1210_Physics ProblemsTechnical Physics

# 1210_Physics ProblemsTechnical Physics - Chapter 43 P43.9 I...

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Chapter 43 551 P43.9 Im r m r =+ 11 2 22 2 where mr = and rr r 12 += . Then r m 1 1 = so m rr 1 2 and r mm 2 1 = + . Also, r m 2 2 = . Thus, r m r 1 2 and r 1 2 = + mmr mmr m m r = + + + = + + = + = 1 2 2 21 2 2 2 2 2 bg µ . P43.10 (a) = + ×= × −− 22 99 35 45 22 99 35 45 1 66 10 2 32 10 27 26 .. af a f ej kg kg Ir == × × = × 6 9 2 45 2 32 10 0 280 10 1 81 10 . kg m kg m 2 (b) hc III I I h I λ π + = = = = = = 2 2 2 2 2 2 22 1 2 11 1 32 2 4 a f a f ×× ×⋅ = cI h 4 2 300 10 4 181 10 26626 10 162 2 82 4 5 34 . . ms kg m Js cm 2 e j P43.11 The energy of a rotational transition is E I J = F H G I K J = 2 where J is the rotational quantum number of the higher energy state (see Equation 43.7). We do not know J from the data. However, E hc ×⋅× × F H G I K J λλ 6626 10 1 10 34 8 19 eV 1.60 J e j . For each observed wavelength, (mm) E (eV) 0.120 4 0.010 32 0.096 4 0.012 88 0.080 4 0.015 44 0.069 0 0.018 00 0.060 4 0.020 56 The
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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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