/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 A helium ion (He \(^{+7}\) ) tha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 85

A helium ion (He \(^{+7}\) ) that comes within about \(10 \mathrm{fm}\) of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of \(3.0 \mathrm{MeV}\) heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than \(10 \mathrm{fm}\) from the center of the atomic nucleus? (I \(\mathrm{fm}=1 \times 10^{-15} \mathrm{~m},\) and \(e\) is the magnitude of the charge of an electron or a proton. \((\mathrm{a}) 2 e ;(\mathrm{b}) 11 e ;(\mathrm{c}) 20 \mathrm{e} ;(\mathrm{d}) 22 e\)

Short Answer

Expert verified
The minimum charge that the nucleus of the atom can possess such that a helium ion \(He^{+7}\) with kinetic energy \(3 MeV\) gets no closer than \(10 fm\) from the center of the atomic nucleus is \(22e\) or option (d).

Step by step solution

01

Understand the given

Firstly, comprehend the problem to understand that we are given: the kinetic energy \(K.E\) of the helium ion as \(3 MeV = 3 * 10^6 eV\), the distance \(r\) which is \(10 fm = 10^-14 cm\), the charge of the helium ion \(Q1\) as \(+7e\), and we are to find the charge of the atomic nucleus \(Q2\).
02

Convert the energy in Joules

The kinetic energy is given in \(eV\), we convert it to Joules for easy calculations considering \(1 eV = 1.6 * 10^-19 J\). So, \(3 * 10^6 eV = 3 * 10^6 * 1.6 * 10^-19 = 4.8 * 10^-13 J\).
03

Calculating the charge

For the helium ion to just come close to the atom and then stop would mean that all its kinetic energy has been used to overcome the electric field of the atom. By this, we mean the kinetic energy of the helium ion is equal to the potential energy of atom due to ion at that point. \(K.E\) = \(U\). On solving, we get \(Q2 = 22e\) as the minimum charge required for the given conditions. This gives us our required solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It's a fundamental concept in physics, which quantitatively describes how much work an object can do because of its velocity. In the context of nuclear reactions, kinetic energy plays a critical role; it determines how particles interact when they come into close proximity with each other.

For instance, a helium ion with a given amount of kinetic energy can induce a nuclear reaction if it collides with another atom's nucleus with sufficient force. However, as in the textbook exercise, if the helium ion's kinetic energy is completely converted into electric potential energy due to the repulsive force of the other nucleus, it will come to a stop without inducing a reaction. This conversion marks the balance point between kinetic energy driving the particle forward and the Coulomb force repelling it.
Coulomb's Law
Coulomb's Law describes the electrostatic interaction between electrically charged particles. It states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This fundamental law is essential for calculating interactions in electrostatics, including those leading to nuclear reactions.

Coulomb's Law is vital for understanding how a nucleus can repel a charged particle like the helium ion in our exercise. The law enables us to calculate the minimum charge an atomic nucleus must have to prevent the helium ion, with its specific kinetic energy, from getting closer than a certain distance, such as the 10 femtometers given in the problem.
Electric Potential Energy
Electric potential energy is the energy a charged particle has due to its position in an electric field. In our textbook problem, the electric potential energy is what the helium ion acquires as it moves against the electric field created by the nucleus of the atom. This energy is directly related to the charge of the ion, the charge of the nucleus, and their separation distance, guided by Coulomb's Law.

The concept of electric potential energy is critical since it determines that a particle will stop moving closer to the nucleus when its kinetic energy is completely transformed into electric potential energy. This conversion is a crucial point in the exercise, as it allows students to solve for the unknown nuclear charge when the distance at which the helium ion stops is known.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle with charge \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed to the left. The charge is released from rest and moves to the left; after it has moved \(6.00 \mathrm{~cm},\) its kinetic energy is \(+2.20 \times 10^{-6} \mathrm{~J}\). What are (a) the work done by the electric force, (b) the potential of the starting point with respect to the end point, and (c) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) ?

Identical point charges \(q_{1}\) and \(q_{2}\) each have positive charge \(+6.00 \mu \mathrm{C}\). Charge \(q_{1}\) is held fixed on the \(x\) -axis at \(x=+0.400 \mathrm{~m}\), and \(q_{2}\) is held fixed on the \(x\) -axis at \(x=-0.400 \mathrm{~m}\). A small sphere has charge \(Q=-0.200 \mu \mathrm{C}\) and mass \(12.0 \mathrm{~g}\). The sphere is initially very far from the origin. It is released from rest and moves along the \(y\) -axis toward the origin. (a) As the sphere moves from very large \(y\) to \(y=0\). how much work is done on it by the resultant force exerted by \(q_{1}\) and \(q_{2} ?\) (b) If the only force acting on the sphere is the force exerted by the point charges, what is its speed when it reaches the origin?

An infinitely long line of charge has linear charge density \(5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton (mass \(1.67 \times 10^{-27} \mathrm{~kg},\) charge \(+1.60 \times 10^{-19} \mathrm{C}\) ) is \(18.0 \mathrm{~cm}\) from the line and moving directly toward the line at \(3.50 \times 10^{3} \mathrm{~m} / \mathrm{s}\). (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?

The maximum voltage at the center of a typical tandem electrostatic accelerator is \(6.0 \mathrm{MV}\). If the distance from one end of the acceleration tube to the midpoint is \(12 \mathrm{~m}\), what is the magnitude of the average electric field in the tube under these conditions? (a) \(41,000 \mathrm{~V} / \mathrm{m}\) (b) \(250.000 \mathrm{~V} / \mathrm{m} ;\) (c) \(500.000 \mathrm{~V} / \mathrm{m} ;\) (d) \(6.000 .000 \mathrm{~V} / \mathrm{m} .\)

A thin insulating rod is bent into a semicircular arc of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the are if the potential is assumed to be zero at infinity.
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Classical Mechanics

Read Explanation

Medical Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Oscillations

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.