Wave function in a hydrogen atom : normalization

In summary, the determined wave function in a hydrogen atom is represented by ## \Psi(r,θ,Φ) = A(x+iy)e^{ \frac{-r}{2a_0}} ## and the constant A can be found by normalization, which is equal to ## A = -\frac{1}{a_0 \sqrt{8 \pi}} (\frac{1}{2a_0})^{3/2} ##. The integral of the wave function squared over all space is equal to 1, and this can be found by changing variables and using the identity ##sin^3(\theta) = \frac{4}{3} cos(\theta) - \frac{1}{3} cos(3
  • #1
Another
104
5

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
 
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  • #2
Another said:
## \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

The angle ##\theta## doesn't go from 0 to ##2\pi##, it only goes from 0 to ##\pi##. To integrate ##sin^3(\theta)##, change variables by letting ##u = cos(\theta)##.
 
  • #3
stevendaryl said:
The angle ##\theta## doesn't go from 0 to ##2\pi##, it only goes from 0 to ##\pi##. To integrate ##sin^3(\theta)##, change variables by letting ##u = cos(\theta)##.

##\int sin^3(\theta) d\theta ## from 0 to ##\pi## =## \frac{4}{3} ##

thank you very much. This is my mistake.
 

1. What is the wave function in a hydrogen atom and why is it important?

The wave function in a hydrogen atom describes the probability of finding an electron in a given location around the nucleus. It is important because it helps us understand the behavior of electrons and their energy levels in the atom.

2. How is the wave function normalized in a hydrogen atom?

The wave function is normalized by ensuring that the total probability of finding an electron in all possible locations around the nucleus is equal to 1. This means that the electron must exist somewhere in the atom at all times.

3. What is the significance of normalizing the wave function in a hydrogen atom?

Normalizing the wave function is important because it allows us to accurately determine the energy levels of the electron in the atom. It also ensures that the wave function is a valid representation of the electron's behavior.

4. How does the normalization of the wave function in a hydrogen atom relate to the Heisenberg Uncertainty Principle?

The normalization of the wave function is related to the Heisenberg Uncertainty Principle because it helps us determine the probability of finding an electron in a specific location. This is a fundamental aspect of the principle, which states that the position and momentum of a particle cannot be known simultaneously.

5. Is the wave function in a hydrogen atom always normalized?

Yes, the wave function in a hydrogen atom is always normalized as it is a requirement for a valid wave function. This ensures that the electron exists somewhere in the atom at all times and that its behavior can be accurately described by the wave function.

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