For Quantum harmonicoscillator. let energy to bring the system from its groundto

Question

For Quantum harmonicoscillator. let energy to bring the system from its groundto the first excited vibrational state correspond to the energy ofan elergy of an electromagnetic wave with a wavelength of 3,300nm.(1nm = 10-9m). This value is quite typical for a C-Hbond stretch vibration as seen in a vibrational spectroscopyexperiment (IR or Raman spectrscopy). h= 6.626*10-34J s,c=2.998*108m/s. a) how many wave amplitude maxima and minima (fulloscillations) are there per cm for this slectromagnetic wave (i.ecalculate the wavenumber) b) Suppose you substitute hydrogen with deuterium : thesffective mass for the vibration changes from 1 amu to 2 amu. Howdoes the corresponding wavenumber of the electromagnetic wavechange? c)what is the corresponding wavelength (in nm) fir thecarbon-deuterium vibration of part (c)?    (hint : it is larger than 3,300nm) For Quantum harmonicoscillator. let energy to bring the system from its groundto the first excited vibrational state correspond to the energy ofan elergy of an electromagnetic wave with a wavelength of 3,300nm.(1nm = 10-9m). This value is quite typical for a C-Hbond stretch vibration as seen in a vibrational spectroscopyexperiment (IR or Raman spectrscopy). h= 6.626*10-34J s,c=2.998*108m/s. a) how many wave amplitude maxima and minima (fulloscillations) are there per cm for this slectromagnetic wave (i.ecalculate the wavenumber) b) Suppose you substitute hydrogen with deuterium : thesffective mass for the vibration changes from 1 amu to 2 amu. Howdoes the corresponding wavenumber of the electromagnetic wavechange? c)what is the corresponding wavelength (in nm) fir thecarbon-deuterium vibration of part (c)?    (hint : it is larger than 3,300nm)

Explanation / Answer

a) Given wave length is 3300nm      3300nm = 3300nm (10^-9m / 1nm) (1^2cm/ 1m) = 3.3*10^-4cm So wave number = 1/ = 1 / 3.3*10^-4cm = 3030.3cm^-1 -------------- (b)   Ev = (v + 1/2)(h/2) = (v +1/2)(h/2) (k / meff)^(1/2)   where , v =0,1,2,3,...... So from the above equation it is clear that as the effectivemass increases , Ev decreases which means the corresponding wavenumber increases. -------------------- (c)   Since you are not provided any forceconstant value for C-D vibration, so calculation of wave numberusing the above formula (Ev = (v + 1/2)(h/2) (k /meff)^(1/2 ) is not possible. But from standard data table it was found that, wave nuber of c-D vibration is 2120cm^-1 So wave length, = 1 / wave number                            =1 / 2120cm^-1                           = 4.7170*10^-4cm                           =  4.7170*10^-4cm(10^-2m / 1cm) (1nm / 10^-9m)                          =4.1770*10^-6 / 10^-9 nm                         = 4177.0nm 4177nm > 3300nm

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