engineering chemistry MCQ #969

The de Broglie wavelength of a particle is λ = h / p. For an electron with momentum 1.89 x 10⁻²⁴ kg m/s, λ will be

engineering chemistry MCQ #969

  1. Question 1

    Q1. The de Broglie wavelength of a particle is λ = h / p. For an electron with momentum 1.89 x 10⁻²⁴ kg m/s, λ will be

    • A) 3.5 Å
    • B) 0.35 Å
    • C) 35 Å
    • D) 350 Å

    Answer: 3.5 Å

    Explanation: Using λ = h / p and h = 6.626 x 10⁻³⁴ J s, λ = 6.626 x 10⁻³⁴ / 1.89 x 10⁻²⁴ = 3.5 x 10⁻¹⁰ m = 3.5 Å.