7716 Views
5406 Views
5674 Views
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, \(I = \mu { r }^{ 2 }\) $$I = \left( 12.398 \times { 10 }^{ -27 } \right) \times { \left( 1.151 \times { 10 }^{ -10 } \right) }^{ 2 }$$ $$I = \left( 12.398 \times { 10 }^{ -27 } \right) \times { \left( 1.325 \times { 10 }^{ -20 } \right) }$$ $$I = 16.425 \times { 10 }^{ -47 } Kg { m }^{ 2 }$$
5704 Views
6508 Views
5830 Views
7085 Views
5026 Views
Apply to join the passionate instructors who share their expertise and knowledge with the world. You'll collaborate with some of the industry's best producers, directors, and editors so that your content is presented in the best possible light..