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A 5.6-MeV alpha particle is shot directly at a uranium atom \((Z=\) 92). About how close will it get to the center of the uranium nucleus? At such high energies the alpha particle will easily penetrate the electron cloud and the effects of the atomic electrons can be ignored. We also assume the uranium atom to be so massive that it does not move appreciably. Then the original KE of the alpha particle will be changed into electrostatic potential energy. This energy, for a charge \(q^{\prime}\) at a distance \(r\) from a point charge \(q\), $$ \text { Potential energy }=q^{\prime} V=k_{0} \frac{q q^{\prime}}{r} $$ Equating the KE of the alpha particle to this potential energy, $$ \left(5.6 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=\left(8.99 \times 10^{9}\right) \frac{(2 e)(92 e)}{r} $$ where \(e=1.60 \times 10^{-19} \mathrm{C}\). We find from this that \(r=4.7 \times 10^{-14}\) \(\mathrm{m} .\)

Step by step solution

01

Understand the Problem

We need to find the distance at which the kinetic energy of a 5.6-MeV alpha particle is entirely converted into electrostatic potential energy as it approaches a uranium nucleus. Given the atomic number of uranium (92), we are to ignore electron cloud effects.

02

Convert Kinetic Energy to Joules

The kinetic energy (KE) of the alpha particle is given in electron volts (MeV). We convert it to joules using the conversion factor 1 eV = 1.60 x 10^-19 J:\[ KE = 5.6 \times 10^6 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 8.96 \times 10^{-13} \text{ J} \]

03

Set Up the Potential Energy Equation

The electrostatic potential energy at distance \( r \) from the nucleus (a point charge) is given by \[ PE = k_0 \frac{q q'}{r} \]where \( k_0 = 8.99 \times 10^9 \), \( q = 92e \) (charge of uranium nucleus), and \( q' = 2e \) (charge of alpha particle). Substituting \( e = 1.60 \times 10^{-19} \text{ C} \), the equation becomes\[ PE = 8.99 \times 10^9 \frac{(92 \times 1.60 \times 10^{-19})(2 \times 1.60 \times 10^{-19})}{r} \]

04

Equate Kinetic and Potential Energies

Equate the converted kinetic energy to potential energy:\[ 8.96 \times 10^{-13} = 8.99 \times 10^9 \frac{(92 \times 1.60 \times 10^{-19})(2 \times 1.60 \times 10^{-19})}{r} \]

05

Solve for r

Rearrange the equation to solve for \( r \):\[ r = \frac{8.99 \times 10^9 \times (92 \times 1.6 \times 10^{-19}) \times (2 \times 1.6 \times 10^{-19})}{8.96 \times 10^{-13}} \]Substitute the numerical values and solve:\[ r = 4.7 \times 10^{-14} \text{ m} \]

06

Conclusion

After solving the equation, the minimum distance (approach distance) the alpha particle will get to the uranium nucleus is approximately \( 4.7 \times 10^{-14} \text{ m} \).

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Key Concepts

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

Alpha Particle

An alpha particle is a type of ionizing radiation ejected by the nuclei of radioactive elements. It consists of two protons and two neutrons, which makes it equivalent to a helium nucleus. Alpha particles are positively charged due to the presence of protons, with a net charge of +2e, where e is the elementary charge. This gives them significant mass relative to other forms of nuclear particles, like beta particles or gamma rays, which makes them less penetrating but quite impactful when interacting with matter.

• Alpha particles are emitted during the decay of heavy elements like uranium and radium.
• Due to their heavier mass, they have lower penetration power compared to beta or gamma radiation.
• Alpha particles can be stopped by a sheet of paper or even the outer layer of human skin.

Understanding alpha particles is crucial in nuclear physics because their interactions, especially with atomic nuclei, provide insights into nuclear structure and energy transformations.

Electrostatic Potential Energy

Electrostatic potential energy is the energy stored due to the positions of charged particles relative to each other. It is a crucial concept for understanding how charged particles, like alpha particles, interact with atomic nuclei. When an alpha particle approaches a nucleus, the repulsive electrostatic forces between their positive charges result in electrostatic potential energy.

• Potential energy is calculated using the formula: \( PE = k_0 \frac{q q'}{r} \)
• Where \( k_0 \) is Coulomb's constant, \( q \) and \( q' \) are the charges, and \( r \) is the separation distance.
• This energy increases as the particle gets closer to the nucleus due to the stronger repulsive force.

In the problem, we analyze how the kinetic energy of an approaching alpha particle is transformed into electrostatic potential energy as it nears the uranium nucleus. This energy conversion ensures that the alpha particle can only approach up to a certain distance, where the potential energy equals the initial kinetic energy.

Kinetic Energy

Kinetic energy (KE) is the energy that a body possesses due to its motion. For an alpha particle moving towards a uranium nucleus, the kinetic energy determines how close it can get before being repelled. In this scenario, the initial KE of the alpha particle is fully converted into electrostatic potential energy.

• Kinetic energy is given by the formula: \( KE = \frac{1}{2} m v^2 \), but here it is originally in the form of MeV (million electron volts).
• Conversion to joules is necessary for calculations involving physical constants such as Coulomb's constant and electron charge.
• The conversion factor from eV to joules is \( 1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J} \).

The initial kinetic energy value of the alpha particle is given in MeV, which is a common unit in nuclear physics due to the high energies involved. When tackling problems of energy conversion between different states, understanding and correctly converting energy units is vital.

Uranium Nucleus

The uranium nucleus is a large and heavy nucleus, with 92 protons giving it a substantial positive charge. This factor plays a significant role in nuclear reactions and energy interactions, as illustrated in this problem where an alpha particle approaches the nucleus. The uranium nucleus remains relatively stationary during these interactions due to its mass.

• The atomic number (Z) of uranium is 92, correlating to its 92 protons.
• The uranium nucleus's large charge leads to strong electrostatic forces when interacting with charged particles.
• Its massiveness means it has negligible recoil in such interactions, altering the path and energy of approaching particles instead.

In nuclear physics, the uranium nucleus is especially notable not only because it appears in natural radioactive decay chains but also due to its use in nuclear reactors and weapons. Its interaction with alpha particles helps to demonstrate key nuclear concepts, such as energy conversion and particle dynamics.

Energy Conversion

Energy conversion refers to the process of transforming energy from one form into another. In nuclear interactions, such as an alpha particle approaching a uranium nucleus, energy conversion is critical. The alpha particle begins with kinetic energy, which, as it nears the opposing charges in the uranium nucleus, is changed into electrostatic potential energy.

• The conversion is guided by the conservation of energy principle.
• Energy transformations help understand how particles interact at the atomic level.
• In this scenario, complete conversion occurs when the alpha particle reaches its closest point of approach to the nucleus.

Understanding this conversion process aids in predicting particle behavior during nuclear reactions and interactions, highlighting the intricate play of forces and the pivotal role of energy in particle physics.