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PC Chapter 03

# PC Chapter 03 - 3 The Quantum Theory of Light 3-1(a dB E2 r...

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29 3 The Quantum Theory of Light 3-1 (a) π π = 2 2 dB E r r dt ( 29 = 2 r dB E dt (b) If r remains constant, then: ( 29 = = 2 r dB E Eq e dt so that ( 29 = = 2 e r dB Fdt dt m dv dt , or +∆ = = = 0 2 2 2 e v v B e v e re dv dB m er dv dB m erB v m B E E r (c) ( 29 ( 29 ϖ - - = = = × × = × 19 31 10 1 T 1.6 10 C 9.1 10 kg 8.8 10 rad sec 2 2 e v eB r m ( 29 ( 29 π ϖ 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 2 2 kA E . The frequency of oscillation will be ( 29 ( 29 ( 29 π π = = = 1 2 1 2 1 1 25 0.56 Hz 2 2 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, ( 29 ( 29 - = × = × 8 2 4 6 2 4 5.7 10 W m 3000 K 4.62 10 W m P K A . (b) = = = × × 2 6 2 6 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 ( 29 ( 29 λ π λ λ = - 5 8 , 1 B hc T hc u T e . Using λ = 0 u and setting λ = max B hc x k T , yields an extremum in ( 29 λ , u T with respect to λ . The result is ( 29 ( 29 λ λ λ - = - + - max max 1 max 0 5 1 B B hc k T hc k T B hc e e k T or ( 29 - = - 5 1 x x e . (b) Solving for x by successive approximations, gives 2245 4.965 x or ( 29 λ - = = × 3 max 4.965 2.90 10 m K B hc T k . 3-6 Planck length - = = × 1 2 35 4.05 10 m hG c Planck time - = = × 1 2 45 5 1.35 10 s hG c
MODERN PHYSICS 31 Planck mass - = = × 1 2 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 ( 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.

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