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Net Potential Energy between two Adjacent Ions

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data

    The net energy is given by:

    [tex]E_N=-\frac{A}{r}+\frac{B}{r^n}[/tex]

    where A, B , and n are constants and r is the interionic separation. Calculate E0 in terms of A, B, and n by the following procedure:

    1. find dEN/dr
    2. set this expression equal to zero and solve for r=ro
    3. substitute ro back into the original equation



    3. The attempt at a solution

    Okay, this is more or less an algebra problem that I am stuck on:

    [tex]E_N=-\frac{A}{r}+\frac{B}{r^n}[/tex]

    [tex]=-Ar^{-1}+Br^{-n}[/tex]

    [tex]\Rightarrow \frac{dE_N}{dr}=Ar^{-2}-nBr^{-n-1}[/tex]

    [tex]0=Ar^{-2}-nBr^{-n-1}[/tex]

    [tex]\Rightarrow 0=\frac{A}{r^2}-\frac{nB}{r^{n+1}}[/tex]

    [tex]\Rightarrow 0=\frac{Ar^{n+1}-nBr^2}{r^2*r^{n+1}}[/tex]

    [tex]\Rightarrow 0=Ar^{n+1}-nBr^2[/tex]

    Here is where my brain melted. Any blatant errors and/or hints?

    Hmmm delicious hints :smile:
     
  2. jcsd
  3. Jan 13, 2009 #2
    factor it out:

    [tex]0=r^2(Ar^{n-1}-nB)[/tex]

    r_0=0 is nonsense, so we have:

    [tex]r_0^{n-1}=\frac{nB}{A}[/tex]

    Now plug this back, and you are done.. :D
     
  4. Jan 13, 2009 #3
    Egads man! That was easy. Nice catch Thaakisfox

    Except that [itex]r_0^{n-1}=\frac{nB}{A}[/itex] is not what I plug back in; I still have to solve explicitly for r which means I need to take the (n-1)th root of nB/A
    right?
     
  5. Jan 13, 2009 #4
    What am I still missing here?

    If:
    [tex]r=(\frac{nB}{A})^{\frac{1}{n-1}}[/tex]

    then:

    [tex]E_0=-\frac{A}{(\frac{nB}{A})^{\frac{1}{n-1}}}+\frac{B}{(\frac{nB}{A})^{\frac{n}{n-1}}}[/tex]

    which is just silly.
     
    Last edited: Jan 14, 2009
  6. Jan 14, 2009 #5
    I am thinking that this just does not clean up any better than this; i am not sure why I assumed that it would:confused:
     
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