ch08-p132

ch08-p132 - E 1 intersects the potential energy curve at a...

This preview shows page 1. Sign up to view the full content.

(d) With M > > m , the kinetic energy is essentially just that of m . Since E = 1 × 10 –19 J, its kinetic energy is K = E U 2.1 × 10 –19 J. (e) Since force is related to the slope of the curve, we must (crudely) estimate F ≈× 11 0 9 N at this point. The sign of the slope is positive, so by Eq. 8-20, the force is negative-valued. This is interpreted to mean that the atoms are attracted to each other. (f) Recalling our remarks in the previous part, we see that the sign of F is positive (meaning it's repulsive) for r < 0.2 nm. (g) And the sign of F is negative (attractive) for r > 0.2 nm. (h) At r = 0.2 nm, the slope (hence, F ) vanishes. 132. The style of reasoning used here is presented in §8-5.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: E 1 intersects the potential energy curve at a value of r ≈ 0.07 nm and seems not to intersect the curve at larger r (though this is somewhat unclear since U ( r ) is graphed only up to r = 0.4 nm). Thus, if m were propelled towards M from large r with energy E 1 it would “turn around” at 0.07 nm and head back in the direction from which it came. (b) The line representing E 2 has two intersection points r 1 ≈ 0.16 nm and r 2 ≈ 0.28 nm with the U ( r ) plot. Thus, if m starts in the region r 1 < r < r 2 with energy E 2 it will bounce back and forth between these two points, presumably forever. (c) At r = 0.3 nm, the potential energy is roughly U = –1.1 × 10 –19 J....
View Full Document

This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online