By Sunil Bhardwaj

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The presence of isotope in a molecule changes reduced mass $$(\mu )$$ of the molecule. Thus changing Moment of Inertia (I) which is inversly proportional to the rotational constant (B). Therfore, the frequency seperation also changes as, $$\Delta \overline { v } = 2B$$ This change in Frequency seperation of rotational spectrum due to presence of an isotope is known as Isotopic effect.

Consider two isotopic molecules A and B. Let $${ I }_{ A }$$ and $${ I }_{ B }$$ be their moment of inertia and $${ \mu }_{ A }$$ and $${ \mu }_{ B }$$ be their reduced masses. Then their rotational constants can be given by, $${ B }_{ A } = \frac { h }{ 8{ \pi }^{ 2 }{ I }_{ A }c }$$ and $${ B }_{ B } = \frac { h }{ 8{ \pi }^{ 2 }{ I }_{ B }c }$$
Taking the ratio of two, $$\frac { { B }_{ A } }{ { B }_{ B } } = \frac { \left( \frac { h }{ 8{ \pi }^{ 2 }{ I }_{ A }c } \right) }{ \left( \frac { h }{ 8{ \pi }^{ 2 }{ I }_{ B }c } \right) } = \frac { { I }_{ B } }{ { I }_{ A } }$$ or $$\frac { { \Delta \overline { v } }_{ A } }{ { \Delta \overline { v } }_{ B } } = \frac { { B }_{ A } }{ { B }_{ B } } = \frac { { I }_{ B } }{ { I }_{ A } }$$
also $$\frac { { \Delta \overline { v } }_{ A } }{ { \Delta \overline { v } }_{ B } } = \frac { { B }_{ A } }{ { B }_{ B } } = \frac { { I }_{ B } }{ { I }_{ A } } = \frac { { \mu }_{ B }{ r }^{ 2 } }{ { \mu }_{ A }{ r }^{ 2 } }$$
As internuclear distance r remain the same. $$\therefore \boxed { \frac { { \Delta \overline { v } }_{ A } }{ { \Delta \overline { v } }_{ B } } = \frac { { B }_{ A } }{ { B }_{ B } } = \frac { { I }_{ B } }{ { I }_{ A } } = \frac { { \mu }_{ B } }{ { \mu }_{ A } } }$$ Thus the ratio of frquency seperation in rotationl spectrum depends on the ratio of masses of different isotopes.

The frequency separation of successive line in the rotation spectrum of $$^{ * }{ C }^{ 16 }O$$ is found to be $$3.842 \times { 10 }^{ 2 } { m }^{ -1 }$$ while that of $$^{ 13 }{ C }^{ 16 }O$$ is $$3.673 { m }^{ -1 }$$. Calculate the isotopic mass of *C.

The frequency separation of $$*{ C }^{ 16 }O = { \Delta \overline { v } }_{ A } = 3.842 \times { 10 }^{ 2 } { m }^{ -1 }$$
the frequency separation of $$^{ 13 }{ C }^{ 16 }O = { \Delta \overline { v } }_{ B } = 3.673 \times { 10 }^{ 2 } { m }^{ -1 }$$
reduced mass of $$^{ 13 }{ C }^{ 16 }O = { \mu }_{ B } = \left( \frac { 13 \times 16 }{ 13 + 16 } \right)$$ $$= \frac { 208 }{ 29 } = 7.172 a.m.u.$$ Therfore from the relation, $$\boxed { \frac { { \Delta \overline { v } }_{ A } }{ { \Delta \overline { v } }_{ B } } = \frac { { B }_{ A } }{ { B }_{ B } } = \frac { { I }_{ B } }{ { I }_{ A } } = \frac { { \mu }_{ B } }{ { \mu }_{ A } } }$$ $$\frac { { \Delta \overline { v } }_{ A } }{ { \Delta \overline { v } }_{ B } } = \frac { { \mu }_{ B } }{ { \mu }_{ A } }$$ let's substitute the values,$$\frac { 3.842 \times { 10 }^{ 2 } { m }^{ -1 } }{ 3.673 \times { 10 }^{ 2 } { m }^{ -1 } } = \frac { 7.172 a.m.u. }{ { \mu }_{ A } }$$ $$\therefore 1.046 = \frac { 7.172 a.m.u. }{ { \mu }_{ A } }$$ $$\therefore { \mu }_{ A } = \frac { 7.172 a.m.u. }{ 1.046 } = 6.857 a.m.u.$$ reduced mass of $$^{ * }{ C }^{ 16 }O$$ $${ \mu }_{ A } = \left( \frac { { m }_{ C } \times { m }_{ O } }{ { m }_{ C } + { m }_{ O } } \right)$$ $$6.857 = \left( \frac { { m }_{ C } \times 16 }{ { m }_{ C } + 16 } \right)$$ $$6.857\left( { m }_{ C } + 16 \right) = 16{ m }_{ C }$$ $$6.857{ m }_{ C } + 109.70 = 16{ m }_{ C }$$ $$16{ m }_{ C } - 6.857{ m }_{ C } = 109.70$$ $$9.143{ m }_{ C } = 109.70$$ $${ m }_{ C } = \frac { 109.70 }{ 9.143 } = 11.998 a.m.u.$$

