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Energy of the Nucleus. How much work is needed to assemble an atomic nucleus containing three protons (such as Be) if we model it as an equilateral triangle of side \(2.00 \times 10^{-13} \mathrm{m}\) with a proton at each vertex? Assume the protons started from very far away.

Short Answer

Expert verified
The work required is approximately \(3.45 \times 10^{-13} \text{ J}\).

Step by step solution

01

Understand the scenario

We need to find the work required to assemble an equilateral triangle with three protons at each vertex. The side of the triangle is given as \(2.00 \times 10^{-13} \text{ m}\). The protons are initially at an infinite distance apart.
02

Use Coulomb's Law

Coulomb's Law for the force between two point charges is \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k = 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \) is the electrostatic constant, \( q \) is the charge of a proton \( (1.6 \times 10^{-19} \text{ C}) \), and \( r \) is the distance between the charges.
03

Calculate the potential energy between two protons

The potential energy \( U \) between two charges in this configuration is given by \( U = k \frac{q^2}{r} \). Substitute \( q = 1.6 \times 10^{-19} \text{ C} \) and \( r = 2.00 \times 10^{-13} \text{ m} \) to find the energy between each pair of protons.
04

Compute the energy for each side

Calculate the potential energy for one pair: \[ U_{pair} = (8.99 \times 10^9) \frac{(1.6 \times 10^{-19})^2}{2.00 \times 10^{-13}} \]. Compute this to find the potential energy stored in one pair of protons.
05

Add the energies for all pairs

For an equilateral triangle, there are three pairs of interactions. Thus, total potential energy is \[ U_{total} = 3 \times U_{pair} \]. Compute this to find the total work needed to assemble the nucleus.

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Key Concepts

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

Atomic Nucleus Formation
The formation of an atomic nucleus involves gathering protons close together, within a tiny space. This is an essential concept in nuclear physics. Protons are positively charged particles. They naturally repel each other due to their like charges. However, in an atomic nucleus, they must be held together through strong interactions.

In our scenario, assembling a nucleus similar to that of beryllium (with three protons), imagine forming an equilateral triangle. You place a proton at each vertex of the triangle. This configuration helps to simplify calculations. The side of the triangle in this exercise is extremely tiny, with a length of about \(2.00 \times 10^{-13} \text{ m}\). This size is typical for nucleons at the atomic level.

Building an atomic nucleus from individual protons requires work against their natural repulsive forces. Understanding how protons are kept together helps us better grasp the concept of nuclear forces. These forces are essential in maintaining the stability of the nucleus.
Coulomb's Law
Coulomb's Law is crucial in understanding interactions between electrically charged particles.

It states that the electric force between two point charges is directly proportional to the product of their charges, and inversely proportional to the square of the distance between them. The equation for the force \( F \) is: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) is the electrostatic constant, \( q_1 \) and \( q_2 \) represent the charges, and \( r \) is the distance between the two charges.

In our context, it helps calculate the force between protons. Since both are positive, they repel each other. The Coulomb force gives us understanding of how much they push apart, and this enables us to calculate the work needed to bring them close enough to form a nucleus.

Moreover, Coulomb's Law assists in deriving the potential energy between charges, which is necessary for understanding energy dynamics when assembling atomic nuclei.
Proton Interactions
Protons are central to nuclear physics, and understanding their interactions is vital.

Each proton carries a positive charge of \(1.6 \times 10^{-19} \text{ C}\). When these protons are placed near each other, like in a nucleus, they experience repulsive forces due to their like charges. This repulsion must be overcome in order to assemble them into a stable formation.

In the example of the equilateral triangle, each pair of protons interacts through electrostatic forces. Using Coulomb's Law, we compute the potential energy created by this repulsion:
  • Potential energy, \( U \), for one pair is determined by: \[ U = k \frac{q^2}{r} \]
  • Here, \( q = 1.6 \times 10^{-19} \text{ C} \) and the distance \( r = 2.00 \times 10^{-13} \text{ m} \)
Since a triangle has three pairs of protons, you calculate the interaction energy for one pair and then multiply it by three to find the total potential energy. This total represents the work needed to overcome the repulsion and assemble the nucleus. Protons interact not only through electric forces like calculated here, but also through nuclear forces that bind them together in a real nucleus.

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Most popular questions from this chapter

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d\) . Two of the point charges are identical and have charge \(q\) . If zerv net work is required to place the three charges at the comers of the triangle, what must the value of the third charge be?

A particle with charge \(+7.60 \mathrm{nC}\) is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved 8.00 \(\mathrm{cm}\) , the additional force has done \(6.50 \times 10^{-5} \mathrm{J}\) of work and the particle has \(4.35 \times 10^{-5} \mathrm{J}\) of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field?

Two stationary point charges \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of 50.0 \(\mathrm{cm}\) . An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 \(\mathrm{cm}\) from the \(+3.00-\mathrm{charge}\) ?

(a) If a spherical raindrop of radius 0.650 \(\mathrm{mm}\) carries a charge of \(-1.20 \mathrm{pC}\) uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.) (b) Two identical raindrops, each with radius and charge specified in part (a), collide and merge into one larger raindrop. What is the radius of this larger drop, and what is the potential at its surface, if its charge is uniformly distributed over its volume?

A total electric charge of 3.50 \(\mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{cm} .\) If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{cm} ;(\mathrm{b}) 24.0 \mathrm{cm}\) (c) \(12.0 \mathrm{cm} .\)

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