Knowledge Bank

We have found a quick and easy way to get the potential energy of a diatomic molecule from the experimental energy $E(v+12,(J+12)2)$ involving vibrational and rotational quantum numbers v and J. At each internuclear spacing r the sum of potential energy from electrons and nuclei, U(r), and centrifugal potential energy. $(J+12)2h2/2\mu r2$ is less than the total energy by the radial kinetic energy: ({J}+\frac{1}{2})^{2}{h}^{2}/2\mu{r}^{2}\leq{E}\left({v}+\frac{1}{2},({J}+\frac{1}{2})^{2}\right). (Here 2$\pi \hslash$ is Planek’s constant and $\nu$ is the reduced mass of the two atoms which make up the diatomic molecule.) Equality holds only when the radial kinetic energy is zero, corresponding to the extrapolated value $v=12$. Under these conditions, the curve $yy1(r)=E(v+12=0,p\varphi /\hslash 2)-p\varphi 2/2\mu r2$ kisses the curve $y2(r)=U(r)$ at the orbiting radius $r2$ = r; for all other values of r, $y1$ lies below $y2$. (Here $p\varphi 2$ is the square of the orbital angular momentum.) Thus $y2$ is the maximum of $y2$ at each radius, and we get the simple formula $U(r)=max(E(v+12=0,p\varphi 2/\hslash 2)-p\varphi 2/2\mu r2$ where . Physically, the square of the orbital angualar momentum is non-negative; its extension above to negative values is an analytical continuation of the experimental energy surface. This formula is right for two soluble problems, and gives results mostly correct to one or two per cent for $HBr81$.