Expression for The Frequency Separation of Rotational Spectral Lines.

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, $${ E }_{ J=1 } = \frac { { \left( 6.626 \times { 10 }^{ -34 } \right) }^{ 2 } }{ 8 \times { \left( 3.14 \right) }^{ 2 } \times \left( 1.68 \times { 10 }^{ -46 } \right) } \times 1\left( 1 + 1 \right) $$ $$ E = \frac { 4.390 \times { 10 }^{ -67 } }{ 8 \times \left( 9.8596 \right) \times \left( 1.68 \times { 10 }^{ -46 } \right) } \times 2 $$ $$ E = \frac { 4.390 \times { 10 }^{ -67 } }{ 132.513 \times { 10 }^{ -46 } } \times 2 $$ $$ E = 6.626 \times { 10 }^{ -23 } J $$

Leave a Reply

© 2022 Edmerls - WordPress Theme by WPEnjoy