PC Chapter 03

PC Chapter 03 - 3 The Quantum Theory of Light 3-1(a dB E2 r = r 2 dt r dB E= 2 dt(b If r remains constant then E = Eq = re dv = dB 2me v v 2r dB e

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29 3 The Quantum Theory of Light 3-1 (a) ππ  =   2 2 dB E rr dt ( 29 = 2 r dB E dt (b) If r remains constant, then: == 2 r dB E E qe dt so that 2 e r dB Fdt dt m dv dt , or +∆ = = ∆= ∫∫ 0 2 2 2 e vv B e v e re dv dB m er dv dB m erB v m B E E r (c) ( 29 ( ϖ -- = = = × × 19 31 10 1 T 1.6 10 C 9.1 10 kg 8.8 10 rad sec 22 e v eB rm ( ( π ϖ λ - × = = = × 8 15 9 3.0 10 m s 2 2 2 3.8 10 rad sec 500 10 m c f ; ϖ ϖ - 5 2.3 10 (d) For the ϖ 0 line the electrons’ plane is parallel to B , therefore, the magnetic flux, Φ B is always zero. This means that F and E are zero and as a consequence, there is no force on the electrons and there will be no v for the electrons. The ϖ ϖ 0 is the case calculated in parts (a)–(c). The ϖ ϖ -∆ 0 will have the same magnitude for F , B , and v as in (a)–(c) but the direction will be opposite. e B
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30 CHAPTER 3 THE QUANTUM THEORY OF LIGHT 3-2 Assume that your skin can be considered a blackbody. One can then use Wien’s displacement law, λ - =×⋅ 2 max 0.289 8 10 m K T with == 0 35 C 308 K T to find λ - - ×⋅ = = ×= 2 6 max 0.289 8 10 m K 9.41 10 m 9 410 nm 308 K . 3-3 (a) The total energy of a simple harmonic oscillator having an amplitude A is 2 2 kA , therefore, ( 29 (29 = 2 2 0.4 m 25 N m 2.0 J 22 kA E . The frequency of oscillation will be ( 29 ππ  ===   12 1 1 25 0.56 Hz 2 k f m . (b) If energy is quantized, it will be given by = n E nhf and from the result of (a) there follows ( 29 ( 29 = = -34 6.63 10 J s 0.56 Hz 2.0 J n E nhf n . Upon solving for n one obtains 33 5.4 10 n . (c) The energy carried away by one quantum of charge in energy will be ( 29 ( 29 -- × 34 34 6.63 10 J s 0.56 Hz 3.7 10 J E hf . 3-4 (a) From Stefan’s law, one has σ = 4 P T A . Therefore, ( - 82 4 62 4 5.7 10 W m 3000 K 4.62 10 W m P K A . (b) ×× 2 75 W 16.2 mm 4.62 10 W m 4.62 10 W m P A . 3-5 (a) Planck’s radiation energy density law as a function of wavelength and temperature is given by λ π λ λ = - 5 8 , 1 B h cT hc uT e . Using λ = 0 u and setting λ = max B hc x kT , yields an extremum in λ , with respect to λ . The result is ( ( λλ λ - = - +- max max 1 max 0 51 BB hc k T hc k T B hc ee or ( 29 - =- x xe . (b) Solving for x by successive approximations, gives 2245 4.965 x or λ - = = 3 max 4.965 2.90 10 m K B hc T k . 3-6 Planck length - = 35 4.05 10 m hG c Planck time - = 45 5 1.35 10 s hG c
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MODERN PHYSICS 31 Planck mass -  =   12 8 5.46 10 kg hG c 3-7 (a) In general, λ = 2 n L where = K 1, 2, 3, n defines a mode or standing wave pattern with a given wavelength. As we wish to find the number of possible values of n between 2.0 and 2.1 cm, we use λ = 2 L n ( 29 (29 ( == ∆= 200 2.0 cm 2 200 2.0 200 2.1 cm 2 190 2.1 10 n n n As n changes by one for each allowed standing wave, there are 10 standing waves of different wavelength between 2.0 and 2.1 cm. 2 m (b) The number of modes per unit wavelength per unit length is (29 λ - 2 10 200 0.5 cm 0.1 n L . (c) For short wavelengths n is almost a continuous function of λ . Thus one may use calculus to approximate ( 29 γλ = 1 n dn L Ld . As λ = 2 L n , λ λ = 2 2 d nL d and λ λ = 2 dn .
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This note was uploaded on 11/05/2009 for the course PHSX 313 taught by Professor Staff during the Fall '09 term at Kansas.

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PC Chapter 03 - 3 The Quantum Theory of Light 3-1(a dB E2 r = r 2 dt r dB E= 2 dt(b If r remains constant then E = Eq = re dv = dB 2me v v 2r dB e

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