Question

The equilibrium internuclear distance in H35Cl molecule is 127.5 pm.
(a) Calculate the reduced mass and moment of inertia of the molecule.
(b) Determine the values of angular momentum L, projection of angular momentum Lz, energy E for the rotational quantum state with J=1.

Answer 1

Page a Given by Internuclear distance in H35cl - Re: 127.5 pm We can calculate reduced mass by using formula, Il me Mar MH + mal where l = reduced mass MH = mass of H. ma - mass of a As we know, M: 1.6735 x 10-4 kg 5.8065 18-26 kg ma ll -27 1.6785 x 10 x 5.8065x10 26 1.6735 X10-27 + 5.8065 x 10 - 26 - 27 16266 X 10 kg i Reduced mass of H35a le1.6266 X 102 kg -27 a) We can calculate moment of inertia by using formula Il re² where. I: moment of inentia I = reduced mass inteonudear distance

Date Page L 1,6266 X 10-27 x (1275 X1000) 2 2. bd4 4.210kg-mol 2,6442 x 10-17 Moment of inertia -47 2. I = 2.6442 x 10 b) Anguloo momentum can be calculated by using formula L = Jelerin h where h= plank's constant с 6.6260 x 10-24m²kg/ As we know, l=1 2. 12 t V 2 x G. 6260 x 10-34 2 TT 1.4914 x 10-34 Angutan momentum -- 1.4914 x 10-34 m² kg 6

CIASSM Date Page Rotational energy can be calculated as E = L² RI where f: Rotational energy 1. Angular momentum I= Moment of inertia (1,4974 x 10-84) 2 x 2.6442 x 10 -47 - 4.20.58 x 10-22 Ener Rotational energy - 4,2058 x 10-22 J.