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Question

An electron moves with a speed of \( 3.0 \times 10^6 \, \text{m/s} \). What is its de Broglie
wavelength? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( m_e = 9.11 \times 10^{-31} \,
\text{kg} \))

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Explanation

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. Momentum p = m v = 9.11 × 10⁻³¹ × 3.0 × 10⁶ = 2.733 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/2.733 × 10⁻²⁴) ≈ 2.425 × 10⁻¹⁰ m = 0.2425 nm . Applying E = h f = h c/λ, p

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