An alpha particle (m=6.64×10−27m=6.64\$$times10^{-27} kg) $$emitted in the radioactive decay of uranium-238238 has an energy of 4.204.20 MeV. What is its de Broglie wavelength?

Ch 39: Particles Behaving as Waves

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 39: Particles Behaving as Waves

문제 4

39장, 문제 4

An alpha particle ( $m = 6.64 \times 10^{- 27}$ kg) emitted in the radioactive decay of uranium- $238$ has an energy of $4.20$ MeV. What is its de Broglie wavelength?

검증된 단계별 안내

Convert the energy of the alpha particle from MeV to joules. Use the conversion factor: 1 MeV = 1.602 × 10^(-13) J. Multiply 4.20 MeV by this factor to get the energy in joules.

Determine the momentum of the alpha particle using the relationship between energy and momentum for a non-relativistic particle: $ E = \frac{p^2}{2m} $, where $ E $ is the energy, $ p $ is the momentum, and $ m $ is the mass. Rearrange the formula to solve for $ p $: $ p = \sqrt{2mE} $. Substitute the mass $ m = 6.64 \times 10^{-27} $ kg and the energy (in joules) into this equation.

Recall the de Broglie wavelength formula: $ \lambda = \frac{h}{p} $, where $ \lambda $ is the wavelength, $ h $ is Planck's constant ($ h = 6.626 \times 10^{-34} $ J·s), and $ p $ is the momentum. Use the momentum calculated in the previous step to find $ \lambda $.

Substitute the values of Planck's constant $ h $ and the momentum $ p $ into the de Broglie wavelength formula to calculate $ \lambda $. Ensure the units are consistent throughout the calculation.

Express the final de Broglie wavelength in meters (m) and, if needed, convert it to a more convenient unit such as nanometers (nm) by using the conversion factor: 1 m = 10^9 nm.

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주요 개념

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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 (6.626 x 10^-34 Js), and p is the momentum of the particle. This concept illustrates the wave-particle duality of matter, indicating that particles like alpha particles exhibit both particle-like and wave-like properties.

Energy and Momentum Relationship

In physics, the energy of a particle is related to its momentum through the equation E = p^2/2m for non-relativistic particles, where E is energy, p is momentum, and m is mass. For an alpha particle, which is a type of helium nucleus, its kinetic energy can be converted into momentum, allowing us to calculate its de Broglie wavelength. Understanding this relationship is crucial for solving problems involving particle behavior in quantum mechanics.

Conversion of Energy Units

In the context of this problem, it is important to convert energy units from MeV (mega-electronvolts) to joules for calculations involving the de Broglie wavelength. The conversion factor is 1 MeV = 1.602 x 10^-13 joules. This conversion is necessary because the standard units in physics equations typically use joules, ensuring consistency and accuracy in calculations involving energy, momentum, and wavelength.