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Topic: Isotope Abundance Determination Via FTIR  (Read 2353 times)

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Offline MrHappy0

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Isotope Abundance Determination Via FTIR
« on: April 15, 2014, 11:22:24 PM »
I'm doing a very fundamental lab exercise where I have to determine isotope abundance for HCl (H35Cl, and H37Cl). When looking at the FTIR spectra fundamental peak region there are doublet splitting patterns for each peak (high res). Consistently for each doublet, the first peak is always smaller which is from a specific isotope of HCl and I'm assuming this is H37Cl due to it's lower natural abundance of 24.5%.

My question lies in the physical chemistry behind this occurrence: why does the heavier isotope have a lower energy absorption peak? ???

Offline mjc123

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Re: Isotope Abundance Determination Via FTIR
« Reply #1 on: April 16, 2014, 06:08:32 AM »
The frequency is proportional to 1/sqrt(μ), where μ is the reduced mass - for a diatomic, μ = m1m2/(m1 + m2). So heavier atoms means lower frequency.
Looking at it classically, mass is inertia - a heavier atom responds more sluggishly to a force. The force constant, which depends on the electron distribution, is virtually identical for the two molecules, so the heavier one will vibrate with lower frequency.
Or think about it in terms of energy. An oscillator is constantly interconverting potential and kinetic energy. The maximum potential energy (at the extrema of the oscillation) is equal to the maximum kinetic energy (at the equilibrium point). A heavier oscillator undergoing an oscillation of the same amplitude and frequency as a lighter one would have the same maximum PE but a higher maximum KE, which is not possible. It must slow down, i.e. oscillate at lower frequency.

Offline Enthalpy

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Re: Isotope Abundance Determination Via FTIR
« Reply #2 on: April 16, 2014, 05:18:48 PM »
In case the question was: why is the absorption weaker at the peak of smaller frequency, then it's because there are fewer 37Cl atoms than 35Cl.

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