Question

8. (32 points) Scientists have studied excited electronic states of molecules by a variety of experimental techniques. The following data were obtained for an excited electronic state of Mgo. rotational constant, B = 0.5014 cm vibrational constant, .= 632.5 cm a) At what energy (in cm) and wavelength (in um, where 1 um = 10 m) will the J = 6 to J - 7 rotational transition occur for this electronic state of Mgo? b) What is the value for le equilibrium bond distance) for a MgO molecule in the above electronic state? Give your final answer in units of nm. c) What is the value for k (force constant) for a MgO molecule in the above electronic state? Give your final answer in units of N/m. The masses for 24Mg and 160 are m(Mg) = 23.9850 amu, and m(160) = 15.9949 amu.

Answer 1

Date: 1 solution: Given B = 0.5o14 cm't values: - vibrational constant . we = 632.5cm difference between two notational (a) formula for energy energy levels Ë z 2B (5+2) T Jo 6 , Jag 1 In this formula energy levels will be No to transition loccur. ! so that, I use from value for rotational lowest value due lowest to highest J - 6 се 22 X 0.5014 2X0.50 14x8 E = 8.0224 an I

wavelength ; 1 (cm) 8.0224 = 0.12465 cm o. 12465 xiou rem = 12u6.5 um (6) equlibrium bond length (red B = h 872.C 872Ć B

Page : Date : I: 6.626X1634 Lax (3114) X3 X.1010x 0:5044 - 2 0.05585 x 1044 I 25.585 x 10-46 Kg m² = 462 formula u= reduce mass A- mim2 mitme mis mass of my m2 = mass of oxygen = 23.9850 x 15.9gug 23.9850 + 15.ggug _323.637677 ___39.9133_ u= 9.5.957638 amu multiply by = 1-67 X 1021 kg which is _ I amy 1 amy a 1.67 X 1024 u. 9.595 7638 X 1.67 X10 27 I 16.0249 X1027 kg

re? I kg m2 = 5.585x10 uu 16.0249.X1027 2 0 :3485 x 1017 Ve 3.485 X10 18 re = 1.8668 X109 m ez 1.8668 nm 2) value of k we= 632.5cm x 3.14 X 3 - -2 16.0249 X10 K = we² x 2 A CX M = (632-5|x. 2 x 3.148 38 / 070 x 15.0249x76* = 120,780, 629 Xiött K = 12.078 x 10 to Nlun TK = 12.078 x1612 Nlend