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Rigid rotators
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- 2. Rigid Rotators Definition: Two rotatingatoms having fixed bond length are called rigid rotators.
- 3. Why we studyrigid rotator: Rigid rotator is studied because it is a model for the diatomic molecule. It helps us to calculate rotational energy of diatomic molecule.
- 4. Classical treatment; Let usconsider two atoms having masses m1 and m2 and fixed bond length r. The center of gravity is fixed. m1r1 ═ m2r2 equation (1) For Rigid rotators r ═ r1 + r2 equation (2) From equation (1) 𝑟2 = 𝑚1𝑟1 𝑚2 Put this value in equation 2; 𝑟 = 𝑟1 + 𝑚1𝑟1 𝑚2 By taking LCM; 𝑟 = 𝑚2𝑟1+𝑚1𝑟1 𝑚2
- 5. 𝑟 = 𝑟1(𝑚2 +𝑚1) 𝑚2 𝑟1 = 𝑚2𝑟 𝑚2+𝑚1 Similarly; 𝑟2 = 𝑚1𝑟 𝑚1+𝑚2 As we know that moment of inertia I= m1r1 2 + m2r2 2 equation (3) Putting the values of r1 and r2 in equation (3) 𝐼 = 𝑚1( 𝑚2𝑟 𝑚2+𝑚1 )2 + 𝑚2( 𝑚1𝑟 𝑚1+𝑚2 )2 𝐼 = 𝑚1𝑚22 𝑟2 (𝑚1+𝑚2)2 + 𝑚12 𝑚2𝑟2 (𝑚1+𝑚2)2 𝐼 = 𝑚1𝑚2𝑟2(𝑚2+𝑚1) (𝑚1+𝑚2)2 𝐼 = 𝑚1𝑚2 𝑚1+𝑚2 𝑟2
- 6. I =µr2 Where µ=reduced mass Derivation of energy; Total energy= K.E + P.E In case of rigid rotator P.E=0 Total energy = K.E 𝐸 = 1 2 𝑚1𝑣12 + 1 2 𝑚2𝑣22 By angular velocity formula 𝑣 = 𝑟𝜔 Put this value in the above equation 𝐸 = 1 2 𝑚1𝑟12 𝜔2 + 1 2 𝑚2𝑟22 𝜔2 𝐸 = 1 2 𝜔2 (𝑚1𝑟12 + 𝑚2𝑟22 )
- 7. 𝐸 = 1 2 𝜔2 𝐼 As angularmomentum is given as 𝐿 = 𝐼𝜔 𝜔 = 𝐿 𝐼 𝐸 = 𝑙2 2𝐼 𝐸 = 𝐿2 2µ𝑟2
- 8. Quantum Treatment; Schrodinger waveequation in polar coordinates is given as; 𝛻2 = 1 𝑟2 𝜕 𝜕𝑟 𝑟2 𝜕 𝜕𝑟 + 1 𝑟2 sin 𝜃 𝜕 𝜕𝜃 𝑠𝑖𝑛𝜃 𝜕 𝜕𝜃 + 1 𝑟2 𝑠𝑖𝑛2 𝜃 𝜕2 𝜕∅2 + 8П2µ ℎ2 𝐸 − 𝑃 𝛹 = 0 For rigid rotators r=constant And P.E =0 So; 𝛻2 = 1 sin 𝜃 𝜕 𝜕𝜃 𝑠𝑖𝑛𝜃 𝜕 𝜕𝜃 + 1 𝑠𝑖𝑛2 𝜃 𝜕2 𝜕∅2 + 8П2µ ℎ2 𝐸𝛹 = 0 As 𝐻 = −ℎ2 𝛻2 8П2µ Putting the value of 𝛻2 in above equation 𝐻 = −ℎ2 8П2µ 1 sin 𝜃 𝜕 𝜕𝜃 𝑠𝑖𝑛𝜃 𝜕 𝜕𝜃 + 1 𝑠𝑖𝑛2 𝜃 𝜕2 𝜕∅2 + 8П2µ ℎ2 𝐸𝛹
- 9. Solution of theabove equation is 𝐻 = ℎ2 8П2 𝐼 𝑙 𝑙 + 1 As 𝐸𝛹 = 𝐻𝛹 For rigid rotator 𝐸𝑌𝑙 𝑚 𝜃, 𝛷 = 𝐻𝑌𝑙 𝑚 𝜃, 𝛷 Putting value of H 𝐸𝑌𝑙 𝑚 𝜃, 𝛷 = ℎ2 8П2 𝐼 𝑙 𝑙 + 1 𝑌𝑙 𝑚 𝜃, 𝛷 𝐸 = ℎ2 8П2 𝐼 𝑙 𝑙 + 1 For rigid rotator l=j 𝐸 = ℎ2 8П2 𝐼 𝐽 𝐽 + 1 Where; J= Rotational quantum number.
- 10. As; 𝐵 = ℎ2 8П2 𝐼 B=rotationalconstant So; 𝐸𝑗 = 𝐵𝐽 𝐽 + 1 It is the relation of rotational energy for rigid rotators having fixed bond length.
- 11. ENERGY TRANSITION; By selectionrule; ∆𝐽 = ±1 ∆𝐸 = 𝐸𝐽+1 − 𝐸𝐽
- 12. Application of rigidrotator; Calculation of bond length; 𝐼 = µ𝑟2 𝑟2 = 𝐼 µ 𝑟 = 𝐼 µ Where; µ=reduced mass µ = 𝑚1𝑚2 𝑚1+𝑚2 × 1 𝑁 𝐴 For one molecule ∆𝐸 = ℎ𝑐𝑣 Or 𝐵′ = ∆𝐸 ℎ𝑐 = 𝐵 ℎ𝑐 = ℎ2 8П2 𝐼 × 1 ℎ𝑐 B’= ℎ 8П2 𝐼𝑐 𝐼 = ℎ 8П2 𝐵′𝑐