Question

12. The equilibrium bond distance of potassium iodide, "K1I is 3.048 Å. (Hint: K-39 mass = 38.963707 amu; 1-127 mass = 126.917871 amu) (a) Calculate its reduced mass and its moment of inertia. (b) Calculate the energy for the state J = 3. (8 pts)

Answer 1

The rotational spectroscopy is useful to determine the energy of the molecule as follows:

E=J(J+1)h^2/8*(3.14)^2*r^2

(a) calculate the reduced mass as follows: el = mim2 in... mit me Reduced mous is lle mass of 88.963707 amu, mass of %, 126.9 198 71 anote k is m, = .: e - KI (38.9637004 omy) x (126.919841 amu) (.38.963707 +126.911787) Jamu 4945.190739 amy xama 165.88157 8 ardy 29.8115428 cm < Convent mass into tg as follows I amy = 1.6605 X1627 kg kg .: 29.8715928 amy x 1.6605 X10 I amu = 4.95021166 X10-26 ta Thus, reduced mass of kI in kilogram is 14.9502 X 10 26 kg

Date Determine the moment of inertia as follows, I allge, 19 (1) A I = moment of inertia, el is the reduced mass and r is the distance of atoms from the solational aseisti er is calcolated as follows: De equilibrium olistances into no 2013.048 X 39.524 Á. Convert distance from Å to metre as follows: Ă I m = 1010 1. 1.524 Å X IMALED 1.524 X10 lotion Ure egin 09, 2 .2 I = er 2. I = (4.9502 X 1020 kg) x (1.524 x 10'° m) 1. I = (4.502 X 1678 19x ( 2.322576 x 10-20 m2) ... I 1.0456 X10745 urgmeer Thue moonent of inertia Beri ise 1.0456 x 1645 kg. m2.

Date Calculate the energy for the state J = 3: o level is The energy Calcolated of as solational energy follows: to E = TUD 62 802 I Here, I is solational quantum number = 3, and his the plank constant, IT = 3.14. Er = 3 (3 + 1 x C6.626 X 10 34 m² tots 8 X (3. 14) X (10456 x 10-45 kg og2) E = 7.9812 810BB 7 :: Ep = 3x4x(4.3903876 x 10-67) [kg.m2/s)" 18 X (3:14)2 x (1.0466 X 1045 kg. m2) E = 5.268.467 X10 66 kg.m2 IS? 8. 247358 x 10-44 L E = 6.38806 X10-23 kg.m2.152. As unit Tolles is equal to ro m² 15² --23 Hence, energy in Joules is 6. 388064 10 J.