De Broglie Wavelength
An electron in a beam is accelerated so that it moves with a momentum of 3.3 × 10⁻²⁴ kg·m/s, and a researcher computes its associated de Broglie wavelength.
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About This Question
- Subject
- physics
- Chapter
- dual nature of matter and radiation
- Topic
- de broglie wavelength
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Approximately 0.2 nm
De Broglie proposed, as NCERT records, that every moving particle has an associated wavelength given by λ=ph, where p is the particle's momentum and h Planck's constant. Substituting h=6.63×10−34 J·s and p=3.3×10−24 kg·m/s gives λ=3.3×10−246.63×10−34=2.0×10−10 m, which equals 0.2 nm. This wavelength is comparable to atomic spacings, which is precisely why electron beams diffract from crystals. The value 2.0 nm is wrong because it misplaces the exponent by one order of magnitude in the division. The value 0.02 nm is wrong because it shifts the exponent the other way, again a power-of-ten slip. The value 20 nm is wrong because it is off by two orders of magnitude, far larger than any electron matter wave at this momentum. As stated in NCERT Class 12, Chapter 11, matter waves have wavelengths inversely proportional to momentum, so faster or heavier particles have shorter wavelengths. The smallness of Planck's constant is what makes matter waves negligible for everyday objects yet significant for electrons, whose tiny momenta yield wavelengths near atomic dimensions. This is exactly why electron beams, but not baseball trajectories, reveal diffraction and interference effects in the laboratory. A magnitude check confirms that an electron of this momentum has a wavelength near atomic dimensions, consistent with the 0.2 nm result.
This medium difficulty physics question is from the chapter dual nature of matter and radiation, covering the topic of de broglie wavelength. It appeared in the 2025 exam.
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