By Sunil Bhardwaj

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The frequency separation between successive line in the rotational spectrum is given by, $$\Delta \overline { \nu } = 2B$$ where B is rotational constant $$= \frac { h }{ 8{ \pi }^{ 2 }Ic }$$ $$\therefore \Delta \overline { \nu } = 2\frac { h }{ 8{ \pi }^{ 2 }Ic } = \frac { h }{ 4{ \pi }^{ 2 }Ic }$$ $$\therefore I = \frac { h }{ 4{ \pi }^{ 2 }c } \frac { 1 }{ \Delta \overline { \nu } }$$ or I is moment of Inertia $$= \mu { r }^{ 2 }$$ $$\therefore \mu { r }^{ 2 } = \frac { h }{ 4{ \pi }^{ 2 }c } \frac { 1 }{ \Delta \overline { \nu } }$$ $$or \ { r }^{ 2 } = \frac { h }{ 4{ \pi }^{ 2 }\mu c } \frac { 1 }{ \Delta \overline { \nu } }$$ Thus, If we know the reduced mass or atomic masses of the molecule, we can find the internuclear distance r from the frequency seperation $$\Delta \overline { \nu }$$ of rotational spectrum.

Calculate the bond length of carbon monoxide molecule, if its first rotational spectrum line appears at $$3.84 \times { 10 }^{ 2 } { m }^{ -1 }$$ (C = 12, O = 16)

We have, $${ \overline { v } }_{ 1 } = 3.84 \times { 10 }^{ 2 }{ m }^{ -1 }$$ Atomic Weight of C = 12
Atomic Weight of O = 16 $$\therefore Reduced \ mass \mu = \left( \frac { { m }_{ 1 } \times { m }_{ 2 } }{ { m }_{ 1 } + { m }_{ 2 } } \right)$$ $$= \left( \frac { 12 \times 16 }{ 12 + 16 } \right) = \frac { 192 }{ 28 } = 6.857 a.m.u.$$ $$= 6.857 \times \left( 1.66 \times { 10 }^{ -27 }Kg \right)$$ $$= 11.383 \times { 10 }^{ -27 }Kg$$ Thus, $$\boxed { { r }^{ 2 } = \frac { h }{ 4{ \pi }^{ 2 }\mu c } \frac { 1 }{ \Delta \overline { v } } }$$

Let's put the values, $${ r }^{ 2 } = \frac { 6.626 \times { 10 }^{ -34 } }{ 4 { \left( 3.14 \right) }^{ 2 } \left( 11.383 \times { 10 }^{ -27 } \right) \left( 3 \times { 10 }^{ 8 } \right) } \frac { 1 }{ \left( 3.84 \times { 10 }^{ 2 }{ m }^{ -1 } \right) }$$ $$= \frac { 6.626 \times { 10 }^{ -34 } }{ 4 \left( 9.87 \right) \left( 11.383 \times { 10 }^{ -27 } \right) \left( 3 \times { 10 }^{ 8 } \right) } \frac { 1 }{ \left( 3.84 \times { 10 }^{ 2 }{ m }^{ -1 } \right) }$$ $$= \frac { 6.626 \times { 10 }^{ -34 } }{ 5.2 \times { 10 }^{ -14 } }$$ $$= 1.274 \times { 10 }^{ -20 }{ m }^{ 2 }$$ $$\therefore r =\sqrt { 1.274 \times { 10 }^{ -20 } } = 1.129 \times { 10 }^{ -10 }m$$

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