Possible Explanation Why τ_(B^(pm))simτ_(B⁰) But τ_(D^(pm))sim τ_(D⁰)

Data show that \tau_{B^{\pm}}\sim\tau_{B^0}, but \tau_{D^{\pm}}\sim 2\tau_{D^0}. The naive interpretation which attributes \tau_{D^{\pm}}\sim 2\tau_{D^0} to a destructive interference between two quark diagrams for D^{\pm} decays, definitely fails in the B-case. We investigate Close and Lipkin's suggestion that the phases for producing radially excited states \psi_{2s} in the decay products of B-mesons can possess an opposite sign to the integrals for \psi_{1s} decay products. Their contributions can partially compensate each other to result in \tau_{B^{\pm}}\sim\tau_{B^0}. Since D-mesons are much lighter than B-mesons, such possibilities do not exist in D-decays.