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

검증된 단계별 안내

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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

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.

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.