Question

Normalize the function Ψ = e^-r

Answer 1

$\int_0^{\infin}a^2e^{-2r}dr=1,$

$\frac{a^2}{-2}e^{-2r}|_0^{\infin}=1,$

$\frac{a^2}2(1-0)=1,$

$a=\sqrt 2,$

$\psi=\sqrt2e^{-r}.$