1192_Physics ProblemsTechnical Physics

# 1192_Physics ProblemsTechnical Physics - Chapter 42 (a)...

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Chapter 42 533 P42.48 (a) I = × ×× L N M O Q P −− 300 10 100 10 150 10 424 10 3 96 2 15 . .. . J s m Wm 2 ej e j π (b) 0600 10 30 0 10 120 10 750 3 9 2 6 2 12 . . . × × × = J m m J M e V P42.49 Et ==× × = P 1 00 10 1 00 10 0 010 0 68 . W s J Eh f hc γ λ == = × 6626 10 694 3 10 286 10 34 8 9 19 . . e j J N E E == × 00100 349 10 19 16 . . . photons *P42.50 (a) N N Ne ee g Ek g EE k h c k B B BB 3 2 300 300 300 300 3 2 32 = −⋅ K K K K bg b g where is the wavelength of light radiated in the transition. N N e N N e 3 2 6 63 10 3 10 632 8 10 1.38 10 300 3 2 75 9 33 34 8 9 23 107 10 = −× × × × . . J s m s m J K K e j e j af (b) N N e u T uB A A = where the subscript u refers to an upper energy state and the subscript A to a lower energy state. Since EEE hc u −= = A photon N N e u hc k T B A = . Thus, we require 102 . = e hc k T B or ln . . 663 10 3 10 632 8 10 1 38 10 34 8 92 3 a f e j e j =− ×⋅ × Js ms m J K T T × × 228 10 115 10 4 6 . ln . . K. A negative-temperature state is not achieved by cooling the system below 0 K, but by
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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