.898 10 m.K −3 =   m : wavelength for peak value T：absolute temperature 火 炉 1000度 2000 800度度 炉火纯青 注意：以上两规律只适用于黑体，对非黑体只 近似成立。 1.0 2.0 3.0 4.0 5.0 20 30 10 40 50 60 (nm) ( . ) 1 1 0( ) − − M T wcm m 2200K 2000K 1800K 1600K

Exp. The sun is a black-body. Its wavelength fom peak valuelm=5100A on the earth. Find the surtace temperature and radiancy 例:太阳可以看成黑体,地球上测出其峰值波长 为xn=5100A,则其表面温度和辐出度为多少? Solve: From a b 772 b2.898×10 T 4m5100×1010≈5700(K M0)=74=567×108×(5700)4 600×1071/m2

Exp. The sun is a black-body. It’s wavelength fom peak valueλm=5100A on the earth. Find the surface temperature and radiancy. 例：太阳可以看成黑体，地球上测出其峰值波长 为 m=5100Å，则其表面温度和辐出度为多少？ From Tm = b 5700( ) 5100 10 2.898 10 1 0 3 K b T m =   = = − −  4 8 4 0( ) = = 5.6710 (5700) − M T T 7 2 = 6.0010 w/ m Solve：

4) Planck’ s Quanti A玉阻古县三 Planck's Radiati o(ar a number of cap/) theories based on cl ^(nm) had only limited sud 1.02.03.04.05.06.0708.09.0 Example:1896, Wiens formula(维恩公式) Wiens formula isn't fit to M 0(T) -e ar the experimental points 5 at long wavelength Where c and c are constants that must be determined by the experimental data

4) Planck’s Quantum Supposition普朗克量子假说 Planck’s Radiation Formula 普朗克辐射公式 a number of capable physicists advanced theories based on classical physics, which, however, had only limited success. Example : 1896, Wien’s formula (维恩公式) T c T e c M    2 5 1 0( ) − = Where C1 and C2 are constants that must be determined by the experimental data. Wien’s formula isn’t fit to the experimental points at long wavelength. 1.02.0 3.04.05.06.0 7.08.09.0 M0(T ) (nm) ) ( . 1 1 − − m wcm 