What is the de Broglie wavelength of an electron (mass = 9.11 10^-31 kg) traveling at a velocity of 2.0 10^6 m/s? (Planck's constant h = 6.63 10^-34 Js)

Answer and explanation

Correct answer: C. 3.64 * 10^-10 m (0.364 nm)

  1. Step 1: Apply the de Broglie wavelength equation: lambda = h / p = h / (m * v).
  2. Step 2: Calculate linear momentum: p = (9.11 10^-31 kg) (2.0 10^6 m/s) = 1.822 10^-24 kg*m/s.
  3. Step 3: Compute wavelength: lambda = (6.63 10^-34 Js) / (1.822 10^-24 kgm/s) = 3.639 * 10^-10 m approx 0.364 nm.
Answer
3.64 * 10^-10 m (0.364 nm).
Why not the others: '7.28 10^-10 m' uses v = 1.0 10^6 m/s; '1.82 10^-10 m' multiplies momentum by 2; '3.64 10^-7 m' makes an order-of-magnitude exponent error.

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