What is the <r> for a hydrogen atom in n=2, l=1, m=0 state?

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SUMMARY

The discussion focuses on calculating the expectation value for a hydrogen atom in the quantum state defined by n=2, l=1, and m=0. The wave function is given as ψ(r,θ,∅)=(1/4(√2pi)ab3/2)(r/ab)(e-r/2ab)(cos(θ). To find , participants emphasize the need to derive the radial function R(r) from the wave function and subsequently use it to compute the probability density P(r) = 4πr²|R(r)|². The integral for is established as = ∫₀^∞ P(r) r³ dr, incorporating the necessary factors for spherical coordinates.

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Homework Statement


Hydrogen is in n=2, l=1, and m=0.
Wave function is ψ(r,θ,∅)=(1/4(√2pi)ab3/2)(r/ab)(e-r/2ab)(cos(θ)

Find <r> for this state.


Homework Equations



P(r) = 4pir2|R(r)|2

<r> is equal to the integral from 0 to ∞ of P(r)dr

The Attempt at a Solution



I understand that you need to go from the wave function to R(r) and then P(r) to put that in the <r> equation. I am just not sure how you do the first step.
I put the equations in the picture below to better visualize it. Thanks in advance!
 

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The equation for ##P(r)## already takes into account the fact that you have integrated the spherical harmonic over ##\theta, \phi##, and this is where the factor ##4 \pi## comes from. So you would have to remove the angular part from the initial wave function.

That said, I believe that you will learn more if you don't take that ready-made equation, but calculate it from first principles:
$$
\langle r \rangle = \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{\infty} \psi^* r \psi r^2 dr \sin \theta d\theta d\phi
$$

Note that you have a factor ##r## missing in your description of the integration step, it should be
$$
\langle r \rangle = \int_{0}^{\infty} P(r) r^3 dr
$$
 

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