By Sunil Bhardwaj

3217 Views


The vibrational energy of a harmonic oscillator is given by Schrodingers relation. i.e. $$ \boxed { { E }_{ v } = \left( v + \frac { 1 }{ 2 } \right) \frac { h }{ 2\pi } \sqrt { \frac { K }{ \mu } } } \qquad ...(1) $$ For lowest energy level, \(v = 0\) $${ E }_{ v } = \frac { 1 }{ 2 } \frac { h }{ 2\pi } \sqrt { \frac { K }{ \mu } } $$ The vibrational energy associated with a molecule under going anharmonic oscillations is given by, $$ \boxed { { E }_{ v } = \left( v + \frac { 1 }{ 2 } \right) h\omega - A } \qquad ...(2) $$ The term A is related to vibrational quantum number and frequency by relation $$ A = { \left( v + \frac { 1 }{ 2 } \right) }^{ 2 }h\omega x \qquad ...(3)$$ where x is called the anharmonicity constant.
Therefore eq (2) becomes, $$ \boxed { { E }_{ v } = \left( v + \frac { 1 }{ 2 } \right) h\omega - { \left( v + \frac { 1 }{ 2 } \right) }^{ 2 }h\omega x } \qquad ...(4)$$

The energy of vibrational level \(v = 0\) can be calculated using equation (4) as,$$ { E }_{ 0 } = \left( 0 + \frac { 1 }{ 2 } \right) h\omega - { \left( 0 + \frac { 1 }{ 2 } \right) }^{ 2 }h\omega x $$ $$ { E }_{ 0 } = \left( \frac { 1 }{ 2 } \right) h\omega - { \left( \frac { 1 }{ 2 } \right) }^{ 2 }h\omega x $$ $$ \boxed { { E }_{ 0 } = \frac { 1 }{ 2 } h\omega - \frac { 1 }{ 4 } h\omega x } \qquad ...(5) $$ The energy of vibrational level \(v = 1\) can be calculated using equation (4) as, $$ { E }_{ 1 } = \left( 1 + \frac { 1 }{ 2 } \right) h\omega - { \left( 1 + \frac { 1 }{ 2 } \right) }^{ 2 }h\omega x $$ $$ { E }_{ 1 } = \left( \frac { 3 }{ 2 } \right) h\omega - { \left( \frac { 3 }{ 2 } \right) }^{ 2 }h\omega x $$ $$ \boxed { { E }_{ 1 } = \frac { 3 }{ 2 } h\omega - \frac { 9 }{ 4 } h\omega x } \qquad ...(6) $$ For transition \(v = 0\) to \(v = 1\), the frequency of spectral line is given by, $$ \Delta { E }_{ v } = { E }_{ 1 } - { E }_{ 0 } $$ $$ \Delta { E }_{ v } = \left[ \frac { 3 }{ 2 } h\omega - \frac { 9 }{ 4 } h\omega x \right] - \left[ \frac { 1 }{ 2 } h\omega - \frac { 1 }{ 4 } h\omega x \right] $$ $$ \Delta { E }_{ v } = \frac { 3 }{ 2 } h\omega - \frac { 9 }{ 4 } h\omega x - \frac { 1 }{ 2 } h\omega + \frac { 1 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = \frac { 3 }{ 2 } h\omega - \frac { 1 }{ 2 } h\omega - \frac { 9 }{ 4 } h\omega x + \frac { 1 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = \frac { 2 }{ 2 } h\omega - \frac { 8 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = h\omega - 2h\omega x $$ $$ \boxed { \Delta { E }_{ v } = h\omega \left( 1 - 2x \right) } \qquad ...(7) $$ Since energy difference is quantised. $$ \Delta { E }_{ v } = hv = hc\overline { v } $$ $$ \therefore hc\overline { v } = h\omega \left( 1 - 2x \right) $$ $$ \therefore \overline { v } = \frac { \omega }{ c } \left( 1 - 2x \right) $$ $$ \therefore \boxed { \overline { v } = \overline { \omega } \left( 1 - 2x \right) } \qquad ...(8) $$ The frequencies of spectral band produced due to transition from \(v = 0\) to \(v = 1\) is called fundamental band or first harmonic band.

