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If, for example, you have a bond on the form:

H - X, where H is hydrogen and X is some other element.

How does the mass of X affect the stretching of the bond during UV spectroscopy? Why do heavier atoms require a less 'energetic' wave to cause them to stretch?

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  • $\begingroup$ en.wikipedia.org/wiki/Quantum_harmonic_oscillator or hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html $\endgroup$
    – orthocresol
    Sep 1, 2016 at 8:43
  • $\begingroup$ UV spectroscopy is not about bond stretching at all. $\endgroup$
    – Ivan Neretin
    Sep 1, 2016 at 9:30
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    $\begingroup$ @Ivan Neretin , but all bound electronic states do have vibrational modes and these will be affected by isotopic substitution. The internuclear separation of excited or ground state is not affected by isotopic substitution, only energy levels so the new spectrum will appear shifted in frequency and changed in intensity. $\endgroup$
    – porphyrin
    Sep 1, 2016 at 9:49
  • $\begingroup$ I didn't even see that, just assumed it was IR spec... seems like the answer below did that too. To be fair UV spec is really not discussed in terms of bond stretching. $\endgroup$
    – orthocresol
    Sep 1, 2016 at 9:52

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In a very crude approximation, every bond can be considered to be a quantum harmonic oscillator. As you might know, the energies of such an oscillator are given by

$$ E_\nu = hc\omega_e \left ( \nu+\tfrac{1}{2}\right )$$

where $h$ is Planck's constant, $c$ the speed of light and $\omega_e=\sqrt{k/\mu}$. The constant $k$ is the force constant of the bond and $\mu$ is the reduced mass $$ \mu=\frac{m_1m_2}{m_1+m_2}$$ So if you increase the mass of one of the particles, while $k$ stays roughly the same, $\omega_e$ becomes smaller and the energy spacing between the vibrational levels decreases.

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  • $\begingroup$ Isotopic substitution effects both ground and excited state vibrational frequencies as described above. However, the effect is very small for all but H to D substitution as can be seem from the reduced mass equation if $m_1 = m_H = 1$, then $\mu=m_2/(1+m_2)$ which is approx 1 unless $m_2$ changes drastically on isotopic substitution. Secondly, the force constant remains $exactly$ the same on isotopic substitution; it is defined in that way since $\mu$ is separated out in the equation for frequency $\omega _e$. $\endgroup$
    – porphyrin
    Sep 1, 2016 at 9:53
  • $\begingroup$ Actually, the fact that the force constant is the same for different isotopologues comes not from the definition but is a consequence of the Born-Oppenheimer Approximation: the nuclei are all assumed to be of infinite mass so the BO curves for H$_2$ and D$_2$ are the same to a good approximation. One can take the finit mass of the nuclei into account and derive adiabatic correction to the BO potential and even interactions between the different potentials to obtain the nonadiabatic corrections. $\endgroup$
    – Paul
    Sep 1, 2016 at 11:46
  • $\begingroup$ The $E_v$ is a consequence of the BO approx. We write down the harmonic osc. energy using a $k$ that does not depend on mass, hence I wrote 'by definition'. In fact in the BO approx the nuclear and electronic motion is separated to allow the Schrodinger eqn. to be solved. It is not necessary to assume infinite mass, the requirements are variously stated as $(m_e/m_a)^{1/4}<<1$ and the terms left by making this approx are of the order of $(m_e/m_a)E_e$ . Considering that relative electron to nuclear mass are typically 1:10000 this means an error of about 1 in $10^4$ to $10^5$ cm$^{-1}$. $\endgroup$
    – porphyrin
    Sep 2, 2016 at 21:00

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