Normalize the function Ψ = e^-r
∫0∞a2e−2rdr=1,\int_0^{\infin}a^2e^{-2r}dr=1,∫0∞a2e−2rdr=1,
a2−2e−2r∣0∞=1,\frac{a^2}{-2}e^{-2r}|_0^{\infin}=1,−2a2e−2r∣0∞=1,
a22(1−0)=1,\frac{a^2}2(1-0)=1,2a2(1−0)=1,
a=2,a=\sqrt 2,a=2,
ψ=2e−r.\psi=\sqrt2e^{-r}.ψ=2e−r.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
