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Nuclear binding energies

  1. Jan 8, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations
    Given in the question


    3. The attempt at a solution
    Using the numbers and equation given in the question, I got a binding energy of Thorium of 1743.64MeV, and a binding energy of Radon (?) of 1720.88MeV

    I've come across a number of this style question before (most notably in the last piece of coursework I handed in before Christmas and haven't gotten back to check feedback for), and I always approach the question in the same way.

    First I plug the various numbers into the equation to get the binding energies of the atoms. Then I say that the binding energy of Th is equal to that of Ra and the alpha particle, and the difference between the two will be the energy released in the decay. However when I set about the question this way I always get a negative value of Q, and I can't see why that is. Surely if the question states that energy is released I'm expecting a positive answer? I seem to remember a peer saying something about why its negative when we were doing the coursework I previously mentioned, but I didn't catch it properly and I can't remember what was said anyway.

    If any of you could explain this to me (in fairly simple terms) then I would appreciate it. If the answer is not supposed to be negative then if you could point me in the direction of the correct answer then that would be appreciated as well.

    Thanks in advance for any help.

    Brewer
     
  2. jcsd
  3. Jan 8, 2008 #2

    malawi_glenn

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    Can you show us how you calculated your Q-value? So we can say what you did wrong or if you did right.

    But I can say that your Q.values look okay, by comparing to experimental ones.

    The Q-value is defined as: Binding energy final - binding energy initial, I think your misstakes lies there.
     
    Last edited: Jan 8, 2008
  4. Jan 8, 2008 #3
    That makes sense I think. I think I was just doing it backwards. I was saying that:

    B(Th) = B(Ra) + B(alpha) + Q, which I can now see is the reverse of what you said. Can you explain why this is like this, or is it just one of those things thats defined like that, and so is that with no questions asked (much like how I perceive the delta function!)

    Thank you for your help though.
     
  5. Jan 8, 2008 #4
    As a result that leaves me with an answer of Q = 5.54MeV. Which is just the modulus of the answer I was getting before.
     
  6. Jan 8, 2008 #5

    malawi_glenn

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    Well the Q-value is defined as the energy relased in a process.

    [tex] Q = \sum M_{initial} - \sum M_{final} [/tex]

    And we know that Mass = energy, if the mass after the reaction is less then initial mass: energy is realseased (Q is positive).

    Now the the binding energy is defined as the difference between the total mass of all parts and the total nucleus (Z,A):
    [tex] B(Z,A) = Z\cdot m_p + (A-Z)\cdot m_n - M(Z,A) [/tex]

    In the reaction for the Q value you just put this into it and viola:
    let the reaction be (Z,A) to (Z-2,A-4) + (2,4)

    [tex] Q = M(Z,A) -(M(Z-2,A-4) + M(2.4)) = Z\cdot m_p + (A-Z)\cdot m_n -B(Z,A) - ((Z-2)\cdot m_p + (A-4-(Z-2))\cdot m_n -B(Z-2,A-4) + [/tex]
    [tex] 2\cdot m_p + 2\cdot m_n - B(2,4)) = \sum B_{final} - \sum B_{initial} [/tex]
     
    Last edited: Jan 8, 2008
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