Question

The inter atomic distance of 14N 16O molecule is 115.1 pm.calculate-
-reduced mass
-its moment of inertia
-the wave number of the line corresponding to lowest absorption in m-1 unit, and
-the energy in m-1 unit for the transition J=2 to J=3.

Answer 1

i)The mass of a nitrogen atom is 14.003 amu; the mass of an oxygen atom is 15.995 amu; and the conversion factor is1.6605×10-27kg/amu.The reduced mass is휇=휇푁휇푂휇푁+휇푂휇=14.003×15.99514.003+15.995=7.4664푎푚푢=7.4664×1.6605×10−27kg=1.24×10−26kg(ii)The moment of inertia is퐼=휇푅2=1.24×10−26×(115.1×10−12)2=1.64×10−46푘푔∙푚2(iii)The rotational constant is23412 8 46 286.626 10 J s170.6 m8 2.998 10 m/s 1.64 10 kg mhBcIBWavenumber is퐹=퐵퐽′(퐽′+1)−퐵퐽(퐽+1)=퐵(1(1+1)−0(0+1))=2퐵=2∗170.6==341.2m−1(iv)퐸=퐵퐽′(퐽′+1)−퐵퐽(퐽+1)=퐵(3(3+1)−2(2+1))=6퐵=6∗170.6m−1퐸=1023.6m−1