hw3 - 3 The Quantum Theory of Light 2 dB E2 r = r dt r dB E...

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3 The Quantum Theory of Light 3-1 (a) E 2 π r = r 2 dB dt E = r 2 dB dt (b) If r remains constant, then: E = Eq = r 2 dB dt e so that Fdt = r 2 dB dt dt = m e dv , or dv = re 2 m e dB dv v v +∆ v = er 2 m e dB 0 B v = erB 2 m e B E E r (c) ω = v r = eB 2 m e = 1.6 × 10 19 C () 1 T 2 9.1 × 10 31 kg = 8.8 × 10 10 rad sec = 2 f = 2 c λ = 2 3.0 × 10 8 ms ( ) 500 × 10 9 m = 3.8 × 10 15 rad sec ; = 2.3 × 10 5 (d) For the 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 0 is the case calculated in parts (a)–(c). The 0 − ∆ will have the same magnitude for F , B , and v as in (a)–(c) but the direction will be opposite. e B 3-2 Assume that your skin can be considered a blackbody. One can then use Wien’s displacement law, max T = 0.289 8 × 10 2 m K with T = 35 0 C = 308 K to find
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λ max = 0.289 8 × 10 2 m K 308 K = 9.41 × 10 6 m = 9 410 nm . 3-4 (a) From Stefan’s law, one has P A = σ T 4 . Therefore, P A = 5.7 × 10 8 Wm 2 K 4 () 3000 K 4 = 4.62 × 10 6 2 . (b) A = P 4.62 × 10 6 2 = 75 W 4.62 × 10 6 2 = 16.2 mm 2 . 3-5 (a) Planck’s radiation energy density law as a function of wavelength and temperature is given by u , T = 8 π hc 5 e hc B T 1 . Using u = 0 and setting x = hc max k B T , yields an extremum in u T with respect to . The result is 0 =− 5 + hc max k B T e hc max k B T e hc max k B T 1 1 or x = 51 e x ( ) . (b) Solving for x by successive approximations, gives x 4.965 or max T = hc k B 4.965 = 2.90 × 10 3 m K . 3-10 The energy per photon, E = hf and the total energy E transmitted in a time t is Pt where power P = 100 kW . Since E = nhf where n is the total number of photons transmitted in the time t , and f = 94 MHz , there results nhf = 100 kW ( ) t = 10 5
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hw3 - 3 The Quantum Theory of Light 2 dB E2 r = r dt r dB E...

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