Question

Prove that the normalization constant of the $1s$radial wave function of the hydrogen atom is localid="1650358688008" ${\left(\pi {a_B}^3\right)}^{-1/2}$, as given in Equations 41.7.

Short answer

The normalization constant of the $1s$radial wave function of the hydrogen atom is$A_{1s}=$${\left(\pi {a_B}^3\right)}^{-1/2}$

Step-by-step solution

Given information

We have to prove that the normalization constant of the $1s$radial wave function of the hydrogen atom is${\left(\pi {a_B}^3\right)}^{-1/2}$.

Simplification

The normalization condition for the three-dimensional hydrogen atom is

${\int }_0^{\infty }{P}_r\left(r\right)dr=4\pi {\int }_0^{\infty }{r}^2{\left|R_{nl}\left(r\right)\right|}^{2}dr=1$($P_r$= probability density , $R_{nl}$= radial wave function ,$a_B$= Bohr radius)

From Equation 41.7, the $1s$radial wave function can be written $R_{1s}\left(r\right)=A_{1s}{e}^{-r/a_B}$. Thus,

$1=4\pi {A^2}_{1s}{\int }_0^{\infty }{r}^2{e}^{-2r/a_B}dr=4\pi {A^2}_{1s}\frac{2!}{{\left(2/a_B\right)}^3}=\pi {a^3}_B{A^2}_{1s}$

$\Rightarrow A_{1s}={\left(\pi {a^3}_B\right)}^{-1/2}$

Hence proved.