De Broglie Wavelength
Calculate the de Broglie wavelength of an electron moving with a velocity of 2.0 × 10^6 m/s. (Given: m_e = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J·s)
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- de broglie wavelength
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
3.64×10−10m
The de Broglie relation λ = h/mv connects the wave character of a particle to its momentum. This duality arises because matter, like light, exhibits both particle and wave properties depending on the experimental setup. Substituting the given values: λ = (6.626 × 10^-34) / (9.11 × 10^-31 × 2.0 × 10^6). Calculating the denominator: 9.11 × 10^-31 × 2.0 × 10^6 = 1.822 × 10^-24 kg·m·s^-1. Therefore λ = 6.626 × 10^-34 / 1.822 × 10^-24 = 3.637 × 10^-10 m ≈ 3.64 × 10^-10 m. Option 2.43 × 10^-10 m corresponds to an electron moving at about 3 × 10^6 m/s. Option 5.28 × 10^-10 m arises from incorrectly halving the velocity. Option 1.82 × 10^-10 m would result if the velocity were doubled to 4 × 10^6 m/s. This is a core de Broglie wavelength calculation problem featured regularly in JEE Main papers. Plausibility check: an electron at 2 × 10^6 m/s should have a wavelength in the angstrom range (~3-4 Å), and 3.64 Å is consistent with this expectation.
This easy difficulty chemistry question is from the chapter atomic structure, covering the topic of de broglie wavelength. It appeared in the 2025 exam.
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