What is Isotopic Effect?

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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