43. The proposed wave function is ψ=−132πaerawhere ais the Bohr radius. Substituting this into the right side of Schrödinger’s equation, our goal is to show that the result is zero. The derivative is ddrae=−−152πso rddrrae22−πand 112112122rddrrddraearaFHGIKJ+LNMOQP+LNMOQP−π. The energy of the ground state is given by Emeh40228εand the Bohr radius is given by ahmeE ea==−208ππε0,.so The potential energy is given by Uer24πε0, so 888488121 1222
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