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Expected value for potential energy (quantum) 
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#1
Nov410, 12:13 PM

P: 11

1. The problem statement, all variables and given/known data
The radial distribution factor for a 1s orbital given: R10 Calculate the expected value for potential energy of a He atom in the ground state. 2. Relevant equations i understand the integral math where I solve down to <1/r> = z/a but now, how do i use the V(r) = Z/(4(pi)E) = k (1/r) equation 3. The attempt at a solution i know Z is 2 for He and E is in coulombs and I need to end with joules, but I am just stumped on this part. 


#2
Nov410, 12:49 PM

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What do you mean by



#3
Nov410, 12:54 PM

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I am trying to find the expected potential energy. The equation I have is:
Z  x (1/r) = V(r) 4*(pi)*E Z is charge which equals 2 for Helium (number of protons) and I would sub in (2/a) for 1/r i am assuming i would use Bohr's radius here where ao= 5.291x10^10m but I don't know how to find expected potential energy after I solved the radial integral for <1/r>.. especially what is E here 


#4
Nov410, 01:14 PM

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Expected value for potential energy (quantum)
OK, you're missing some pretty basic stuff, which is why your confusion seemed so strange to me. First, the potential is
[tex]V(r) = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}[/tex] You should recognize that expression from basic from basic electromagnetics. It's not E in the expression, but ϵ_{0}, the permittivity of free space. Second, the expectation value of the potential energy is [tex]\langle V(r) \rangle = \langle \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}\rangle[/itex] If you already have <1/r>, you pretty much have the answer. 


#5
Nov410, 01:24 PM

P: 11

i am having trouble following the units. aren't you missing an e^2 term?
the units should go: V(r) = Ze^2/(4(pi)e0) * <1/r> Ze^2 = C^2 e0 = C/(V*m) 1/r = 1/m So, V(r) = C*V = J if my <1/r> is actually (2/a), i think i substitute in the following: Z= 2 (no units for Helium) a= 5.29 x 10^10 m what is eo? i have a constant that is written eo= 8.854 x 10^12 F/m and i know F is 9.648C/mol can you please walk me through the substitution with eo? 


#6
Nov410, 01:38 PM

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You're thinking of the Faraday constant F, the amount of charge in one mole of electrons (ignoring the sign), which is not the same thing as a farad, also denoted by F, the unit of capacitance. 


#7
Nov410, 01:49 PM

P: 11

thank you, i understand now, i appreciate the help



#8
Jan1811, 05:25 PM

P: 5

Can somebody set this straight for me (I have the same problem)? All you need to do is substitute in values? If so, where does the value of r come from, the expectation value for r, <r>?



#9
Jan1911, 12:17 AM

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To find <1/r>, you have to do the integral
[tex]\int \psi^*(\vec{r})\left(\frac{1}{z}\right)\psi(\vec{r})\,d^3\vec{r}[/tex] where ψ is the wavefunction for the state. 


#10
Mar3011, 05:31 AM

P: 7

I have a question about the last expression, I mean the integral for the expectation value. If we are given potential V(x) and ground state energy E_{0}, and corresponding eigenvector U_{0} , is it possible to calculate <V(x)> without knowing wavefunction?
Another question is: because V(x) is a linear operator can we assume that [tex] \int\psi^*V(x)\psi dx=\left\psi\right^2\int{V(x)dx} ? [/tex] 


#11
Mar3011, 06:34 AM

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#12
Mar3011, 07:02 AM

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#13
Mar3011, 02:04 PM

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You should start a new thread with your problem and showing your work so far.



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