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Discuss the vibrational spectra of simple diatomic molecules vibrating anharmonically.SpectroscopyPhysical Chemistry - Cepek Media

By Sunil Bhardwaj

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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.

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 $$1: \frac { 1 }{ 2 } : \frac { 1 }{ 3 }$$ .

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