# Quantum Mechanics - Quick integral probability questionWatch

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#1
Okay so my book says (the context is electron hydrogen energy levels) the probability of finding the electron within dv=dv(r,∅,θ) is :

ψψ*r^2sinθdθd∅ [1]

where ψ is the wavefunction

And to find within dr of r is : ∫∫ψψ*r^2sinθdrdθd∅,[2] where the inner integral ranges over 0 to PI and the outter from 0 to 2PI

I don't really understand this. Why doesn't the first integral [1] for dv not require dr? As isnt the volume element given by r^2sinθdrdθd∅?

I also dont really understand why we don't need to include some integral limits corresponding to dr in [2]

If anyone can help explain things this would be greatly appreciated . Ta in advance !!
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5 years ago
#2
I guess that in the first integral it's a mistake and a dr should be there or we want to find the probability as a function of r.

In the second recall that the probability of finding a particle in some small is given by
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#3
(Original post by soup)
I guess that in the first integral it's a mistake and a dr should be there or we want to find the probability as a function of r.

In the second recall that the probability of finding a particle in some small is given by
For [2] I'm still struggling to see why we would not include limits
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