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Chapter 7: Problem 68

The force between the deuterons is zero at approximately a. \(3 \mathrm{fm}\) b. \(4 \mathrm{fm}\) c. \(5 \mathrm{fm}\) d. the force is never zero.

Short Answer

Expert verified
The answer is, therefore, option b. \(4 \, \mathrm{fm}\)

Step by step solution

01

Understanding the question

This physics problem asks you to determine the point at which the attractive nuclear force between two deuterons becomes equal to the repulsive electromagnetic force. This distance is the point where the overall force becomes zero.
02

Analysis of Mattauch's isobar rule

According to Mattauch's isobar rule, the repulsion due to electrostatic force and the attraction due to nuclear force both act between the two nucleons. They balance each other at a distance of approximately \(2 \, \mathrm{fm}\). Since a deuteron consists of two nucleons (proton and neutron), the distance should be doubled, which is \(4 \, \mathrm{fm}\). This is because the neutron in the deuteron doesn't contribute to the electric force, but only to the nuclear force.
03

Conclusion

Han therefore, the distance at which the force becomes zero between the two deuterons is about \(4 \, \mathrm{fm}\).

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Most popular questions from this chapter

In order for initially two widely separated deuterons to get close enough to fuse, their kinetic energy must be about a. \(0.1 \mathrm{MeV}\) b. 3 MeV. c. -3 MeV. d. \(0.3 \mathrm{MeV}\)

A particle slides back and forth on a frictionless track whose height as a function of horizontal position \(x\) is \(y=a x^{2},\) where \(a=0.92 \mathrm{m}^{-1} .\) If the particle's maximum speed is \(8.5 \mathrm{m} / \mathrm{s},\) find its turning points.

Current automotive standards call for bumpers that sustain essentially no damage in a \(4-\mathrm{km} / \mathrm{h}\) collision with a stationary object. As an automotive engineer, you'd like to improve on that. You've developed a spring-mounted bumper with effective spring constant \(1.3 \mathrm{MN} / \mathrm{m} .\) The springs can compress up to \(5.0 \mathrm{~cm}\) before damage occurs. For a \(1400-\mathrm{kg}\) car, what do you claim as the maximum collision speed?

If the force is zero at a given point, must the potential energy also be zero at that point? Give an example.

In ionic solids such as \(\mathrm{NaCl}\) (salt), the potential energy of a pair of ions takes the form \(U=b / r^{n}-a / r,\) where \(r\) is the separation of the ions. For \(\mathrm{NaCl}, a\) and \(b\) have the SI values \(4.04 \times 10^{-28}\) and \(5.52 \times 10^{-98},\) respectively, and \(n=8.22 .\) Find the equilibrium separation in NaCl.
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