Rigid rotor

November 9th, 2016

Let’s consider the model of diatomic molecules in two material points m_1 and m_2, attached to the ends of weightless rigid rod (Fig. 1).

 Fig.1 Rigid rotator model

The energy of rotational motion around the center of mass of the molecule depends on the moment of inertia and angular velocity rotator:

E=\frac{I \omega^2}{2}   (1)

The moment of inertia of the system of material points is the sum of moments of inertia of  points:

I= m_1r_1^2+m_2r_2^2 (2)

Given that r=r_1+r_2 and m_1r_1=m_2r_2  we get to:

r_1= \frac{m_2 r}{m_1+m_2} ;  r_2= \frac{m_1 r}{m_1+m_2}   (3)

Taking into account (3) we get to the moment of inertia:

I=\mu r^2 , (4)

where \mu=\frac{m_1m_2}{m_1+m_2}.

 Writing expression for the moment of inertia of the molecule in the form of (4) indicates that the problem may simplify by considering the rotation not two points is at the center of mass, but the rotation of of one point with the mass \mu.

This approach makes it possible to consider the Schrödinger equation for the mass \mu on condition U=0 that stems from rotator rigidity condition:

\Delta\psi+\frac{8\pi^2\mu}{h^2}E\psi=0.    (5)

Writing Laplacian in spherical coordinates and demanding that the wave function is finite, continuous and unambiguous quantum-chemical transformation after getting a simple expression for the energy spectrum of the rigid rotator:

E_J=\frac{h^2}{8 \pi^2 I}J(J+1),   (6)

where J is the rotational quantum number, which is set to J=0,1,2,3,…. Energy transition of the rotator from energy state E_2 to the energy state E_1 according to Bohr frequency condition is:

h\nu=E_2-E_1.   (7)

Let’s divide (7) on hc and introduce B=\frac {h}{8 \pi^2 c I}, we receive:

     \frac {\nu}{c}= B(J+1)(J+2)-BJ(J+1)=2B(J+1) ,   (9)

where  J=0,1,2,3….

From equation (9) follows that the difference of two neighboring wavenumbers in the pure rotational spectrum is 2B what makes it possible to find the rotational constant  B, moment of inertia of the molecule I=\frac{h}{8 \pi^2 c B} and interatomic distance  r=\sqrt \frac {I}{\mu}.

(The information has been taken from article: Shvets V.D. Application of  results modern spectroscopy investigations at training of physics and physical chemistry . – Hidh Education. – № 2-3. – 2002. – P. 41–48. )


Task 1. The distance between the nuclear of the molecule N2 is equal to 1.095A0. Define: the moment of inertia J of the molecule N2; angular momentum corresponding to the lower excited state; rotational constant B; energy change during the transition from the third rotational level to the second; the distance between lines of the pure rotational spectrum of the molecule.

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Task 2.  Find the rotational constant and the moment of inertia I of the molecule CO, if the distance between adjacent lines of the pure rotational spectrum of the molecule emission CO is equal to 0.48 mеV.

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Task 3. For the molecule О2 find: the reduced mass μ; interatomic distance d, if the rotational constant is B=0.178 mеV.

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Task 4.  Find for the molecule NO: momentum of inertia of the molecule if the interatomic distance d=1.15 A^0; rotational constant B of the molecule; temperature T, where the average kinetic energy of translational motion is equal to the energy required to excite the molecule for the first rotational energy level.

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Task 5. Find the distance d between the adjacent nuclei of the molecule CH, if the interval between adjacent lines of pure rotational spectrum of emission of the molecule is equal to 29 cm^{-1}

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(The information has been taken from book «Elements of Quantum Mechanics” (author V.D. Shvets), the part of which was published in journal “Problems in Education” (article: “Programming of students’ educational activity in training of elements of quantum mechanics”), 2006. – № 43. – P. 50–64. This book has Recommendation of Ministry Education of Ukraine № 44 from 14.01.1999. )


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