3. Why we study rigid 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 us consider 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
)