By Dr. Shahid Ali Yousafzai

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In spectroscopy we plot absorption or transmission with respect to different variables for example we plot in UV-vis the absorption with respect to wavelength and it is from lower wavelength to the higher wavelength. Similarly, in PL spectroscopy we plot the emission intensity with respect to wavelength and in some cases when we want to calculate the band gap from the UV-vis spectroscopy then we calculate energy and we draw a slope on it.

Why in IR spectra we need the wave number? This is a very important question, in order to understand this thing we take the equation of a harmonic oscillator because we are understanding the oscillations of atom we want to understand in IR spectra the atomic excitations inside a molecule where atoms are attached to each other like two masses attached to a spring. So we use the harmonic oscillator equation $$F=-Kx$$ where K is the spring constant. Now we can write $$F=ms=\frac { { d }^{ 2 }y }{ d{ t }^{ 2 } } =-Kx$$ From this we can derive that the $$v=\frac { 1 }{ 2\pi } \sqrt { \frac { K }{ m } }$$ Now I can change this equation by $$\frac { v }{ c } =k=\frac { 1 }{ 2\pi c } \sqrt { \frac { K }{ m } }$$then in order to convert the masses into atomic mass unit and the mass is two atoms means I will be discussing the bonds between two atoms so the mass will be replaced by the reduced mass $$k=\frac { 1 }{ 2\pi c } \sqrt { \frac { K{ N }_{ A } }{ \mu } }$$ $$\mu =\frac { { m }_{ 1 }+{ m }_{ 2 } }{ { m }_{ 1 }{ m }_{ 2 } }$$ $$=\left( 2911\sqrt { g } /cm \right) \sqrt { \frac { 1 }{ \mu } }$$ Wavenumber actually resolves the closely spaced peaks and this is the benefit of doing this.

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