=Odpowiedź:
a0=0,5A
R(20)=(1/2a0)^(3/2)*2*(1-[r/2a0])*e^(-[r/2a0]) gdzie:
r=3,6a0 zatem:
R(20)=(1/2a0)^(3/2)*2*(1-[3,6a0/2a0])*e^(-[3,6a0/2a0])=
=[0,3536a0^(3/2)]*2*(-4/5)*e^(-1,8)=[0,3536a0^(3/2)]*(-0,743845)=
=[0,3536*0,5^(3/2)]*(-0,743845)=0,125*(-0,743845)=(-0,09298)