By Sunil Bhardwaj

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The rotational energy \({ E }_{ J }\) of diatomic molecule is given by Schrodinger's relation$${ E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J\left( J + 1 \right)$$ where \(h\) is Plank's constant,
\(I\) is the moment of inertia of the molecule,
and \(J\) is the rotational quantum number which can have value 0, 1, 2, 3, ... etc.

Consider a transition from a rotational energy level \(J\) to another rotational energy level \(J'\),
Therefore Energy at different levels is, $$ { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J\left( J + 1 \right)$$ $${ E }_{ J }' = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J'\left( J' + 1 \right) $$ and change in energy is given by,$$\Delta { E }_{ J } = { E }_{ J } - { E }_{ J }' $$ $$ \Delta { E }_{ J } = \left[ \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J\left( J + 1 \right) \right] - \left[ \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J'\left( J' + 1 \right) \right] $$ $$ \Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ J\left( J + 1 \right) - J'\left( J' + 1 \right) \right] \qquad ...(1) $$ But, according to the selection rule, \(\Delta J = \pm 1\) i.e. \(\Delta J = J - J' = 1\) or \(J' = J - 1\).

Substituting the value of J' in equation (1) we get, $$\Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ J\left( J + 1 \right) - \left( J - 1 \right) \left( J - 1 + 1 \right) \right] $$ $$ \Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ J\left( J + 1 \right) - J\left( J - 1 \right) \right] $$ $$ \Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ J\left( J + 1 - J + 1 \right) \right] $$ $$\boxed { \Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ 2J \right] } $$ This equation gives the energy of the absorbed radiation for rotational transition.

According to Planck's quantum theory, energy changes are quantized, \(\Delta E = h\nu = hc\overline { \nu } \)
where \(\nu)\) is frequency,
\(\overline { \nu }\) is the wave number and
\(c\) is the velocity of light. $$\therefore \Delta { E }_{ J } = hc\overline { v } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ 2J \right]$$ $$or \ \boxed { \overline { v } = \frac { h }{ 8{ \pi }^{ 2 }Ic } \left[ 2J \right] }$$ The term \(\left( \frac { h }{ 8{ \pi }^{ 2 }Ic } \right)\) is constant for a given molecule and it is called rotational constant or Bjerrum's constant (B). $$\therefore \boxed { \overline { v } = 2BJ }$$ Thus, the frequency in wave numbers of lines in the rotational spectrum corresponding to the different rotational transitions can be found using above equation.

When \(J = 0\), the rotational energy is zero and the molecule doesnot rotate at all. This is called the ground rotational state of the molecule.

When \(J = 1\), $$\overline { { v }_{ 1 } } = 2B(1) = 2B$$ This gives the frequency of the first absorption line.

When \(J = 2\), $$\overline { { v }_{ 2 } } = 2B(2) = 4B$$ This gives the frequency of the second absorption line.

When \(J = 3\), $$\overline { { v }_{ 3 } } = 2B(3) = 6B$$ This gives the frequency of the third absorption line.

For transition \(J = 1\) to \(J = 2\) $$\Delta \overline { v } = 4B - 2B = 2B$$ For transition \(J = 2\) to \(J = 3\) $$\Delta \overline { v } = 6B - 4B = 2B$$

Thus, the frequency separation between successive lines in the rotational spectrum is given by $$\boxed { \Delta \overline { v } = 2B = \frac { h }{ 4{ \pi }^{ 2 }Ic } }$$

The rotational energy \({ E }_{ J }\) of diatomic molecule is given by Schrodinger's relation $${ E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J\left( J + 1 \right)$$ where h is Plank's constant, I is the moment of inertia of the molecule, and J is the rotational quantum number which can have value 0, 1, 2, 3, ... etc.

Consider a transition from a rotational energy level J to another rotational energy level J', Therefore Energy at different levels is, \({ E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J\left( J + 1 \right) \) $$ { E }_{ J }' = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J'\left( J' + 1 \right) $$ and change in energy is given by, \(\Delta { E }_{ J } = { E }_{ J } - { E }_{ J }' \) $$ \Delta { E }_{ J } = \left[ \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J\left( J + 1 \right) \right] - \left[ \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J'\left( J' + 1 \right) \right]$$ $$\Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ J\left( J + 1 \right) - J'\left( J' + 1 \right) \right] \qquad ...(1) $$ But, according to the selection rule, \(\Delta J = \pm 1\) i.e. \(\Delta J = J - J' = 1\) or \(J' = J - 1\).

Substituting the value of J' in equation (1) we get, \( \Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ J\left( J + 1 \right) - \left( J - 1 \right) \left( J - 1 + 1 \right) \right]\) $$ \Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ J\left( J + 1 \right) - J\left( J - 1 \right) \right]$$ $$\Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ J\left( J + 1 - J + 1 \right) \right]$$ $$\boxed { \Delta { E }_{ J } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ 2J \right] }$$ This equation gives the energy of the absorbed radiation for rotational transition.

