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Suppose a force between two particles decreases distance according to F=k/rb. What is the limit on b if the energy required to separate the particlesInfinitely far is not to be infinite?

Short Answer

Expert verified

The limit on b is b>1.

Step by step solution

01

Given data

A force between two particles decreases with distance, according to F=k/rb.

02

Concept of Potential energy

The force Fx is the negative of the derivative of the potential energy U such that,

Fx=-dUdx

Therefore, the differential form of the potential energy can be written as,

dU=-Fxdx

03

Calculation of the potential energy

Integrate the above equation on both sides, andgetthe expression for the total potential energy as:

Ux=-∫Fxdx

For force,the formula can be written as:

F=krb

The energy of this force would be given as,

U=-∫Fdr

Substitute F=krbin the above equation as:

U=∫krbdrU=-∫kr-bdrU=-kr-b+1-b+1

When b is not 1 , we need -b+1<0to make sure the energy is not infinite at infinite separation, which is b>1. Thus, the limit on is b>1

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Most popular questions from this chapter

In the following exercises, two protons are smashed together in an attempt to convert kinetic energy into mass and new particles. Indicate whether the proposed reaction is possible. If not, indicate which rules are violated. Consider only those for charge, angular momentum, and baryon number If the reaction is possible, calculate the minimum kinetic energy required of the colliding protons.

p+p→Σ++p+K0

Although not truly fundamental, the residual strong force shared by nucleons can also be represented by a feynmandiagram, with pion filling the role of mediating boson.The accompanying diagram represents a force between a neutron and a proton mediated by aπ0 boson.Suggest a diagram in which they exchange a π-particle.

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(a)

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ddt∫Ψ*(x,t)Ψ'(x,t)dx=∫(Ψ(x,t)∂∂tΨ*(x,t)+Ψ*(x,t)∂∂tΨ(x,t))dx

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