The magnitude of the force of attraction between a sodium ion and a chloride ion

Question

The magnitude of the force of attraction between a sodium ion and a chloride ion is F=ke^2/r^2 , where r is the separation of the centers of the ions and ke^2 = 2.31*10^-28 Nm 2 .

(a) Determine the work done by the electric force as a sodium ion is moved far away from a chloride ion. Use the equilibrium separation of the ions in a crystal, 0.236 nm as the starting separation and let the final separation go to infinity. (Hint: Draw a picture with the chlorine ion at the origin and the sodium ion moving along the x-axis.) (b) Your answer to part (a) should be negative meaning that energy must be input from another source to break the ionic bond. Compare the absolute value of your answer to the dissociation energy for NaCl, 6.82*10-19 J.

Explanation / Answer

Here ,

a) work done by electric force

work done = -change in potential energy

work done = -9*10^9 * (1.602 *10^-19)^2/(0.236 *10^-9)

work done = -9.787 *10^-19 J

the work done needed is -9.787 *10^-19 J

b)

ratio of energy = 6.82 *10^-19/(9.787 *10^-19)

ratio of energy = 0.697

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