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Calculate binding energy & compare to Electrostatic Repulsion

  • 18-04-2012 9:57pm
    #1
    Smythe
    Registered Users, Registered Users 2 Posts: 434 ✭✭


    4He2 has size of ~ 2 x 10^-15 m and a mass of 4.0015u. Calculate the binding energy of this nucleus, and compare it to the electrostatic repulsion of its constituents



    For the first part I have calculated B.E. = 4.5 x 10^-12 J This agrees with my notes.

    For the second part do I use the equation E = (q1 x q2) / [4∏(ε0)r^2]

    Which would be;

    (1.6 x 10^-19)^2 / [4∏ x 8.85 x 10^-12 x (1 x 10^-15)^2]

    = 230 J

    The reason I ask is because in the notes I have the r = 1 x 10^-15 value is not squared in the same calculation therefore giving an answer of 2.3 x 10^-13 J.

    So I wasn't sure which is correct. Thanks for any advice!


Comments

  • #2
    Delphi91
    Registered Users, Registered Users 2 Posts: 1,501 ✭✭✭Delphi91


    Smythe wrote: »
    For the first part I have calculated B.E. = 4.5 x 10^-12 J This agrees with my notes.

    For the second part do I use the equation E = (q1 x q2) / [4∏(ε0)r^2]

    Which would be;

    (1.6 x 10^-19)^2 / [4∏ x 8.85 x 10^-12 x (1 x 10^-15)^2]

    = 230 J

    The reason I ask is because in the notes I have the r = 1 x 10^-15 value is not squared in the same calculation therefore giving an answer of 2.3 x 10^-13 J.

    So I wasn't sure which is correct. Thanks for any advice!

    The first answer is correct.

    The equation in which r is not squared is the equation for the potential energy between the two charged particles.


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