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Hydrogen bonding force problem

  1. Jan 19, 2015 #1
    1. The problem statement, all variables and given/known data

    In a DNA molecule, the base pair adenine and thymine is held together by two hydrogen bonds (see figure below).

    16-figure-05.gif

    Let's model one of these hydrogen bonds as four point charges arranged along a straight line. Using the information in the figure below, calculate the magnitude of the net electric force along this hydrogen bond.

    16-p-019.gif

    2. Relevant equations

    F=(k*|q1|*|q2|)/r^2

    3. The attempt at a solution

    Since we are looking at 4 point charges, we need 4 equations. They are:

    k = 8.99x10^9 (N*m^2)/C^2
    N = -3x10^-10 C
    O = -4x10^-10 C
    H = 3x10^-10 C
    C = 4x10^-10 C

    F_no = (k*|N|*|O|) / ((3x10^-10 m)^2) = -1.20x10^10 N -> negative as repelling force
    F_nc = (k*|N|*|C|) / ((4.2x10^-10 m)^2) = 6.12x10^9 N
    F_ho = (k*|H|*|O|) / ((1.8x10^-10 m)^2) = 3.33x10^10 N
    F_hc = (k*|H|*|C|) / ((3x10^-10 m)^2) = -1.2x10^10 N -> negative as repelling force

    Since we are asked for find the magnitude of the net electric force, we need to sum all 4 forces together.

    ΣF = F_no + F_nc + F_ho + F_hc = 1.54x10^10 N

    .... Opps. So I must have done something wrong.

    The hint given was "You need 4 eqns: pair N&O, N&C, H&O, H&C and use the charges & distances appropriate for each pair to get a net magnitude between the bases." ... which to my knowledge I have here.

    So not sure what I did wrong. Any help would be much appreciated.
     
  2. jcsd
  3. Jan 19, 2015 #2

    TSny

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    Check these numbers. What does "e" stand for in the picture?
     
  4. Jan 19, 2015 #3

    haruspex

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    I'm not following your arithmetic.
    It would be clearer to leave out k and the unit charge of an electron as common factors to bring in later. E.g. for N::O write -0.3*0.4/(0.12+0.18)2.
     
  5. Jan 19, 2015 #4
    Ah! Small but critical oversight on my part. "e" meant electron ... so that each base pair is a factor .3 or .4 charge of an electron.

    This would make:

    N = (-.3)(1.602x10^-19 C) = -4.806x10^-20 C
    H = (.3)(1.602x10^-19 C) = 4.806x10^-20 C
    O = (-.4)(1.602x10^-19 C) = -6.408x10^-20 C
    C = (.4)(1.602x10^-19 C) = 6.408x10^-20 C

    Using these new values for N, H, O, C I get

    F_no = -3.08x10^-10 N -> repelling force
    F_nc = 1.57x10^-10 N
    F_ho = 8.55x10^-10 N
    F_hc = -3.08x10^-10 N -> repelling force

    Such that,

    ΣF = 2*(-3.08x10^-10 N) + 1.57x10^-10 N + 8.55x10^-10 N = 3.99x10^-10 N which ended up be the correct answer.

    Thanks a lot.
     
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