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In 1935 , Yukawa proposed an early theory of the force that held the neutrons and protons together in the nucleus. His equation for the potential energy of two such particles, at a center-tocenter distance \(r\), was \(P E(r)=g r^{-1} e^{-r / a}\), where \(g\) parametrizes the strength of the interaction, \(e\) is the base of natural logarithms, and \(a\) is about \(10^{-15} \mathrm{~m}\). Find the force between two nucleons that would be consistent with this equation for the potential energy.

Short Answer

Expert verified
The force is given by \( F(r) = g e^{-r/a} \left( \frac{1}{r^2} + \frac{1}{ar} \right) \).

Step by step solution

01

Understand the Problem

We are given Yukawa's equation for potential energy between two nucleons, which is \( PE(r) = g r^{-1} e^{-r/a} \). We are tasked with finding the force between these two nucleons, which requires differentiating the potential energy with respect to \( r \).
02

Differentiate Potential Energy

The force can be determined by finding the negative derivative of the potential energy with respect to \( r \). Thus, the force \( F(r) \) is given by \( F(r) = -\frac{d}{dr}(PE(r)) \).
03

Apply the Product Rule

The potential energy function involves a product of the terms \( gr^{-1} \) and \( e^{-r/a} \). Using the product rule, \( \frac{d}{dr} (uv) = u'v + uv' \), we compute:- Let \( u = gr^{-1} \), then \( u' = -gr^{-2} \).- Let \( v = e^{-r/a} \), then \( v' = -\frac{1}{a} e^{-r/a} \).Thus, the derivative is \( -\left( -gr^{-2} e^{-r/a} + gr^{-1}(-\frac{1}{a}) e^{-r/a} \right) \).
04

Simplify the Derivative Expression

Simplifying the expression from the product rule gives:\[ F(r) = g r^{-2} e^{-r/a} + \frac{g}{a} r^{-1} e^{-r/a} \]Therefore, \[ F(r) = g e^{-r/a} \left( r^{-2} + \frac{1}{a} r^{-1} \right) \].
05

Write the Final Expression for the Force

The expression for the force between the two nucleons is given by:\[ F(r) = g e^{-r/a} \left( \frac{1}{r^2} + \frac{1}{ar} \right) \].This equation gives the force based on Yukawa's potential when considering both exponential decay and distance-dependent terms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nucleons
Nucleons are the building blocks of atomic nuclei, consisting of protons and neutrons. They are held together by a fundamental force known as the strong nuclear force, which is one of the four fundamental forces in physics. This force is incredibly powerful over short distances and allows protons and neutrons to stay bonded within the nucleus, despite the repulsive electromagnetic force between the positively charged protons.
This nuclear force is not like everyday forces such as gravity or electromagnetism. It operates on a short range, roughly the size of a typical atomic nucleus, or approximately one femtometer ( 1 imes 10^{-15} ext{ meters} ). To understand the interactions between nucleons, physicists represent these forces with mathematical models like the Yukawa potential, which describes how nucleons influence each other at the subatomic level.
Differentiation
Differentiation is a mathematical process used to find the rate at which one quantity changes with respect to another. In the context of physics and potential energy, differentiation helps determine the force exerted by a potential field.
For Yukawa's potential energy equation, expressed as \( PE(r) = g r^{-1} e^{-r/a} \), we can find the force by differentiating this expression with respect to the distance \( r \). The force is the negative derivative of potential energy, as a negative gradient indicates the direction in which the potential energy decreases most rapidly.
Applying the product rule in differentiation, which states that the derivative of a product \( uv \) is \( u'v + uv' \), allows us to break down complex expressions like Yukawa's potential into simpler components that can be differentiated more easily.
Force Calculation
The force between particles in a potential energy field is derived by differentiating the potential energy with respect to distance, and then taking the negative of the result. For the Yukawa potential, this force calculation involves applying the product rule to differentiate the potential energy expression.
For Yukawa's potential energy, we have:
  • \( u = gr^{-1} \) and its derivative \( u' = -gr^{-2} \)
  • \( v = e^{-r/a} \) and its derivative \( v' = -\frac{1}{a} e^{-r/a} \)
By substituting these into the product rule, and remembering that the force is the negative of this derivative, you find the expression \( F(r) = g e^{-r/a} ig( r^{-2} + \frac{1}{a} r^{-1} \big) \).
This expression tells us how the force varies with the distance \( r \) between two nucleons. The force decreases both with increasing distance and due to the exponential decay term. These components reflect the short-range nature of nuclear forces.
Potential Energy
Potential energy is the stored energy in a system that has the potential to do work as the system changes states. In the context of Yukawa's equation, potential energy arises from the interactions between nucleons.
Yukawa's potential energy function of the form \( PE(r) = g r^{-1} e^{-r/a} \) accounts for two factors:
  • The inverse relationship with distance \( r \) highlights how potential energy decreases as nucleons move further apart.
  • The exponential decay term \( e^{-r/a} \) captures how the force's influence sharply diminishes over short ranges.
Together, these terms model the strong nuclear force's behavior, reflecting how it binds nucleons at very short distances while rapidly decreasing beyond the typical size of a nucleus. Understanding the potential energy allows physicists to predict the force dynamics that hold atomic nuclei together.

