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A helium ion (He\(^{++}\)) that comes within about 10 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 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 fm from the center of the atomic nucleus? (1 fm = 1 \(\times\) 10\(^{-15}\) m, and \(e\) is the magnitude of the charge of an electron or a proton.) (a) 2\(e\); (b) 11\(e\); (c) 20\(e\); (d) 22\(e\).

Short Answer

Expert verified
The minimum charge is 22\(e\). Option (d).

Step by step solution

01

Understanding the Problem

You need to find the minimum charge (in terms of the elementary charge \(e\)) of an atomic nucleus such that a helium ion with a kinetic energy of 3.0 MeV does not come closer than 10 fm. A helium ion \( \text{He}^{++} \) has a charge of \(2e\).
02

Calculate the Coulomb Potential Energy

The potential energy \(U\) between the helium ion and the atomic nucleus, given their charges and separation distance, is given by the formula: \[ U = \frac{k \cdot (Z \cdot e) \cdot (2e)}{r} \] where \(k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\) is the Coulomb constant, \(Z\) is the charge of the nucleus, \(e\) is the elementary charge \(1.6 \times 10^{-19} \text{ C}\), and \(r = 10 \text{ fm} = 10 \times 10^{-15} \text{ m}\).
03

Equate Potential Energy to Kinetic Energy

To find the minimum charge \(Z\), equate the calculated potential energy \(U\) to the given kinetic energy. Convert the kinetic energy from MeV to joules: \[ 3.0 \text{ MeV} = 3.0 \times 1.6 \times 10^{-13} \text{ J} = 4.8 \times 10^{-13} \text{ J} \]. Set \( U = 4.8 \times 10^{-13} \text{ J} \).
04

Solve for the Charge \(Z\)

Substitute the known values and solve for \(Z\):\[ \frac{(8.99 \times 10^9)(Z \cdot 1.6 \times 10^{-19}) (2 \cdot 1.6 \times 10^{-19})}{10 \times 10^{-15}} = 4.8 \times 10^{-13} \] Simplifying gives \(Z \approx 22\).
05

Select the Minimum Charge Option

From the given options: (a) 2\(e\), (b) 11\(e\), (c) 20\(e\), (d) 22\(e\), the calculated \(Z = 22\) matches the final option (d).

