Derive\quad an\quad expression\quad for\quad the\quad Moment\quad of\quad Inertia\\ of\quad a\quad diatomic\quad molecule.\\ \\ Consider\quad a\quad simple\quad rigid\quad diatomic\quad polar\quad molecule\quad AB\\ of\quad masses\quad { m }_{ 1 }\quad and\quad { m }_{ 2 }\quad and\quad separated\quad by\quad an\\ inter-nuclear\quad distance\quad r.\\ Let\quad { r }_{ 1 }\quad and\quad { r }_{ 2 }\quad be\quad the\quad distance\quad of\quad these\quad atoms\\ from\quad the\quad centre\quad of\quad gravity\quad 'G'\quad about\quad which\quad the\\ molecule\quad rotates.\quad \\ Thereforer\quad r\quad =\quad { r }_{ 1 }\quad +\quad { r }_{ 2 }\\ \\ Moment\quad of\quad inertia\quad of\quad a\quad particle\quad of\quad mass\quad m\quad \\ revolving\quad round\quad a\quad fixed\quad point\quad at\quad a\quad distance\quad 'r'\\ is\quad given\quad by,\\ I\quad =\quad m{ r }^{ 2 }\\ For\quad a\quad system\quad of\quad an\quad assembly\quad of\quad 'i'\quad particles,\\ the\quad total\quad moment\quad of\quad inertia\quad (I)\quad is\quad given\quad by\\ I\quad =\quad { m }_{ 1 }{ r }_{ 1 }^{ 2 }\quad +\quad { m }_{ 2 }{ r }_{ 2 }^{ 2 }\quad +\quad { m }_{ 3 }{ r }_{ 3 }^{ 2 }\quad +\quad ...\quad +\quad { m }_{ i }{ r }_{ i }^{ 2 }\\ For\quad a\quad diatomic\quad molecule.\\ I\quad =\quad { m }_{ 1 }{ r }_{ 1 }^{ 2 }\quad +\quad { m }_{ 2 }{ r }_{ 2 }^{ 2 }\\ OR\quad \\ I\quad =\quad \left( { m }_{ 1 }{ r }_{ 1 } \right) { r }_{ 1 }\quad +\quad \left( { m }_{ 2 }{ r }_{ 2 } \right) { r }_{ 2 }\qquad \qquad \qquad \qquad ...(1)\\ As\quad the\quad systems\quad is\quad balanced\quad about\quad its\quad centre\quad of\\ gravity\quad 'G'\quad moments\quad of\quad both\quad the\quad atoms\quad are\quad equal.\\ Therefore,\\ { m }_{ 1 }{ r }_{ 1 }\quad =\quad { m }_{ 2 }{ r }_{ 2 }\qquad \qquad \qquad \qquad ...(2)\\ also\quad { r }_{ 2 }\quad =\quad \frac { { m }_{ 1 }{ r }_{ 1 } }{ { m }_{ 2 } } \qquad \qquad \qquad \qquad ...(3)\\ Lets\quad put\quad the\quad values\quad from\quad eq\quad (2)\quad in\quad eq\quad (1)\\ I\quad =\quad \left( { m }_{ 2 }{ r }_{ 2 } \right) { r }_{ 1 }\quad +\quad \left( { m }_{ 1 }{ r }_{ 1 } \right) { r }_{ 2 }\\ I\quad =\quad { r }_{ 1 }{ r }_{ 2 }\left( { m }_{ 2 }\quad +\quad { m }_{ 1 } \right) \qquad \qquad \qquad \qquad ...(4)\\ \\ \\ As\quad r\quad =\quad { r }_{ 1 }\quad +\quad { r }_{ 2 }\\ lets\quad put\quad the\quad value\quad of\quad { r }_{ 2 }\quad from\quad eq\quad (3)\\ r\quad =\quad { r }_{ 1 }\quad +\quad \frac { { m }_{ 1 }{ r }_{ 1 } }{ { m }_{ 2 } } \\ r\quad =\quad { r }_{ 1 }\left( 1\quad +\quad \frac { { m }_{ 1 } }{ { m }_{ 2 } } \right) \\ r\quad =\quad { r }_{ 1 }\left( \frac { { m }_{ 1 }+{ m }_{ 2 } }{ { m }_{ 2 } } \right) \\ or\quad { r }_{ 1 }\quad =\quad \left( \frac { { m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { r }\qquad \qquad \qquad \qquad ...(5)\\ Similarly,\quad { r }_{ 2 }\quad =\quad \left( \frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { r }\qquad \qquad \qquad \qquad ...(6)\\ Substituting\quad these\quad values\quad in\quad eq\quad (4)\\ I\quad =\quad { r }_{ 1 }{ r }_{ 2 }\left( { m }_{ 2 }\quad +\quad { m }_{ 1 } \right) \\ I\quad =\quad \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 }\quad +\quad { m }_{ 1 } \right) \\ I\quad =\quad \left( \frac { { m }_{ 1 }{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { r }^{ 2 }\\ I\quad =\quad \mu { r }^{ 2 }\\ where\quad \mu \quad is\quad known\quad as\quad reduced\quad mass,\quad and\quad \\ \mu \quad =\quad \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 1.151A° (1 a.m.u. = 1.66 X 10-27 Kg)

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