Quantum Numbers
An electron is accelerated through a potential difference of 150 V. Its de Broglie wavelength is closest to (in nm): (h = 6.626×10⁻³⁴ J s, m_e = 9.11×10⁻³¹ kg, 1 eV = 1.602×10⁻¹⁹ J)
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- quantum numbers
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.10 nm
- Wavelength Formula: λ=ph=2mEkh.
- Accelerated Electron: Energy Ek=qV, where q is electronic charge.
- Shortcut for Electron: λ(in A˚)≈V150.
- Apply Shortcut:
- V=150 V.
- λ≈150150=1.0 Å.
- Unit Conversion: 1.0 Å =10−10 m =0.1 nm.
- Result: The wavelength is 0.10 nm.
This hard difficulty chemistry question is from the chapter atomic structure, covering the topic of quantum numbers. It appeared in the 2025 exam.
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