# Discuss the vibrational spectra of simple diatomic molecules vibrating Harmonically.

Consider a diatomic molecule AB. Two atoms A and B are connected by an elastic spring. A is fixed at one end and B is kept hanging at their normal equilibrium distance $$r$$. When the spring is stretched by an external force $$f$$ and then released, the equilibrium distance is disturbed, and its length increases. The opposing force restores back the spring to its original equilibrium distance. Thus, the atom B performs simple harmonic motion along the axis of the bond.

When bond atoms A and B are vibrating and oscillating simultaneously then the vibrational frequency $$(\mu )$$ is equal to oscillating frequency $$(\omega )$$, is given by the relation, $$\omega = v = \frac { 1 }{ 2\pi } \sqrt { \frac { K }{ \mu } }$$

On the basis of quantum mechanical treatment and application of Schrondinger wave equation ,it is found that the energy associated with vibration of a molecule is quantised and the vibrational energy of a harmonic oscillator is given by, $${ E }_{ v } = \left( v + \frac { 1 }{ 2 } \right) h\omega$$ where $$v$$ = vibrational quantum number having values 0,1,2,…..etc. $$h$$ = Plancks constant.

For lowest energy level, $$v = 0$$ $${ E }_{ v } = \frac { 1 }{ 2 } h\omega$$ For different vibrational quantum number, corresponding energy levels will have value as

For $$v = 1$$, $${ E }_{ 1 } = \left( 1 + \frac { 1 }{ 2 } \right) h\omega \qquad { E }_{ 1 } = \frac { 3 }{ 2 } h\omega$$ For $$v = 2$$, $${ E }_{ 2 } = \left( 2 + \frac { 1 }{ 2 } \right) h\omega \qquad { E }_{ 2 } = \frac { 5 }{ 2 } h\omega$$ For $$v = 3$$, $${ E }_{ 3 } = \left( 3 + \frac { 1 }{ 2 } \right) h\omega \qquad { E }_{ 3 } = \frac { 7 }{ 2 } h\omega$$ For $$v = 4$$, $${ E }_{ 4 } = \left( 4 + \frac { 1 }{ 2 } \right) h\omega \qquad { E }_{ 4 } = \frac { 9 }{ 2 } h\omega$$ and so on.

The difference between two successive vibrational energy levels can be obtained as follows.
For the transition $$v$$ to $$v’$$ we get, $$\Delta { E }_{ vib } = { E }_{ v’ } – { E }_{ v }$$ $$= \left( v’ + \frac { 1 }{ 2 } \right) h\omega – \left( v + \frac { 1 }{ 2 } \right) h\omega$$ $$= \left( v’ + \frac { 1 }{ 2 } – v – \frac { 1 }{ 2 } \right) h\omega$$ $$= \left( v’ – v \right) h\omega$$ for the transition $$v = 1$$ to $$v = 2$$ $$\Delta { E }_{ vib } = \left( v’ – v \right) h\omega$$ $$= \left( 2 – 1 \right) h\omega = h\omega$$ $$= \frac { h }{ 2\pi } \sqrt { \frac { K }{ \mu } }$$ It can be understood from this relation that the vibrational energy levels are also equally spaced having a constant energy difference of $$h\omega$$ between any two successive energy levels and the spacing between the levels depends on the force constant $$(K)$$ and the reduced mass $$(\mu )$$.

Calculate zero point energy and force constant of a molecule whose reduced mass is $$1.2 \times { 10 }^{ -27 }$$ kg. The wave number of origin of the band in IR spectrum is $$37 \times { 10 }^{ 4 } { m }^{ -1 }$$.

Solution: We have reduced mass $$\mu = 1.2 \times { 10 }^{ -27 }$$ kg
wave number $$\overline { \upsilon } = 37 \times { 10 }^{ 4 } { m }^{ -1 }$$
Force constant K can be calculated from equation, $$\overline { \upsilon } = \frac { 1 }{ 2\pi c } \sqrt { \frac { K }{ \mu } }$$ $$\left( 37 \times { 10 }^{ 4 } { m }^{ -1 } \right) =\frac { 1 }{ 2 \times 3.14 \times \left( 3 \times { 10 }^{ 8 } m/s \right) } \sqrt { \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } }$$ $$\left( 37 \times { 10 }^{ 4 } { m }^{ -1 } \right) = \frac { 1 }{ 18.84 \times { 10 }^{ 8 } } \sqrt { \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } }$$ $$\sqrt { \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } } = \left( 37 \times { 10 }^{ 4 } { m }^{ -1 } \right) \times \left( 18.84 \times { 10 }^{ 8 } \right)$$ $$\sqrt { \frac { K }{ 1.2 \times { 10 }^{ -27 } kg } } = 6.971 \times { 10 }^{ 14 }$$ $$\frac { K }{ 1.2 \times { 10 }^{ -27 } kg } = { \left( 6.971 \times { 10 }^{ 14 } \right) }^{ 2 } = 4.859 \times { 10 }^{ 29 }$$ $$K = \left( 4.859 \times { 10 }^{ 29 } \right) \times \left( 1.2 \times { 10 }^{ -27 } \right) = 583.105 N{ m }^{ -1 }$$ Zero Point Energy is the vibrational energy at $$v = 0$$ $$\therefore { E }_{ 0 } = \frac { 1 }{ 2 } h\omega = \frac { 1 }{ 2 } h\upsilon = \frac { 1 }{ 2 } h\overline { \upsilon } c$$ $$\qquad = \frac { 1 }{ 2 } \left( 6.626 \times { 10 }^{ -34 } Js \right) \left( 37 \times { 10 }^{ 4 } { m }^{ -1 } \right) \left( 3 \times { 10 }^{ 8 } m/s \right)$$ $$\qquad = \frac { 1 }{ 2 } \left( 7.355 \times { 10 }^{ -20 } \right)$$ $$= 3.677 \times { 10 }^{ -20 } J$$

The vibrational frequency of HCl is $$2.988 \times { 10 }^{ 5 } { m }^{ -1 }$$. Calculate the zero point energy of the molecule.

Solution: We have wave number $$\overline { \upsilon } = 2.988 \times { 10 }^{ 5 } { m }^{ -1 }$$
Zero Point Energy is the vibrational energy at $$v = 0$$ $$\therefore { E }_{ 0 } = \frac { 1 }{ 2 } h\omega = \frac { 1 }{ 2 } h\upsilon = \frac { 1 }{ 2 } h\overline { \upsilon } c$$ $$\qquad = \frac { 1 }{ 2 } \left( 6.626 \times { 10 }^{ -34 } Js \right) \left( 2.988 \times { 10 }^{ 5 } { m }^{ -1 } \right) \left( 3 \times { 10 }^{ 8 } m/s \right)$$ $$\qquad = \frac { 1 }{ 2 } \left( 5.94 \times { 10 }^{ -20 } \right)$$ $$\qquad = 2.97 \times { 10 }^{ -20 } J$$

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