This problem is similar to Problem 1.4 except that the externally applied force

Question

This problem is similar to Problem 1.4 except that the externally applied force comes from a spring (force proportional to spring deflection or displacement) rather than remaining constant (independent of spring displacement or position). Consider the approach of two atoms where, as illustrated in Figure 1.5, the lower is part of a solid surface and the other is at the end of a fine tip that is slowly brought down vertically. We may model this system as if the top atom is suspended from the end of a spring of effective stiffness K. If K = 0.1 N/m, calculate the value of r at which an instability occurs and the tip “jumps” into contact with the surface. Will another instability—and an outward “jump”—occur
when the surfaces are separated? In reality, will the top atom or the whole tip
tend to move sideways during the approach and separation process? Assume that
the atoms interact with the same Lennard-Jones potential as in Problem 1.4 or
Worked Example 1.2.
[Hint: As in the case of Problem 1.4, to understand this problem fully, it is
instructive to be able to solve it graphically as well as numerically. First, plot the
force F(r) against r. Next, find at what point (or points) on the curve the slope is
þ0.1 N/m. Draw a line through the point that has this slope, and find where it
cuts the curve again at another point. If you think about it, these two points on
the line give the start and end points of a “jump.” Alternatively, this problem can
also be solved by considering the full energy-distance plots.]

For full text book see: http://www.iust.ac.ir/files/fnst/ssadeghzadeh_52bb7/files/Intermolecular_and_Surface_Forces.pdf

Explanation / Answer

As, Force , F = 1/2kx2

=> 8.98 * 109 * (1.6 * 10-19)2/(1 * 10-10)2 = 1/2 * 0.1 * x2

=> x = 6.78 * 10-4 m

Thus, value of r at which an instability occurs = 6.78 * 10-4 m

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