Similarly, The energy of vibrational level \(v = 2\) can be calculated using equation (4) as, $$ { E }_{ 2 } = \left( 2 + \frac { 1 }{ 2 } \right) h\omega - { \left( 2 + \frac { 1 }{ 2 } \right) }^{ 2 }h\omega x $$ $$ { E }_{ 2 } = \left( \frac { 5 }{ 2 } \right) h\omega - { \left( \frac { 5 }{ 2 } \right) }^{ 2 }h\omega x $$ $$ \boxed { { E }_{ 2 } = \frac { 5 }{ 2 } h\omega - \frac { 25 }{ 4 } h\omega x } \qquad ...(9) $$ For transition \(v = 0\) to \(v = 2\), the frequency of spectral line is given by, $$ \Delta { E }_{ v } = { E }_{ 2 } - { E }_{ 0 }$$ $$ \Delta { E }_{ v } = \left[ \frac { 5 }{ 2 } h\omega - \frac { 25 }{ 4 } h\omega x \right] - \left[ \frac { 1 }{ 2 } h\omega - \frac { 1 }{ 4 } h\omega x \right] $$ $$ \Delta { E }_{ v } = \frac { 5 }{ 2 } h\omega - \frac { 25 }{ 4 } h\omega x - \frac { 1 }{ 2 } h\omega + \frac { 1 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = \frac { 5 }{ 2 } h\omega - \frac { 1 }{ 2 } h\omega - \frac { 25 }{ 4 } h\omega x + \frac { 1 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = \frac { 4 }{ 2 } h\omega - \frac { 24 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = 2h\omega - 6h\omega x $$ $$ \boxed { \Delta { E }_{ v } = 2h\omega \left( 1 - 3x \right) } \qquad ...(10) $$ Since energy difference is quantised. $$ \Delta { E }_{ v } = hv = hc\overline { v } $$ $$ \therefore hc\overline { v } = 2h\omega \left( 1 - 3x \right) $$ $$ \therefore \overline { v } = 2\frac { \omega }{ c } \left( 1 - 3x \right) $$ $$ \therefore \boxed { \overline { v } = 2\overline { \omega } \left( 1 - 3x \right) } \qquad ...(11) $$ The frequencies of spectral band produced due to transition from \(v = 0\) to \(v = 2\) is called first overtone or second harmonic band.

Also, The energy of vibrational level \(v = 3\) can be calculated using equation (4) as, $$ { E }_{ 3 } = \left( 3 + \frac { 1 }{ 2 } \right) h\omega - { \left( 3 + \frac { 1 }{ 2 } \right) }^{ 2 }h\omega x $$ $$ { E }_{ 3 } = \left( \frac { 7 }{ 2 } \right) h\omega - { \left( \frac { 7 }{ 2 } \right) }^{ 2 }h\omega x $$ $$ \boxed { { E }_{ 3 } = \frac { 7 }{ 2 } h\omega - \frac { 49 }{ 4 } h\omega x } \qquad ...(12) $$ For transition \(v = 0\) to \(v = 3\), the frequency of spectral line is given by, $$ \Delta { E }_{ v } = { E }_{ 3 } - { E }_{ 0 }$$ $$ \Delta { E }_{ v } = \left[ \frac { 7 }{ 2 } h\omega - \frac { 49 }{ 4 } h\omega x \right] - \left[ \frac { 1 }{ 2 } h\omega - \frac { 1 }{ 4 } h\omega x \right] $$ $$ \Delta { E }_{ v } = \frac { 7 }{ 2 } h\omega - \frac { 49 }{ 4 } h\omega x - \frac { 1 }{ 2 } h\omega + \frac { 1 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = \frac { 7 }{ 2 } h\omega - \frac { 1 }{ 2 } h\omega - \frac { 49 }{ 4 } h\omega x + \frac { 1 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = \frac { 6 }{ 2 } h\omega - \frac { 48 }{ 4 } h\omega x $$ $$ \Delta { E }_{ v } = 3h\omega - 12h\omega x $$ $$ \boxed { \Delta { E }_{ v } = 3h\omega \left( 1 - 4x \right) } \qquad ...(13) $$ Since energy difference is quantised. $$ \Delta { E }_{ v } = hv = hc\overline { v } $$ $$ \therefore hc\overline { v } = 3h\omega \left( 1 - 4x \right) $$ $$ \therefore \overline { v } = 3\frac { \omega }{ c } \left( 1 - 4x \right) $$ $$ \therefore \boxed { \overline { v } = 3\overline { \omega } \left( 1 - 4x \right) } \qquad \qquad \qquad ...(14)$$ The frequencies of spectral band produced due to transition from \(v = 0\) to \(v = 3\) is called second overtone or third harmonic band.

TLC

So if we take the ratio of frequencies we get, $$ { \overline { \omega } }\left( 1 - 2x \right) : 2{ \overline { \omega } }\left( 1 - 3x \right) : 3{ \overline { \omega } }\left( 1 - 4x \right) $$ As anharmonicity constant \(x\) is very small we can say, $$ \left( 1 - 2x \right) \approx \left( 1 - 2x \right) \approx \left( 1 - 2x \right) \approx 1 $$ Therefore, equation becomes, $$ { \overline { \omega } }: 2{ \overline { \omega } }: 3{ \overline { \omega } } = 1: 2: 3 $$ Thus, the frequencies of the origin of fundamental, first and second overtone bands are approximately in the ratio of 1: 2: 3 and in terms of wavelengths they are \(1: \frac { 1 }{ 2 } : \frac { 1 }{ 3 }\) .