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Key Concepts

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

Coulomb potential energy
Coulomb potential energy is a crucial concept in nuclear physics and relates to the energy between two charged particles due to their electric fields. It is calculated using the formula:
  • \(U = \frac{k \cdot (q_1) \cdot (q_2)}{r}\)
Here, \(U\) represents the potential energy, \(k\) is Coulomb's constant \((8.99 \times 10^9 \text{ N m}^2/\text{C}^2)\), \(q_1\) and \(q_2\) are the charges of the particles, and \(r\) is the distance separating them.
This energy is important when studying nuclear reactions, especially when charged particles such as ions approach each other. The interaction energy increases as the particles get closer, making it essential for determining the conditions under which nuclear reactions occur. Understanding Coulomb potential energy helps predict whether a reaction will be elastic (bouncing apart) or inelastic (leading to nuclear reaction or fusion).
In our exercise, we need to calculate the Coulomb potential energy between a helium ion and an atomic nucleus to ensure that the ion does not come closer than a specific distance. This calculation will allow us to understand how much energy is needed to overcome the repulsive force and result in a potential nuclear reaction.
Helium ions
Helium ions, also known as alpha particles, are composed of two protons and two neutrons, making them quite massive and positively charged. They are often denoted as \( \text{He}^{++} \) due to their double positive charge, which means their charge is equal to \(2e\), where \(e\) is the elementary charge \(1.6 \times 10^{-19}\) C.
These ions are utilized in various nuclear physics applications, particularly in experiments investigating nuclear reactions and radiation interactions. When helium ions are introduced to other atoms, the interactions can lead to different scenarios, depending on the energy levels and conditions, such as scattering or triggering a reaction.
In this specific exercise, a helium ion approaches an atomic nucleus. The understanding of the interactions is pivotal to knowing how close the ion can approach without initiating a nuclear reaction. This distance is heavily influenced by the ion's kinetic energy and the surrounding potential energy field, determined by the atomic charges.
Nuclear reactions
Nuclear reactions are processes where two nuclei or nuclear particles collide, leading to a new formation of particles or nuclei. These reactions release or absorb considerable amounts of energy due to changes in the nuclei's composition.
There are various types of nuclear reactions, with the most prominent being:
  • Fission - where a heavy nucleus splits into smaller nuclei.
  • Fusion - where two light nuclei combine to form a heavier nucleus.
  • Collision - can be elastic, leading to scattering, or inelastic, causing changes within the nuclei.
Helium ions are often used to study such reactions due to their straightforward structure, and significant positive charge, making them excellent candidates for initiating nuclear events when interacting with a target nucleus.
In our exercise, the focus is on preventing the helium ion from coming closer than a certain distance (10 fm) to another nucleus, to avoid a nuclear reaction, which could potentially alter the nucleus's structure or release energy.
Elementary charge
The elementary charge is a fundamental constant in physics, symbolized as \(e\), and its value is \(1.6 \times 10^{-19} \text{ C}\). It represents the smallest unit of electric charge, found in electrons and protons.
In calculations concerning electric forces, fields, and potential energies, the elementary charge serves as the standard unit for expressing charge equally for both positive and negative particles. This means that understanding and correctly using \(e\) is crucial for dealing with any electron or proton-related interactions in physics.
Our exercise involves
  • Helium ion (\( \text{He}^{++} \)), which has a charge of \(2e\)
  • Nucleus charge assumed as \(Ze\),
where \(Z\) represents an integer indicating the number of elemental charges in the nucleus.
In context, applying the elementary charge helps calculate the force interactions and potential energy of ions approaching each other, providing a clear understanding of whether conditions will permit a nuclear reaction or lead to scattering. The charge interactions outlined in the exercise are fundamental to predicting how close particles may approach each other in nuclear physics settings.

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Most popular questions from this chapter

A positive point charge \(q_1 = +5.00 \times 10^{-4}\) C is held at a fixed position. A small object with mass 4.00 \(\times 10^{-3}\) kg and charge \(q_2 = -3.00 \times 10^{-4}\) C is projected directly at \(q_1\) . Ignore gravity. When \(q_2\) is 0.400 m away, its speed is 800 m\(/\)s. What is its speed when it is 0.200 m from \(q_1\) ?

In a certain region of space the electric potential is given by \(V = +Ax^2y - Bxy^2,\) where \(A =\) 5.00 \(V/m^3\) and \(B =\) 8.00 \(V/m^3\). Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x =\) 2.00 m, \(y =\) 0.400 m, and \(z = 0\).

A metal sphere with radius \(R_1\) has a charge \(Q_1\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_2\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

Identical charges \(q = +\)5.00 \(\mu\)C are placed at opposite corners of a square that has sides of length 8.00 cm. Point \(A\) is at one of the empty corners, and point \(B\) is at the center of the square. A charge \(q_0 = -\)3.00 \(\mu\)C is placed at point \(A\) and moves along the diagonal of the square to point \(B\). (a) What is the magnitude of the net electric force on \(q_0\) when it is at point \(A\)? Sketch the placement of the charges and the direction of the net force. (b) What is the magnitude of the net electric force on \(q_0\) when it is at point \(B\)? (c) How much work does the electric force do on \(q_0\) during its motion from \(A\) to \(B\)? Is this work positive or negative? When it goes from \(A\) to \(B\), does \(q_0\) move to higher potential or to lower potential?

A point charge \(q_1 = +\)2.40 \(\mu\)C is held stationary at the origin. A second point charge \(q_2 = -\)4.30 \(\mu\)C moves from the point \(x =\) 0.150 m, \(y =\) 0 to the point \(x =\) 0.250 m, \(y =\) 0.250 m. How much work is done by the electric force on \(q_2\)?

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