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## Homework Statement

Determined wave function in a hydrogen atom.

## Ψ(r,θ,Φ) = A(x+iy)e^{ \frac{-r}{2a_0}}## << find A by normalization

Answer of a question in my book is ## A = -\frac{1}{a_0 \sqrt{8 \pi}} (\frac{1}{2a_0})^{3/2} ##

## Homework Equations

## \int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = \int \int \int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) r^2 \sin (θ) drdθdΦ =1 ##

##x = r \sinθ \cos Φ ##

##y = r \sinθ \sin Φ##

##z = r \cosθ ##

## The Attempt at a Solution

r [0→∞] ,θ[-##\pi##→##\pi##] ,Φ[0→##2 \pi##][/B]

## \int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = \int \int \int e^{ \frac{-r}{a_0}} r^2(\sin^2θ \cos^2Φ + \sin^2θ \sin^2Φ) r^2 \sin θ drdθdΦ##

## \int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = \int \int \int e^{ \frac{-r}{a_0}} r^4\sin^3θ drdθdΦ##

## \int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = 2\pi \int e^{ \frac{-r}{a_0}} r^4 dr \int \sin^3θ dθ##

## \int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = 2\pi a_0^5 4! \int \sin^3θ dθ##

## \int \sin^3θ dθ = 0 ## ; θ[-##\pi##→##\pi##] because ##sin^3θ## is odd function

## \int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = 0 ## My answer is wrong , please help