A) \[{{R}^{-1}}\]
B) \[{{R}^{-2}}\]
C) \[{{R}^{-4}}\]
D) \[{{R}^{-6}}\]
Correct Answer: D
Solution :
\[{{E}_{dis}}=\frac{-3}{4}\frac{\alpha _{0}^{2}I}{{{(4\pi {{\varepsilon }_{0}})}^{2}}{{r}^{6}}}\] where \[{{\alpha }_{0}}\] is the polarizability, is the first ionization potential, \[{{\varepsilon }_{0}}\]is the vacuum permittivity and r is the distance of the two atoms. The potential is proportional to \[1/{{r}^{6}}\].
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