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Ch 42: Nuclear Physics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 42: Nuclear PhysicsProblema 47
Capitolo 42, Problema 47

What energy (in MeV) alpha particle has a de Broglie wavelength equal to the diameter of a ²³⁸U nucleus?

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Step 1: Understand the problem. The de Broglie wavelength of the alpha particle is given to be equal to the diameter of the ²³⁸U nucleus. The de Broglie wavelength formula is \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle.
Step 2: Estimate the diameter of the ²³⁸U nucleus. The diameter of a nucleus can be approximated using the formula \( d \approx 2r \), where \( r \) is the nuclear radius. The nuclear radius can be calculated using \( r = r_0 A^{1/3} \), where \( r_0 \approx 1.2 \, \text{fm} \) (femtometers) and \( A \) is the mass number of the nucleus. For ²³⁸U, \( A = 238 \).
Step 3: Relate the momentum \( p \) of the alpha particle to its kinetic energy \( E \). The momentum is given by \( p = \sqrt{2mE} \), where \( m \) is the mass of the alpha particle and \( E \) is its kinetic energy. The mass of the alpha particle is approximately \( m \approx 3727 \, \text{MeV}/c^2 \).
Step 4: Substitute \( \lambda = d \) into the de Broglie wavelength formula \( \lambda = \frac{h}{p} \). Rearrange to solve for \( E \): \( E = \frac{h^2}{2m \lambda^2} \). Use \( h \approx 4.1357 \times 10^{-15} \, \text{eV·s} \) and \( c \approx 3 \times 10^8 \, \text{m/s} \) to convert constants appropriately.
Step 5: Perform unit conversions to ensure consistency. Convert the nuclear diameter \( d \) from femtometers to meters, and ensure all constants are in compatible units (e.g., \( h \) in \( \text{MeV·s} \), \( m \) in \( \text{MeV}/c^2 \), and \( \lambda \) in meters). Plug in the values to calculate the energy \( E \) in MeV.

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de Broglie Wavelength

The de Broglie wavelength is a fundamental concept in quantum mechanics that relates a particle's momentum to its wavelength. It is given by the formula λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. This concept is crucial for understanding the wave-particle duality of matter, particularly for particles like alpha particles in nuclear physics.
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Alpha Particle

An alpha particle is a type of nuclear particle consisting of two protons and two neutrons, essentially a helium nucleus. It is emitted during radioactive decay processes and carries a positive charge. Understanding the properties of alpha particles, including their mass and charge, is essential for calculating their energy and interactions with matter, particularly in the context of nuclear reactions.
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Energy-Momentum Relation

The energy-momentum relation in relativistic physics describes how the energy of a particle is related to its momentum and mass. For a particle like an alpha particle, the total energy can be expressed as E = √(p²c² + m₀²c⁴), where E is energy, p is momentum, m₀ is rest mass, and c is the speed of light. This relationship is vital for converting between kinetic energy and momentum, especially when determining the energy of particles with specific wavelengths.
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