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Ch 39: Particles Behaving as Waves
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 39: Particles Behaving as WavesProblem 4
Chapter 38, Problem 4

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

Verified step by step guidance
1
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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.
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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.
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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.
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Related Practice
Textbook Question

Through what potential difference must electrons be accelerated if they are to have:

(a) the same wavelength as an x ray of wavelength 0.2200.220 nm; and

(b) the same energy as the x ray in part (a)?

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Textbook Question

An electron has a de Broglie wavelength of 2.80×10−102.80\(\times\)10^{-10} m. Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).

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Textbook Question

Calculate the de Broglie wavelength of a 5.005.00-g bullet that is moving at 340340 m/s. Will the bullet exhibit wavelike properties?

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Textbook Question

For crystal diffraction experiments (discussed in Section 39.139.1), wavelengths on the order of 0.200.20 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is a photon.

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Textbook Question

(a) An electron moves with a speed of 4.70×1064.70\(\times\)10^6 m/s. What is its de Broglie wavelength?

(b) A proton moves with the same speed. Determine its de Broglie wavelength.

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Textbook Question

An electron is moving with a speed of 8.00×1068.00\(\times\)10^6 m/s. What is the speed of a proton that has the same de Broglie wavelength as this electron?

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