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Kinetic Energy and Uncertainty Principle

  1. Mar 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A proton is confined in a uranium nucleus of radius 7.43 fm. Determine the proton’s minimum kinetic energy K ≥ ∆K according to the uncertainty principle if the proton is well approximated by a Gaussian wave packet confined by the nuclear diameter.

    2. Relevant equations

    [tex]K_{min}\geq\frac{\hbar^2}{2ml^2}[/tex]

    [tex]l=2*7.43fm=14.86*10^-15m[/tex]

    [tex]m=m_{p}=1.6726*10^-27kg[/tex]

    [tex]\hbar=1.0546*10^-34J*s[/tex]


    3. The attempt at a solution

    I think my problem is that i might be using incorrect equations or i'm messing up with units somewhere. I put in the values above into the first equation and got

    [tex]K_{min}\geq\frac{\hbar^2}{2ml^2}=\frac{(1.0546*10^-34J*s)^2}{2*1.6726*10^-27kg*(14.86*10^-15m)^2}=1.5056*10^-14J[/tex]

    and my web homework program is telling me the answer is incorrect. can anyone lead me in the correct direction please?
     
    Last edited: Mar 26, 2007
  2. jcsd
  3. Mar 26, 2007 #2

    Dick

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    The uncertainty principle says delta(x)*delta(p)>=hbar/2. Are you missing that factor of 2?
     
  4. Mar 26, 2007 #3

    Mentz114

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    There's something wrong with the powers of ten. I make the result about 10^-11, not 10^-14. i.e. -68 + 27 + 30 = -11.

    [Edited : OK, you could be in the right ballpark ...]
     
    Last edited: Mar 26, 2007
  5. Mar 26, 2007 #4
    the equation was provided in my book and before i found it in the book i found the relationship between momentum and kinetic energy and put energy in for momentum in delta(x)delta(p)>=hbar/2 and solved for energy and got the same result

    i think that's because the answer is one digit followed by a decimal with powers of ten, but some of the numbers have two digits in my equation and are multiplied by 2, etc.
    you might be right though, if anyone wants to check my work more thoroughly... i quite often make simple mistakes with powers ot ten, unit conversions, etc.
     
  6. Mar 26, 2007 #5

    Dick

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    Your formula for K_min does not contain that factor of two.
     
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