# Expression for the Moment of Inertia of a diatomic molecule.

Consider a simple rigid diatomic polar molecule AB of masses $${ m }_{ 1 }$$ and $${ m }_{ 2 }$$ and separated by an inter-nuclear distance r. Let $${ r }_{ 1 }$$ and $${ r }_{ 2 }$$ be the distance of these atoms from the centre of gravity G about which the molecule rotates. Therefore $$r = { r }_{ 1 } + { r }_{ 2 }$$ Moment of inertia of a particle of mass m revolving round a fixed point at a distance $$r$$ is given by, $$I = m{ r }^{ 2 }$$ For a system of an assembly of $$i$$ particles, the total moment of inertia $$(I)$$ is given by $$I = { m }_{ 1 }{ r }_{ 1 }^{ 2 } + { m }_{ 2 }{ r }_{ 2 }^{ 2 } + { m }_{ 3 }{ r }_{ 3 }^{ 2 } + … + { m }_{ i }{ r }_{ i }^{ 2 }$$ For a diatomic molecule. $$I = { m }_{ 1 }{ r }_{ 1 }^{ 2 } + { m }_{ 2 }{ r }_{ 2 }^{ 2 }$$ OR $$I = \left( { m }_{ 1 }{ r }_{ 1 } \right) { r }_{ 1 } + \left( { m }_{ 2 }{ r }_{ 2 } \right) { r }_{ 2 } \qquad …(1)$$
As the systems is balanced about its centre of gravity $$‘G’$$ moments of both the atoms are equal. Therefore, $${ m }_{ 1 }{ r }_{ 1 } = { m }_{ 2 }{ r }_{ 2 }\qquad …(2)$$ also $${ r }_{ 2 } = \frac { { m }_{ 1 }{ r }_{ 1 } }{ { m }_{ 2 } } \qquad …(3)$$

Lets put the values from eq (2) in eq (1) $$I = \left( { m }_{ 2 }{ r }_{ 2 } \right) { r }_{ 1 } + \left( { m }_{ 1 }{ r }_{ 1 } \right) { r }_{ 2 }$$ $$I = { r }_{ 1 }{ r }_{ 2 }\left( { m }_{ 2 } + { m }_{ 1 } \right) \qquad …(4)$$ As $$r = { r }_{ 1 } + { r }_{ 2 }$$ lets put the value of $${ r }_{ 2 }$$ from eq (3) $$r = { r }_{ 1 } + \frac { { m }_{ 1 }{ r }_{ 1 } }{ { m }_{ 2 } }$$ $$r = { r }_{ 1 }\left( 1 + \frac { { m }_{ 1 } }{ { m }_{ 2 } } \right)$$ $$r = { r }_{ 1 }\left( \frac { { m }_{ 1 }+{ m }_{ 2 } }{ { m }_{ 2 } } \right)$$ $$or \ { r }_{ 1 } = \left( \frac { { m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { r }\qquad …(5)$$ Similarly, $${ r }_{ 2 } = \left( \frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { r }\qquad …(6)$$

Substituting these values in eq (4) $$I = { r }_{ 1 }{ r }_{ 2 }\left( { m }_{ 2 } + { m }_{ 1 } \right)$$ $$I = \left( \frac { { m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { r }\times \left( \frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { r }\times \left( { m }_{ 2 } + { m }_{ 1 } \right)$$ $$I = \left( \frac { { m }_{ 1 }{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { r }^{ 2 }$$ $$I = \mu { r }^{ 2 }$$ where $$\mu$$ is known as reduced mass, and $$\mu = \left( \frac { { m }_{ 1 }{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right)$$

The atomic masses of nitrogen and oxygen are 14.0067 and 15.9996 respectively. Calculate moment of inertia of nitric oxide molecule if the inter-nuclear distance is 1.151Å $$(1 a.m.u. = 1.66 X { 10 }^{ -27 } Kg)$$

We have
mass of nitrogen = 14.0067 a.m.u.
mass of oxygen = 15.9996 a.m.u.
inter-nuclear distance = 1.151Å $$= 1.151 \times { 10 }^{ -10 }m$$
$$\therefore reduced \ mass \ \mu = \left( \frac { { m }_{ 1 }{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right)$$ $$\mu = \left( \frac { 14.0067 \times 15.9996 }{ 14.0067 + 15.9996 } \right)$$ $$= \frac { 224.1016 }{ 30.0063 }$$ $$\mu = 7.468 a.m.u.$$ $$= 7.468 \times 1.66 \times { 10 }^{ -27 } Kg$$ $$\mu = 12.398 \times { 10 }^{ -27 } Kg$$Moment of Inertia is given by, $$I = \mu { r }^{ 2 }$$ $$I = \left( 12.398 \times { 10 }^{ -27 } \right) \times { \left( 1.151 \times { 10 }^{ -10 } \right) }^{ 2 }$$ $$I = \left( 12.398 \times { 10 }^{ -27 } \right) \times { \left( 1.325 \times { 10 }^{ -20 } \right) }$$ $$I = 16.425 \times { 10 }^{ -47 } Kg { m }^{ 2 }$$

© 2022 Edmerls - WordPress Theme by WPEnjoy