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1. Determine the ratio of populations in the lowest and first excited energy levels, Po/P1, for...

1. Determine the ratio of populations in the lowest and first excited energy levels, Po/P1, for the indicated systems at two temperatures, 50 K and 298 K: a) The 1700 cm−1 stretching mode for the carbonyl group in the peptide bond of a polypeptide b) the 150 cm−1 torsional mode for the C=C bond in trans-2-butadiene

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Expert Solution

Ans :

a) Here wavenumber = 1700cm-1

speed of light = 2.9x1010 cm/s

Now harmonic frequency = (2.9x1010 cm/s)(1700 cm-1)

                                          =4.93x1013 /sec

At 50 K , the population ratio

  

       

Population ratio at 298K

b) Here wavenumber = 150cm-1

speed of light = 2.9x1010 cm/s

Now, harmonic frequency =(2.9x1010 cm/s)(150 /cm)

                                          =4.35x1012 /sec

At 50K , the population ratio

At 298K , the population ratio

queen_honey_blossom answered 2 years ago
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