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, \(I = \mu { r }^{ 2 }\) $$I = \left( 12.398 \times { 10 }^{ -27 } \right) \times { \left( 1.151 \times { 10 }^{ -10 } \right) }^{ 2 }$$ $$I = \left( 12.398 \times { 10 }^{ -27 } \right) \times { \left( 1.325 \times { 10 }^{ -20 } \right) }$$ $$I = 16.425 \times { 10 }^{ -47 } Kg { m }^{ 2 }$$

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