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Most popular questions from this chapter

A mass moving in one dimension is attached to a horizontal spring. It slides on the surface below it, with equal coefficients of static and kinetic friction, \(\mu_{k}=\mu_{s} .\) The equilibrium position is \(x=0 .\) If the mass is pulled to some initial position and released from rest, it will complete some number of oscillations before friction brings it to a stop. When released from \(x=a(a>0)\), it completes exactly \(1 / 4\) of an oscillation, i.e., it stops precisely at \(x=0 .\) Similarly, define \(b>0\) as the greatest \(x\) from which it could be released and comlete \(1 / 2\) of an oscillation, stopping on the far side and not coming back toward equilibrium. Find \(b / a\). Hint: To keep the algebra simple, set every fixed parameter of the system equal to \(1 .\)

Two speedboats are identical, but one has more people aboard than the other. Although the total masses of the two boats are unequal, suppose that they happen to have the same kinetic energy. In a boat, as in a car, it's important to be able to stop in time to avoid hitting things. (a) If the frictional force from the water is the same in both cases, how will the boats' stopping distances compare? Explain. (b) Compare the times required for the boats to stop.

In the power stroke of a car's gasoline engine, the fuel-air mixture is ignited by the spark plug, explodes, and pushes the piston out. The exploding mixture's force on the piston head is greatest at the beginning of the explosion, and decreases as the mixture expands. It can be approximated by \(F=a / x\), where \(x\) is the distance from the cylinder to the piston head, and \(a\) is a constant with units of \(\mathrm{N} \cdot \mathrm{m}\). (Actually \(a / x^{1.4}\) would be more accurate, but the problem works out more nicely with \(a / x !\) ) The piston begins its stroke at \(x=x_{1}\), and ends at \(x=x_{2}\). The 1965 Rambler had six cylinders, each with \(a=220 \mathrm{~N} \cdot \mathrm{m}, x_{1}=1.2 \mathrm{~cm}\), and \(x_{2}=10.2 \mathrm{~cm}\) (a) Draw a neat, accurate graph of \(F\) vs \(x\), on graph paper. (b) From the area under the curve, derive the amount of work done in one stroke by one cylinder. (c) Assume the engine is running at 4800 r.p.m., so that during one minute, each of the six cylinders performs 2400 power strokes. (Power strokes only happen every other revolution.) Find the engine's power, in units of horsepower \((1 \mathrm{hp}=746 \mathrm{~W})\) (d) The compression ratio of an engine is defined as \(x_{2} / x_{1}\). Explain in words why the car's power would be exactly the same if \(x_{1}\) and \(x_{2}\) were, say, halved or tripled, maintaining the same compression ratio of \(8.5 .\) Explain why this would not quite be true with the more realistic force equation \(F=a / x^{1.4}\).

A rail gun is a device like a train on a track, with the train propelled by a powerful electrical pulse. Very high speeds have been demonstrated in test models, and rail guns have been proposed as an alternative to rockets for sending into outer space any object that would be strong enough to survive the extreme accelerations. Suppose that the rail gun capsule is launched straight up, and that the force of air friction acting on it is given by \(F=b e^{-c x}\), where \(x\) is the altitude, \(b\) and \(c\) are constants, and \(e\) is the base of natural logarithms. The exponential decay occurs because the atmosphere gets thinner with increasing altitude. (In reality, the force would probably drop off even faster than an exponential, because the capsule would be slowing down somewhat.) Find the amount of kinetic energy lost by the capsule due to air friction between when it is launched and when it is completely beyond the atmosphere. (Gravity is negligible, since the air friction force is much greater than the gravitational force.)

A soccer ball of mass \(m\) is moving at speed \(v\) when you kick it in the same direction it is moving. You kick it with constant force \(F\), and you want to triple the ball's speed. Over what distance must your foot be in contact with the ball?

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