Calculate the net force that thymine exerts on adenine. Is it attractive or repu

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Explanation / Answer

We know the hydrogen ion (H+) has a charge of +1.6 * 10^(-19) C, the oxygen ion (O-) has a charge of 2 * (-1.6 * 10^(-19)) C and the nitrogen ion(N-) has a charge of 3 * (-1.6 * 10^(-19)) C. From given figure, the total force of attraction between O and H on the upper set is given by (considering the resultant distance between O and H from given figure and data). Foh = 9 * 10^9 * 2 * 1.6 * 10^(-19) * 1.6 * 10^(-19) / [(0.28-0.11) * 10(-9)]^2 => Foh = -1594.46 * 10^(-11) N (we included -ve sign to say that the force is attractive) Similarly, the total force of repulsion between O and N on the upper set is given by (considering the resultant distance between O and N from given figure and data). Fon = 9 * 10^9 * 2 * 1.6 * 10^(-19) * 3 * 1.6 * 10^(-19) / [(0.28) * 10(-9)]^2 => Fon = 1763.27 * 10^(-11) N Hence the resultant force due the upper set = Fo = [-1594.46 * 10^(-11)] + [1763.27 * 10^(-11)] = 168.81 * 10^(-11) N (i.e., repulsive). ---> (1) Similarly, applying the same method for the lower set, the total force of repulsion between N and N on the lower set is given by (considering the resultant distance between N and N from given figure and data). Fnn = 9 * 10^9 * 3 * 1.6 * 10^(-19) * 3 * 1.6 * 10^(-19) / [(0.3) * 10(-9)]^2 => Fnn = -2304 * 10^(-11) N the total force of attraction between H and N on the lower set is given by (considering the resultant distance between H and N from given figure and data). Fhn = 9 * 10^9 * 1.6 * 10^(-19) * 3 * 1.6 * 10^(-19) / [(0.3 - 0.11) * 10(-9)]^2 => Fhn = 1914.68 * 10^(-11) N Hence the resultant force due the lower set = Fn = [-2304 * 10^(-11)] + [1914.68 * 10^(-11)] = -389.32 * 10^(-11) N (i.e., attractive). ---> (2) Hence combining equations (1) and (2), the resultant force by Thymine on Adenine is given by F = Fo + Fn = [168.81 * 10^(-11)] + [-389.32 * 10^(-11)] = -2.2051 * 10^(-9) N Hence Thymine exerts an attractive force of -2.2051 * 10^(-9) N on Adenine.

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