The position for the absorption band for carbonyl stretching vibration (frequency = 5.1 x1013 vibrations /second) is a) 1700 cm-1 b) 1000 cm-1 c) 3000 cm-1 d) 2500 cm-1
Question
The position for the absorption band for carbonyl stretching vibration (frequency = 5.1 x1013 vibrations /second) is a) 1700 cm-1 b) 1000 cm-1 c) 3000 cm-1 d) 2500 cm-1
Solution
The position for the absorption band for carbonyl stretching vibration can be calculated using the formula for the frequency of a wave, which is:
v = c/λ
where: v is the frequency, c is the speed of light (which is approximately 3 x 10^10 cm/s), and λ is the wavelength.
We are given that the frequency (v) is 5.1 x 10^13 vibrations/second. We can rearrange the formula to solve for the wavelength (λ):
λ = c/v
Substituting the given values:
λ = (3 x 10^10 cm/s) / (5.1 x 10^13 s^-1) λ = 0.000588 cm
However, in spectroscopy, it's common to use wavenumbers (ν̃) instead of wavelength for convenience. The wavenumber is the reciprocal of the wavelength, and it's given in cm^-1. So, to find the wavenumber, we take the reciprocal of the wavelength:
ν̃ = 1/λ ν̃ = 1 / 0.000588 cm ν̃ = 1700 cm^-1
So, the position for the absorption band for carbonyl stretching vibration is approximately 1700 cm^-1. Therefore, the correct answer is a) 1700 cm^-1.
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