  Using the method of QCD Sum Rules, we estimate the energy of the lowest
hybrid Charmonium state, and find it to be at the energy of the $\Psi'(2S)$
state, about 600 Mev above the $J/\Psi(1S)$ state. Since our solution is not
consistent with a pure hybrid at this energy, we conclude that the $\Psi'(2S)$
state is probably an admixed $c \bar{c}$ and hybrid $c \bar{c}g$ state. From
this conjecture we find a possible explanation of the famous $\rho-\pi$ puzzle.
