Question

The moment of inertia of a diatomic molecule about an axis passing through its center of mass and perpendicular to the line joining the two atoms will be: (= Reduced mass of the system)

• A. $\mu r^2$
• B. $\frac{\mu r^2}{2}$
• C. $0$
• D. $\frac{3}{4}\mu r^2$

Answer 1

The correct option is A $\mu r^2$

In a stable equilibrium position, the two atoms of a diatomic molecule are separated by a certain distance $r$. The distance is called intermolecular distance or bond length.

Let us imagine the diatomic molecule as a system of two tiny spheres at either end of a thin weightless rod.

Let C be the center of mass of the molecule. Let $r_1$ and $r_2$ be the distances of the two atoms from the center of mass C of the molecule.

Now $r_1+r_2=r\to \left(1\right)$

and $m_1{r}_1=m_2{r}_2\to \left(2\right)$

Where $m_1$ and $m_2$ are the masses of two atoms

From equation $\left(1\right)$

$r_2=r-r_1$

From equation $\left(2\right)$

$m_1{r}_1=m_2(r-r_1)$ or $m_1{r}_1=m_2{r}_2-m_2{r}_1$

$r_1(m_1+m_2)=m_{2}r{r}_1=\frac{m_2}{m_1+m_2}r{r}_2=\frac{m_1}{m_1+m_2}r$

Let I be moment of inertial of diatomic molecules about an axis passing through center of mass of molecule and perpendicular to bond length

$I+m_1{r}_1^2+m_2{r}_2^2$

or $I=m_1[\frac{m_2}{m_1+m_2}{r}_0{]}^2+m_2[\frac{m_1}{m_1+m_2}{r}_0{]}^2$

or $I=\frac{m_1{m}_2^2}{(m_1+m_2)}{r}^2+\frac{m_2{m}_1^2}{(m_1+m_2{)}^2}{r}^2$

or $I=\frac{m_1{m}_2(m_2+m_1)}{(m_1+m_2{)}^2}{r}^2$

or $I=\frac{m_1{m}_2}{m_1+m_2}{r}^2\frac{m_1{m}_2}{m_1+m_2}=\mu$(reduced mass of molecule)

$I=\mu r^2$

Therefore moment of inertia of a diatomic molecule about an axis passing through the center of the diatomic molecule and perpendicular to bond length is product of mass of molecule and square of bond length.