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How close can a \(5.00-\mathrm{MeV}\) alpha particle get to a uranium- 238 nucleus, assuming the only interaction is Coulomb?

Short Answer

Expert verified
The closest distance an alpha particle with a kinetic energy of 5.00 MeV can approach a uranium-238 nucleus is approximately 7.71 x 10^(-14) meters.

Step by step solution

01

1. Identifying the given information and defining the variables

We are given: - The kinetic energy of the alpha particle: \(E_k = 5.00 \ \mathrm{MeV}\) - The alpha particle has charge \(q_\alpha = 2e\), where \(e = 1.60 \times 10^{-19} \ \mathrm{C}\) is the elementary charge. - The charge of the uranium-238 nucleus, which has atomic number \(Z = 92\), and hence charge \(q_U = Ze\). We want to find the closest distance of approach, \(r_{min}\), between the alpha particle and the uranium-238 nucleus.
02

2. Convert energy from MeV to Joules

First, we should convert the given kinetic energy of the alpha particle from MeV (Mega-electronvolt) to Joules: $$ E_k = 5.00 \ \mathrm{MeV} \cdot \frac{1.60 \times 10^{-13} \ \mathrm{J}}{1 \ \mathrm{MeV}} = 8.00 \times 10^{-13} \ \mathrm{J} $$
03

3. Using conservation of energy

At the closest approach, the kinetic energy of the alpha particle will be converted entirely into potential energy. Thus, we can set up an equation for the conservation of energy: $$ E_k = \frac{kq_\alpha q_U}{r_{min}} $$ where \(k = 8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2}\) is the electrostatic constant.
04

4. Find the closest distance of approach

We want to find \(r_{min}\), so we can rearrange the conservation of energy equation to solve for it: $$ r_{min} = \frac{kq_\alpha q_U}{E_k} $$ Now, we can substitute the known values into the equation and calculate the closest distance of approach: $$ r_{min} = \frac{(8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2})(2 \cdot 1.60 \times 10^{-19} \ \mathrm{C})(92 \cdot 1.60 \times 10^{-19} \ \mathrm{C})}{8.00 \times 10^{-13} \ \mathrm{J}} $$ Finally, after performing the calculation, we find the closest distance between the alpha particle and the uranium-238 nucleus: $$ r_{min} \approx 7.71 \times 10^{-14} \ \mathrm{m} $$

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Key Concepts

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

Alpha Particle
The alpha particle is a type of nuclear radiation consisting of two protons and two neutrons. This composition makes it identical to the nucleus of a helium atom.
  • It carries a positive charge of +2e, where e is the elementary charge.
  • Because of its positive charge, it interacts with other charged particles via Coulomb forces.
Alpha particles are relatively large compared to other types of nuclear radiation, such as beta or gamma particles. This size results in them having a low penetration ability, but they can cause significant damage to materials they collide with due to their high kinetic energy.
Uranium-238
Uranium-238 is a heavy element known for its use in nuclear reactors and as a starting point for producing other radioactive elements. It is one of the most common isotopes of uranium. - The nucleus of uranium-238 consists of 92 protons and 146 neutrons. - Due to its high atomic number, it has a large positive charge, which is responsible for its ability to repel positively charged particles, such as alpha particles. The interaction between an alpha particle and a uranium-238 nucleus is primarily governed by the Coulomb force, leading to phenomena such as the scattering and potential capture of the alpha particle.
Kinetic Energy Conversion
Kinetic energy conversion is a critical concept when discussing the interaction between charged particles. In the context of our exercise, we begin with an alpha particle that has a specific kinetic energy as it approaches a uranium-238 nucleus. - As the alpha particle gets closer to the uranium nucleus, its kinetic energy decreases. - At the point of closest approach, the entire kinetic energy is converted into electrostatic potential energy. This process highlights the conservation of energy principle, where the total energy remains constant, merely switching forms between kinetic and potential as dictated by the interaction of charged particles.
Electrostatic Constant
The electrostatic constant, represented by the symbol \( k \), is a crucial parameter in Coulomb's Law, which describes the force between two point charges. Its value is approximately \( 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \).- This constant quantifies the strength of the electrostatic force in vacuum.- It is instrumental in calculating the potential energy between two charged particles, which is essential in determining interactions such as those between an alpha particle and a uranium-238 nucleus.Understanding the electrostatic constant helps us apply the principles of energy conservation and determine the effects of electric charge interactions on particles in nuclear and atomic physics.

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Most popular questions from this chapter

Before you look it up, make a prediction of the spin (intrinsic spin, i.e., actual angular momentum) of the deuteron, \({ }^{2} \mathrm{H}\). Explain your reasoning.

In a simple case of chain radioactive decay, a parent radioactive species of nuclei, A, decays with a decay constant \(\lambda_{1}\) into a daughter radioactive species of nuclei, B, which then decays with a decay constant \(\lambda_{2}\) to a stable element C. a) Write the equations describing the number of nuclei in each of the three species as a function of time, and derive an expression for the number of daughter nuclei, \(N_{2}\), as a function of time, and for the activity of the daughter nuclei, \(A_{2},\) as a function of time. b) Discuss the results in the case when \(\lambda_{2}>\lambda_{1}\left(\lambda_{2} \approx 10 \lambda_{1}\right)\) and when \(\lambda_{2}>>\lambda_{1}\left(\lambda_{2} \approx 100 \lambda_{1}\right)\).

In neutron stars, which are roughly \(90 \%\) neutrons and supported almost entirely by nuclear forces, which of the following binding-energy terms becomes relatively dominant compared to ordinary nuclei? a) the Coulomb term b) the asymmetry term c) the pairing term d) all of the above e) none of the above

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The most common isotope of uranium, \({ }_{92}^{238} \mathrm{U},\) produces radon \({ }_{86}^{222} \mathrm{Rn}\) through the following sequence of decays: $$\begin{array}{c}{ }^{238} \mathrm{U} \rightarrow{ }^{234} \mathrm{Th}+\alpha,{ }^{234} \mathrm{Th} \rightarrow{ }^{234} \mathrm{~Pa}+\beta^{-}+\bar{\nu}_{e}, \\\\{ }_{91}^{234} \mathrm{~Pa} \rightarrow{ }_{92}^{234} \mathrm{U}+\beta+\bar{\nu}_{e},{ }^{234} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+\alpha ,\\\\{ }_{91}^{230} \mathrm{Th} \rightarrow{ }_{90}^{226} \mathrm{Ra}+\alpha,{ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+\alpha,\end{array}$$. A sample of \({ }_{92}^{238} \mathrm{U}\) will build up equilibrium concentrations of its daughter nuclei down to \({ }_{88}^{226} \mathrm{Ra} ;\) the concentrations of each are such that each daughter is produced as fast as it decays. The \({ }_{88}^{226} \mathrm{Ra}\) decays to \({ }_{86}^{222} \mathrm{Rn},\) which escapes as a gas. (The \(\alpha\) particles also escape, as helium; this is a source of much of the helium found on Earth.) In high concentrations, the radon is a health hazard in buildings built on soil or foundations containing uranium ores, as it can be inhaled. a) Look up the necessary data, and calculate the rate at which \(1.00 \mathrm{~kg}\) of an equilibrium mixture of \({ }_{92}^{238} \mathrm{U}\) and its first five daughters produces \({ }_{86}^{222} \mathrm{Rn}\) (mass per unit time). b) What activity (in curies per unit time) of radon does this represent?

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