Question #128099

At t = 0 the hydrogen atom is in the superposition state Ψ(𝑟⃗, 0) =1√3𝜓100 + 𝐴𝜓210 where A is a real positive constant. Find A by normalization


1
Expert's answer
2020-08-02T15:04:32-0400

Given wave function is

ψ(r,0)=13ψ100+Aψ210\psi(r,0) = \frac{1}{\sqrt {3}} \psi _{100}+ A \psi_{210}

According to Normalization condition,

ψ(r,0)2dr=1\int |\psi(r,0)|^2 dr=1

13ψ100+Aψ2102dr=1\int |\frac{1}{\sqrt {3}} \psi _{100}+ A \psi_{210}|^2 dr=1

13ψ1002dr+A2ψ2102dr=1\frac{1}{3}\int |\psi _{100}|^2 dr + A^2 \int |\psi _{210}|^2 dr=1

13+A2=1\frac{1}{3}+A^2=1

Since ψ1002dr=1and\int |\psi _{100}|^2 dr =1 and ψ2102dr=1\int |\psi _{210}|^2 dr=1

On simplifying above equation, we get

A=23A=\sqrt\frac{2}{3}


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