The frequency separation of successive line in the rotation spectrum of $$^{ 1 }{ H }^{ 35 }Cl$$ is $$2 \times { 10 }^{ 3 } { m }^{ -1 }$$ while that of $$^{ * }{ H }^{ 35 }Cl$$ is $$1.089 \times { 10 }^{ 3 } { m }^{ -1 }$$. Calculate the isotopic mass of *H.

The frequency separation of $$^{ 1 }H^{ 35 }Cl = { \Delta \overline { v } }_{ A } = 2 \times { 10 }^{ 3 } { m }^{ -1 }$$
the frequency separation of $$*{ H }^{ 35 }Cl = { \Delta \overline { v } }_{ B } = 1.089 \times { 10 }^{ 3 } { m }^{ -1 }$$
reduced mass of $$^{ 1 }H^{ 35 }Cl = { \mu }_{ A }$$ $$= \left( \frac { 1 \times 35 }{ 1 + 35 } \right)$$ $$= \frac { 35 }{ 36 } = 0.972 a.m.u.$$ Therfore from the relation, $$\boxed { \frac { { \Delta \overline { v } }_{ A } }{ { \Delta \overline { v } }_{ B } } = \frac { { B }_{ A } }{ { B }_{ B } } = \frac { { I }_{ B } }{ { I }_{ A } } = \frac { { \mu }_{ B } }{ { \mu }_{ A } } }$$ $$\frac { { \Delta \overline { v } }_{ A } }{ { \Delta \overline { v } }_{ B } } = \frac { { \mu }_{ B } }{ { \mu }_{ A } }$$ let's substitute the values, $$\frac { 2 \times { 10 }^{ 3 } { m }^{ -1 } }{ 1.089 \times { 10 }^{ 3 } { m }^{ -1 } } = \frac { { \mu }_{ B } }{ 0.972 a.m.u. }$$ $$\therefore 1.837 = \frac { { \mu }_{ B } }{ 0.972 a.m.u. }$$ $$\therefore { \mu }_{ B } = 1.837 \times 0.972 a.m.u. = 1.785 a.m.u.$$ reduced mass of $$*{ H }^{ 35 }Cl$$ $$= { \mu }_{ B } = \left( \frac { { m }_{ H } \times { m }_{ Cl } }{ { m }_{ H } + { m }_{ Cl } } \right)$$ $$1.785 = \left( \frac { { m }_{ H } \times 35 }{ { m }_{ H } + 35 } \right)$$ $$1.785\left( { m }_{ H } + 35 \right) = 35{ m }_{ H }$$ $$1.785{ m }_{ H } + 62.479 = 35{ m }_{ H }$$ $$35{ m }_{ H } - 1.785{ m }_{ H } = 62.479$$ $$33.215{ m }_{ H } = 62.479$$ $${ m }_{ H } = \frac { 62.479 }{ 33.215 } = 1.881 a.m.u.$$

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