Hydrogen bonding force problem

  • #1
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Homework Statement



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


Homework Equations



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

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.
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Answers and Replies

  • #2
TSny
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N = -3x10^-10 C
O = -4x10^-10 C
H = 3x10^-10 C
C = 4x10^-10 C
Check these numbers. What does "e" stand for in the picture?
 
  • #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.
 
  • #4
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Check these numbers. What does "e" stand for in the picture?
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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