Question

What is the reduced mass of ¹H⁸¹Br? What is the moment of inertia if R=141.4 pm?

Calculate the frequency corresponding the the transition between the first and the second,

and between the second and third rotational energy levels.

Answer 1

reduced mass of 1H81Br = 1 x 81/1 + 81 = 0.988 amu/6.023 x 10^-26 = 1.64 x 10^-27 kg

R = 141.4 pm = 1.414 x 10^-10 m

moment of inertia (I) = reduced mass x R^2 = 1.64 x 10^-27 x (1.414 x 10^-10)^2 = 3.28 x 10^-47 kg.m^2

rotattional constant Be = h/8pi^2CI = 6.626 x 10^-34/8 x 3.14^2 x 3 x 10^8 x 3.28 x 10^-47 x 100 = 8.54 cm-1

Frequency corresonding to first (J=0) to second (J=1) and second (J=1) to third (J=3) levels

For J=0 to J=1 rotational energy level transiiton = cBeJ(J+1) = 3 x 10^8 x 854 x 2 = 5.124 x 10^11 s-1

For J=1 to J=2 rotational energy level transiiton = cBeJ(J+1) = 3 x 10^8 x 854 x 6 - 5.124 x 10^11 = 1.025 x 10^12 s-1