engineering chemistry MCQ #991

The de Broglie wavelength of a particle is given by λ = h / p. What is the de Broglie wavelength of an electron with momentum 1.05 × 10⁻²⁴ kg m/s?

engineering chemistry MCQ #991

  1. Question 1

    Q1. The de Broglie wavelength of a particle is given by λ = h / p. What is the de Broglie wavelength of an electron with momentum 1.05 × 10⁻²⁴ kg m/s?

    • A) 6.26 × 10⁻¹⁰ m
    • B) 6.26 × 10⁻⁸ m
    • C) 6.26 × 10⁻¹² m
    • D) 6.26 × 10⁻⁶ m

    Answer: 6.26 × 10⁻¹⁰ m

    Explanation: Using λ = h / p, we get λ = 6.626 × 10⁻³⁴ / 1.05 × 10⁻²⁴ = 6.31 × 10⁻¹⁰ m ≈ 6.26 × 10⁻¹⁰ m.