According to Planck's quantum theory, energy changes are quantized, \(\Delta E = hv = hc\overline { v } \) where \(\nu)\) is frequency, \(\overline { \nu }\) is the wave number and \(c\) is the velocity of light. $$\therefore \Delta { E }_{ J } = hc\overline { v } = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } \left[ 2J \right]$$ $$or \ \boxed { \overline { v } = \frac { h }{ 8{ \pi }^{ 2 }Ic } \left[ 2J \right] } $$ The term \(\left( \frac { h }{ 8{ \pi }^{ 2 }Ic } \right)\) is constant for a given molecule and it is called rotational constant or Bjerrum's constant (B). $$ \therefore \boxed { \overline { v } = 2BJ } $$ Thus, the frequency in wave numbers of lines in the rotational spectrum corresponding to the different rotational transitions can be found using above equation.

When \(J = 0\), the rotational energy is zero and the molecule doesnot rotate at all. This is called the ground rotational state of the molecule.

When \(J = 1\), $$\overline { { v }_{ 1 } } = 2B(1) = 2B$$ This gives the frequency of the first absorption line.

When \(J = 2\), $$ \overline { { v }_{ 2 } } = 2B(2) = 4B $$ This gives the frequency of the second absorption line.

When \(J = 3\), $$\overline { { v }_{ 3 } } = 2B(3) = 6B$$ This gives the frequency of the third absorption line.

For transition \(J = 1\) to \(J = 2\) $$\Delta \overline { v } = 4B - 2B = 2B$$

For transition \(J = 2\) to \(J = 3\) $$\Delta \overline { v } = 6B - 4B = 2B$$ Thus, the frequency separation between successive lines in the rotational spectrum is given by $$\boxed { \Delta \overline { v } = 2B = \frac { h }{ 4{ \pi }^{ 2 }Ic } } $$

Moment of inertia of the NH radical is \(1.68 \times { 10 }^{ -46 } kg \ { m }^{ 2 }\). At what frequency in the microwave region would you expect the transition J = 2 to J = 3?

We have,
Moment of Inertia \(I = 1.68 \times { 10 }^{ -46 }kg{ m }^{ 2 } \)
The transition from J = 2 to J = 3 corresponds to $$\boxed { \Delta \overline { v } = 2B = \frac { h }{ 4{ \pi }^{ 2 }Ic } } $$ Let's substitute the values, $$\Delta \overline { v } = \frac { \left( 6.626 \times { 10 }^{ -34 } \right) }{ 4 \times { \left( 3.14 \right) }^{ 2 } \times \left( 1.68 \times { 10 }^{ -46 } \right) \times \left( 3 \times { 10 }^{ 8 } \right) } $$ $$ \Delta \overline { v } = \frac { 6.626 \times { 10 }^{ -34 } }{ 4 \times \left( 9.8596 \right) \times \left( 1.68 \times { 10 }^{ -46 } \right) \times \left( 3 \times { 10 }^{ 8 } \right) } $$ $$ \Delta \overline { v } = \frac { 6.626 \times { 10 }^{ -34 } }{ 198.769536 \times { 10 }^{ -38 } } $$ $$= 3.334 \times { 10 }^{ 2 } m^{ -1 } $$ \(\therefore\) The frequency for the transition J = 2 to J = 3 is \(3.334 \times { 10 }^{ 2 } m^{ -1 }\)

The inter-nuclear distance of a diatomic molecule NO is 1.15 Ao and moment of inertia is \(1.64 \times { 10 }^{ -46 } kg { m }^{ 2 }\). Calculate the energy of first rotational energy level. (J = 1) \((h = 6.626 \times { 10 }^{ -34 } Js)\)

We have, Internuclear distance \(r = 1.15 A°\)
Moment of Inertia \(I = 1.64 \times { 10 }^{ -46 }kg{ m }^{ 2 } \)
Energy of fisrt rotation \({ E }_{ J=1 } = ?\) Energy is given by, $$\boxed { E = \frac { { h }^{ 2 } }{ 8{ \pi }^{ 2 }I } J\left( J + 1 \right) } $$ Let's put the values, $${ E }_{ J=1 } = \frac { { \left( 6.626 \times { 10 }^{ -34 } \right) }^{ 2 } }{ 8 \times { \left( 3.14 \right) }^{ 2 } \times \left( 1.68 \times { 10 }^{ -46 } \right) } \times 1\left( 1 + 1 \right) $$ $$ E = \frac { 4.390 \times { 10 }^{ -67 } }{ 8 \times \left( 9.8596 \right) \times \left( 1.68 \times { 10 }^{ -46 } \right) } \times 2 $$ $$ E = \frac { 4.390 \times { 10 }^{ -67 } }{ 132.513 \times { 10 }^{ -46 } } \times 2 $$ $$ E = 6.626 \times { 10 }^{ -23 } J $$