In rotational spectroscopy, the energy differences are measured typically in uni
ID: 492637 • Letter: I
Question
In rotational spectroscopy, the energy differences are measured typically in units of the rotational constant B_e, E = B_e J (J + 1). Most often, B_e is provided in units of cm^-1 (conversion factor h middot c ! [sometimes B is given in MHz]) Thus, E = hcBJ(J + 1), with B = h/8 pi^2 mu c r^2 or B[cm^-1] = 16.858/mu[amu] r^2 [A^2] The last conversion combines all constants and computes (quickly) B in units of cm^-1 if the reduced mass is given in units of amu and the equilibrium bond distance in units of A. Alternatively, if B is extracted from experiment, then the bond distance can be analyzed if the reduced mass is known. For cationic neutral, and anionic^14 N^16 O the rotational constant has been measured to 1.9982, 1.7043, 1.427 cm^-1, respectively. What are the bond distances in these molecules?Explanation / Answer
Calculate the reduced mass as µ = mN*mO/(mN + mO) = (14)*(16)/(14 + 16) = 7.4667
Next use the relation 1 Aº = 1.0*10-8 cm
Plug in the values and write
1) Cationic 14N16O: 1.9982 cm-1 = 16.858/(7.4667 amu)*r2
===> r2 = 16.858/(7.4667)*(1.9982) = 1.12788
===> r =1.0620
The bond distance is 1.062 Aº (ans).
2) Neutral 14N16O: 1.7043 = 16.858/(7.4667)*r2
===> r2 = 16.858/(7.4667)*(1.7043) = 1.3247
===> r = 1.1509 1.151
The bond distance is 1.151 Aº (ans).
3) Anionic 14N16O: 1.427 = 16.858/(7.4667)*r2
===> r2 = 16.858/(7.4667)*(1.427) = 1.5821
===> r = 1.2578 1.258
The bond distance is 1.258 Aº (ans).
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