- #1

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Ive determined the potential energy operator to be V=-e

^{2}/4∏ε

_{0}r

and a wave function of

ψ= (1/4∏)

^{1/2}

therefore i get

<V> = ∫∫∫ψ*Vψr

^{2}sin∅drd∅dphi

integrals from 0 to r, 0 to pi, 0 to 2pi

not sure where to go from here.

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- Thread starter a.11en
- Start date

- #1

- 2

- 0

Ive determined the potential energy operator to be V=-e

and a wave function of

ψ= (1/4∏)

therefore i get

<V> = ∫∫∫ψ*Vψr

integrals from 0 to r, 0 to pi, 0 to 2pi

not sure where to go from here.

- #2

- 2

- 0

but if anyone is interested ill explain.

For a 1s orbital of a hydrogen the wavefunction is ψ=root(1/∏ao

this gives

∫∫∫ψVψr

integrals are zero to 2pi, zero to pi, and zero to infinity.

then factor out any terms that are not a function of r,θ, or∅.

This gives several terms outside of the integral: ∫∫∫e

then you can separate the integrals and evaluate. they were pretty easy to do.

the final answer was <v> = -e

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