By Sunil Bhardwaj

6354 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 \(1: \frac { 1 }{ 2 } : \frac { 1 }{ 3 }\) .

Latest News

  • Become an Instructor 4 March, 2018

    Apply to join the passionate instructors who share their expertise and knowledge with the world. You'll collaborate with some of the industry's best producers, directors, and editors so that your content is presented in the best possible light..

More Chapters

  • Spectroscopy
  • Basic Quantum Chemistry
  • Phase Rule
  • Electrochemistry
  • Colloidal State
  • Chemical Thermodynamics
  • Gaseous State
  • Applied Electrochemistry
  • Ionic Equilibria
  • Nuclear Chemistry
  • Solid State Chemistry
  • Chemical Kinetics

    • Other Subjects

  • English
  • Applied Physics
  • Environmental Studies
  • Physical Chemistry
  • Analytical Chemistry
  • Organic Chemistry
  • Soft Skills
  • Engineering Drawing
  • General Medicine
  • Mathematics
  • Patente B Italia
  • Adult Education
  • Engineering Chemistry
  • Conceptual Physics
  • Patente (ITALIA)