# how to obtain potential curves from database?

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!

• You could try the method in this paper. sciencedirect.com/science/article/abs/pii/009784859500016L Oct 3, 2020 at 7:45
• 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?
– dpdp
Oct 4, 2020 at 7:11
• 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 Oct 4, 2020 at 7:21
• 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.
– dpdp
Oct 5, 2020 at 17:19

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. 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.