The J=3---J=2 pure rotational transtion for the CS molecule occur at 146.831 GHZ

Question

The J=3---J=2 pure rotational transtion for the CS molecule occur at 146.831 GHZ. Calculate the bond length(in pm) for the CS molecule, assuming that there is negligible centrifugal distortion.

Explanation / Answer

Ghz --> 1e-9 s in frequency (see B in equation below) r = sqrt(i/reduced mu) i = h/8(pi^2)B i = 6.626e-34 / (8*3.14^2)(146.831e-9) = 1.233e-42 mu of C MM/avogadro's constant 12 g/mol / (6.02e23/mol) = 1.993e-23 g = 1.993e-26 kg mu of S 32 / 6.02e23 = 5.316 e-26 kg reduced mu muC*muS/(muC+muS) = ((1.993e-26)*(5.316e-26))/(5.316e-26 + 1.993e-26) = 1.45e-26 r = sqrt(1.233e-42/1.45e-26) = 9.22e-9 m = 9220 pm Source: https://docs.google.com/viewer?a=v&q=cache:VXdNd1xtg-AJ:www.oup.com/uk/orc/bin/9780199277896/01student/solutions/ch11_student.pdf+&hl=en&gl=us&pid=bl&srcid=ADGEESgtR8_a0yeRaGDY9nKZ5EVdmXOk4OSj0v7N-V3ZZ-GWSHoMXu4ahVKIj40GCTtv0sZdMiStxvC_QvQsOAJ8-DkufM17jrXvnLXCNP6cOVnP5fbBmQiabnVlcbjvxewl3Fw3w0QC&sig=AHIEtbSXyJH12y8EUuEzs40diJ13b-tSJA&pli=1

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