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Data show that τB± ∼ τB0 , but τD± ∼ 2τD0 . The naive interpretation which attributes τD± ∼ 2τD0 to a destructive interference between two quark diagrams for D decays, definitely fails in the B-case. We investigate Close and Lipkin’s suggestion that the phases for producing radially excited states ψ2s in the decay products of B-mesons can possess an opposite sign to the integrals for ψ1s decay products. Their contributions can partially compensate each other to result in τB± ∼ τB0 . Since D-mesons are much lighter than B-mesons, such possibilities do not exist in D-decays. PACS numbers: 13.20.Fc, 13.20.He, 12.39.-x, 13.25.-k A Possible Explanation Why τB± ∼ τB0 But τD± ∼ 2τD0