BLACKB ODY RADIATION LAWS. Stefan-Boltzmann. Blackbody E = 1. T is Temperature (Kelvin) R(T) = EaT. R(T) = at4. a = 5.67x10.

Transcription

1 LAWS a = 5.67x10 8 (m2 1() T is Temperature (Kelvin) R(T) = at4 Blackbody E = 1 R(T) = EaT 4 Stefan-Boltzmann BLACKB ODY RADIATION

2 Wien Displacement AmaxT = b b = 2.898x10 3m K

3 CONSIDER TWO OPAQUE QBJECTS N THERMAL EQUILIBRIUM PROOF: GOOD EMITTER is GOOD ABSORBER GOOD ABSORBER is GOOD EMITTER BLACKBODY RADIATION

4 Consider time interval At r = reflectance = fraction of radiation reflected a = absorptance = fraction of radiation absorbed I is the Intensity of radiation in the region A2 is Area of object 2 A1 is Area of object 1 W2 is Emittance of object 2 W1 is Emittance of object 1 2

5 Object #1 Total energy radiated = W1 A1 At Total energy absorbed = a1 I A1 At Thermal equilibrium dictates Total energy radiated = Total energy absorbed Therefore W1A1At =a1ia1at Or W1 = a1 Do same calculation for object # 2 3

6 W2 = a21 Divide first by second to get W1 = a1 a1 a21 a2 or w1=w2 a1 a2 or for many objects 4

7 reflected light emitted light a GOOD EMITTER is a GOOD ABSORBER Therefore a1 Wi w2 w w4 a2 a3 a4 3 Cold (bi 5

8 Radiation from a Black Body 1:/I \4OOOK ì \<3000K 2000K Brooks/Cole - Thomson 6

9 Wien attempt V ($) I Sc-. I I I I I I 0 2%5 0.ooiCdi Thon,.oci -e s I /\ S j exponentia1 ts Wiens 5 A points I law a. a, Experimental Curve fitting Must explain the data

10 But how many modes? Average Energy = kt Each mode (wavelength present) has System of standing waves wavelengths present. but not all Energy reflected back and forth Consider Black Body Chamber Lord Rayleigh attempt:

11 %da (kt)xc8d (Average Energy) x (Number of Modes) Energy Density Wx will be dn 2 cia and Number of modes with wavelength between 9

12 Emittance is related to Energy Density by: C 4 So WdX = 22r(kT)cCD 1D

13 4 sc,, V.. Rayleigh Jeans Law 2005 Brooks/Cole -Thomson o i Wavelength (jim) This is the Ultraviolet Catastrophe 11

14 =fl X 2a Thus where no,],2,3 = 2.,ira Therefore E=O at x=oandx=a E. re sin sin 2in x 0 Equation for E vector of wave moving in x direction is Consider hollow metal cube with sides a 12

15 2 2 - =n 2a 2a n n,, and n We want to form a special space with coordinates (A) or for v since Xf== c 2 2a 2 2 gflx+ +flz For radiation moving in random direction Then for the other directions 13

16 There are two polarizations for eachfthus two modes per point. gives one point in space and represents one allowed frequency f n n and n Each set of integers IC 14

17 The total number of modes is 2 times this. volume= I rn +n +n.,) X y (B) 1r are only positive the volume of region shown is n n and n Since Volume of sphere = 43 fi = figure). Now consider sphere with radius n (see the 15

18 dn =8z A d2 volume between X and A + da differentiate to get the number of modes per unit V 3c N 8nf 3 3 Number of modes per unit volume is The volume of the cube is a3 83 c } 3 c3 N=2 2af181a 3f3 Thus Cube Equation A to get 16

19 = fen 0ekTdE Total Energy = fnode Total number of modes Why Average Energy = kt? Catastrophe. This is what we used to show Ultraviolet W2d2=8lckT6U between? and ) + da. then will be The energy density of waves with wavelength mode kt Rayleigh used for the average energy for each 17

20 E ave TotalEnergy Total number 18 The energy in the container is quantized. ENERGY IS QUANTIZED PLANCK S SOLUTION continuous and thus you can use the integrals. This result is due to considering the energy kt This gives for Average Energy for each mode j n 0e ktde Therefore Average Energy will be

21 for Total Energy Thus =mun 0e kt mu = (mu,)n, The total energy of tim modes will be kt =n 0e kt Em mu Number of oscillators with Em Let E, be the energy where m is an interger. You can add energy or take energy out only in quantized amounts example amount = ii 19

22 = Z mun 0eU Ea,e = flu This gives 0e 1c7 >Zn Z mun 0e Divide to get average energy per mode and Total number of modes mu = mu mu ekt 1 20

23 Letu=hv=hc/A Then E ave he he 2 exkt 1 Use the number of modes we obtained 8,r da Energy density 21

24 u(2,t)da = W2d2 8ch2 5 ch eakt 1 or Emittance PJ7 2d2 2,rc 2h25 ch eakt 1 dx Jaw en s law Wavelength, microns 22

25 BLACKBODY RADIATION LAWS Stefan-Boltzmann R(T) = EaT 4 Blackbody E = 1 R(T) 4 crt T is Temperature (Kelvin) a = 5.67x10 8 (m2 1()

26 b = 1898x10 3m K 2max = b Wien Displacement

27 Electrons are observed coming off with various velocities. L a) J flel MV #,Jia4s. PL4 eisck.e Light (ultraviolet) falls on a metal surface: PHOTOELECTRIC EFFECT 23

28 02005 Brooks!Cole - Thomson 10-V supply + ammeter Sc n si tive Collector emitter Light Metallic To study the effect we need 24

29 called the stopping voltage, V The voltage necessary to stop all electrons is o 25 &ootsfco4e - linmon (a) 0 Applied voltage 1 2 Photocurren t What we find is 25

30 Plot V times e for various frequencies of light,f 26

31 Metal 2 Metal 1 Slope = h 0 / 102, I jco1 SmoksiCole - Thomson Cannot explain with electromagnetic wave. 27

32 F ir b 4 F 4 Pk!L K 3. Each photon has energy hf more. 2. Photons appear or disappear in units of 1 or called photons 1. Light made up of pulses or quanta of energy Einstein Solution 28

33 or hf çb=ev The energy needed to escape will be the work function. The electrons with the highest velocity will have come from the top of the energy levels in the metal near the Fermi level. hf energy=ke=ev =mv2 left over energy goes into kinetic energy electron uses certain amount of energy to excape one electron in metal. Photon has energy hf and gives this energy to 29

34 ev =hf-ø 30

35 Metal 2 Metal 1 Slope = h 0 IIoy I I / 2005 BrooksICo(e - Thomson Also: 31

36 No time delay one electron. photon gives all of energy to Increase frequency greater and give more energy to electrons and need larger V. energy of photons is Increase intensity of electrons increases and current increases number 32

37 WHAT IS LIGHT WAVE? PARTICLE? 33