Question

If the wave number of J=3 arrow 2 rotational transition of 1H35Cl considered as a rigid rotator is 63.56.calculate the moment of inertia of molecule and bond length​

Answer 1

Answer:

a. If the wavenumber of the J = 3←2 rotational transition of 1H35Cl considered as a rigid rotator is 63.56 cm−1, what is (a) the moment of inertia of the molecule, (b) the bond length?

b. If the wavenumber of the J = 1←0 rotational transition of 1H81Br considered as a rigid rotator is 16.93 cm−1, what is (a) the moment of inertia of the molecule, (b) the bond length?

Solutions:

Explanation:

If the wavenumber of the J = 3?2 rotational transition of 1H35Cl considered as a rigid rotator is 63.56 cm-1, what is (a) the moment of inertia of the molecule, (b) the bond length? Answer Moment of Inertia For calculating the moment of Inertia we, need to calculate the wavenumber of the transition first hc? ¯= ?E=hcB[J(J+1)-(J-1)J]=2hcBJ[13.25,13.27] Since the value of J which is the upper state 3 as per the question and B is the rotational