De Broglie Wavelength From Momentum
In a diffraction setup an electron is found to travel with a linear momentum of 3.3×10−24 kg m s−1. What de Broglie wavelength should be associated with this electron?
Select the correct option:
Solution
2 A˚
Louis de Broglie proposed that every moving material particle has a wave character, with a wavelength λ=h/p inversely proportional to its momentum p. This matter-wave idea unifies the particle and wave descriptions and is the basis of electron diffraction. Here the momentum is supplied directly, so no kinetic-energy conversion is needed. Substituting h=6.6×10−34 J s and p=3.3×10−24 kg m s−1 gives λ=3.3×10−246.6×10−34=2×10−10 m=2 A˚. The option 1 A˚ would require double the momentum, while 4 A˚ assumes half the momentum. The option 0.5 A˚ misplaces the powers of ten by a factor of four. The result, of the order of one angstrom, is comparable to atomic spacings in crystals, which is exactly why electrons of this momentum produce observable diffraction in the Davisson-Germer experiment. The magnitude check confirms the answer is physically sensible. It is worth stressing that the matter-wave wavelength is governed entirely by momentum, so two particles of very different mass moving with the same momentum share the same wavelength. This momentum-only dependence is what lets electron microscopes achieve resolution far beyond optical instruments, because the electron wavelength can be made hundreds of times smaller than that of visible light by raising its momentum.
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About This Question
- Subject
- physics
- Chapter
- dual nature of radiation and matter
- Topic
- de broglie wavelength from momentum
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
2 A˚
Louis de Broglie proposed that every moving material particle has a wave character, with a wavelength λ=h/p inversely proportional to its momentum p. This matter-wave idea unifies the particle and wave descriptions and is the basis of electron diffraction. Here the momentum is supplied directly, so no kinetic-energy conversion is needed. Substituting h=6.6×10−34 J s and p=3.3×10−24 kg m s−1 gives λ=3.3×10−246.6×10−34=2×10−10 m=2 A˚. The option 1 A˚ would require double the momentum, while 4 A˚ assumes half the momentum. The option 0.5 A˚ misplaces the powers of ten by a factor of four. The result, of the order of one angstrom, is comparable to atomic spacings in crystals, which is exactly why electrons of this momentum produce observable diffraction in the Davisson-Germer experiment. The magnitude check confirms the answer is physically sensible. It is worth stressing that the matter-wave wavelength is governed entirely by momentum, so two particles of very different mass moving with the same momentum share the same wavelength. This momentum-only dependence is what lets electron microscopes achieve resolution far beyond optical instruments, because the electron wavelength can be made hundreds of times smaller than that of visible light by raising its momentum.
This easy difficulty physics question is from the chapter dual nature of radiation and matter, covering the topic of de broglie wavelength from momentum. It appeared in the 2025 exam.
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