De Broglie Wavelength
The de Broglie wavelength of an electron accelerated through a potential difference of 150 V is approximately (mass of electron = 9.1 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C):
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Solution
1.0 Å
The kinetic energy gained by the electron is KE = eV = 1.6 × 10⁻¹⁹ × 150 = 2.4 × 10⁻¹⁷ J. Since KE = ½mv², we get v = √(2KE/m) = √(2 × 2.4 × 10⁻¹⁷ / 9.1 × 10⁻³¹) = √(5.27 × 10¹³) ≈ 7.26 × 10⁶ m/s. The de Broglie wavelength λ = h/mv = 6.63 × 10⁻³⁴ / (9.1 × 10⁻³¹ × 7.26 × 10⁶) = 6.63 × 10⁻³⁴ / 6.61 × 10⁻²⁴ ≈ 1.0 × 10⁻¹⁰ m = 1.0 Å. This confirms the wave nature of electrons as proposed by de Broglie.
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- de broglie wavelength
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
1.0 Å
The kinetic energy gained by the electron is KE = eV = 1.6 × 10⁻¹⁹ × 150 = 2.4 × 10⁻¹⁷ J. Since KE = ½mv², we get v = √(2KE/m) = √(2 × 2.4 × 10⁻¹⁷ / 9.1 × 10⁻³¹) = √(5.27 × 10¹³) ≈ 7.26 × 10⁶ m/s. The de Broglie wavelength λ = h/mv = 6.63 × 10⁻³⁴ / (9.1 × 10⁻³¹ × 7.26 × 10⁶) = 6.63 × 10⁻³⁴ / 6.61 × 10⁻²⁴ ≈ 1.0 × 10⁻¹⁰ m = 1.0 Å. This confirms the wave nature of electrons as proposed by de Broglie.
This hard 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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