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I'm sorry if this is a really super basic question. I am looking at the potential curves of a diatomic molecule in a book , and to plot them. I google the molecule and got to the NIST website, there I get diatomic mass and constants for each state, such as:

  • Te - minimum electronic energy (cm-1)
  • ωe - vibrational constant – first term (cm-1)
  • ωexe - vibrational constant – second term (cm-1)
  • ωeye - vibrational constant – third term (cm-1)
  • Be - rotational constant in equilibrium position (cm-1)
  • αe - rotational constant – first term (cm-1)
  • γe - rotation-vibration interaction constant (cm-1)
  • De - centrifugal distortion constant (cm-1)
  • βe - rotational constant – first term, centrifugal force (cm-1)
  • re - internuclear distance (Å)
  • Trans. - observed transition(s) corresponding to electronic state

Great, I should be able to use these to plot the potential energy curves for each electronic state somehow right? there is no formula i can see that here, is it just morse potential? what happens if the state is unbound? Please help me understand how to plot the potential curves. Thanks!

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    $\begingroup$ You could try the method in this paper. sciencedirect.com/science/article/abs/pii/009784859500016L $\endgroup$
    – porphyrin
    Oct 3 '20 at 7:45
  • $\begingroup$ Im sorry, that doesn't answer the question. and that paper is from 1995. mathematica (which I dont have) is probably very different today. the paper related to RKR potentials, why is that appropriate for for this question? $\endgroup$
    – dpdp
    Oct 4 '20 at 7:11
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    $\begingroup$ The fact that it is Mathematica is unimportant. The equations for the classical turning points of the potential for the potential are given in terms of the parameters you quote. I coded this up in Python (free) quite easily. If you cannot see equations due to pay wall I will put them in an answer $\endgroup$
    – porphyrin
    Oct 4 '20 at 7:21
  • $\begingroup$ yes, I can't access that paper. Thanks for the clarification. I'd very appreciate if you can help with posting the equation in your answer. $\endgroup$
    – dpdp
    Oct 5 '20 at 17:19
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The RKR method determines the classical turning points $r_{\pm}$ of the potential energy based on knowing experimentally determined spectroscopic constants. ( Rydberg, 1931, Z. Physik. v73,376, Klein, 1932 Z. Physik. v76, 226, Rees Proc. Phys. Soc.(Lond.) 1947, v59, 998.)

The energy equation takes the usual form. Parameters all in cm$^{-1}$, distances in m, reduced mass $\mu$ in u, $h$ in cm$^{-1}$s.

$$E_v=\omega_e(v+1/2) - x_e\omega_e(v+1/2)^2 + y_e\omega_e(v+1/2)^3 + z_e\omega_e(v+1/2)^4 +\cdots$$

The vibrational quantum number $v$ should be considered as a variable that has values from $-1/2$ to whatever positive value is required.

The derivative of the rotational energy wrt. $J$ at $J=0$ is also needed,

$$B'(v)= B_e-\alpha_e(v+1/2)+\gamma_e(v+1/2)^2+\cdots$$

The function representing the turning points is

$$r_\pm= \frac{f(v)}{2}\left (\sqrt{1+\frac{4}{f(v)g(v)}}\pm 1 \right)$$

where functions $f, \, g$ are integrals. These functions 'blow up' at the upper integration limit $x \to v$ so a correction is added to prevent this. The basic equation is

$$f(v)=\frac{h}{\pi\sqrt{2\mu}}\int_{-1/2}^v\frac{dx}{ \sqrt{E(v)-E(x)}}$$

and the modified one

$$f(v)=\frac{h}{\pi\sqrt{2\mu}}\left(\int_{-1/2}^{v-\delta}\frac{dx}{ \sqrt{E(v)-E(x)} }+ 2\sqrt{\frac{\delta}{Q_v}}\right)$$

where $\delta$ is small $\approx 10^{-5}$ and

$$Q_v=\omega_e - 2x_e\omega_e(v+1/2) + 3y_e\omega_e(v+1/2)^2 + 4z_e\omega_e(v+1/2)^3 +\cdots$$

The function $g$ with correction is

$$g(v)=\frac{4\pi\sqrt{2\mu}}{h} \left( \int_{-1/2}^{v-\delta} \frac{B'(x)}{\sqrt{E(v)-E(x)} }dx+2B'(v)\sqrt{\frac{\delta}{Q_v} } \right) $$

A set of values for CO is
$\mu = 6.85841,\; \omega_e= 2169.8135,\; x_e\omega_e=13.2883,\; y_e\omega_e=0.010511,\; z_e\omega_e= 0.000057$ $ B_e= 1.9312808,\; \alpha_e= 0.0175044,\; \gamma_e= 0.000000548$.

Plot energy $E(x)$(cm$^{-1}) $ vs $r_\pm(x)$ (pm) is shown. The value at $x=-1/2$ should be $r_e=103.14$ pm the equilibrium bond length with energy of zero. As values are based on cm$^{-1}$ a scaling of $10^{14}$ is needed to go to pm.

RKr-CO

I used Jupyter notebooks (via Anaconda) and python 3 to do the calculation using built in integrator from numpy/scipy and matplotlib to draw the graph. All are free and very easy to use.

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