University:
Widener University
Course:
PHYS-121 | Fundamentals of Physics I
Academic year:

2023
Views:

374

Pages:

2
Author:

Wyatt Hays

Examination #2 Following is a complete list of observed transitions involving levels J = 0, 1, and 2 for two isotopomers of formaldehyde in their vibrational ground states: H2 12 C16 O H2 13 C16 O 71.14 MHz 4829.66 14488.65 72837.97 140839.54 145602.98 150498.36 — 4593.09 13778.86 71024.80 137449.97 141983.75 146635.69 Note that in a real spectrum you could not arrange to see only transitions involving J = 0, 1, and 2. In fact, there are more than 54 observed H2 12 C16 O transitions below 150498.36 MHz. I. (50 points) Assign the above microwave transitions for both isotopic molecules. It is distinctly to your ad vantage to assume that H2 CO belongs to the C2v point group and then show that the observed spectrum is consistent with this assumption. You will ﬁnd the energy level expressions on the next page (from page 192 of Bernath) to be useful, but I doubt whether the spectrum can be assigned without knowledge of the dipole moment direction relative to the inertial axis system, asymmetric top rotational selection rules, and a good guess of the trial geometry. II. (10 points) Explain why no other transitions involving J = 0, 1, and 2 were observed. III. (10 points) Obtain values of A, B, and C for the two isotopic species. Since we have neglected centrifugal dis tortion, it will not be possible to ﬁt all transitions exactly with only 3 rotational constants. Devise a procedure that gives a “best ﬁt” to all lines. IV. (10 points) Explain why the inertial defect Δ = Ic − I a − I b is a good test for planarity. Why does H2 CO appear NOT to be planar from the microwave spectrum? page 2 V. (20 points) Obtain a best possible geometry for H2 CO using your A, B, and C values for the two isotopomers. If you have not yet discovered the value of iterative calculations on a pocket calculator, now is the time! JKa ,Kc 000 110 111 101 220 221 211 212 202 330 331 321 322 312 313 303 Rotational energy 0 A+B A+C B+C � 2A + 2B + 2C + 2 (B − C)2 + (A − C)(A − B) 4A + B + C A + 4B + C A + B + 4C � 2A + 2B + 2C − 2 �(B − C)2 + (A − C)(A − B) 5A + 5B + 2C + 2 �4(A − B)2 + (A − C)(B − C) 5A + 2B + 5C + 2 �4(A − C)2 − (A − B)(B − C) 2A + 5B + 5C + 2 4(B − C)2 + (A − B)(A − C) 4A + 4B + 4C � 5A + 5B + 2C − 2 �4(A − B)2 + (A − C)(B − C) 5A + 2B + 5C − 2 �4(A − C)2 − (A − B)(B − C) 2A + 5B + 5C − 2 4(B − C)2 + (A − B)(A − C)