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Calculate the de Broglie wavelength of an electron with a mass of 9.1 x 10⁻³¹ kg moving at a velocity of 42 m/s.

Physics

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Answer

1.7×10−5 m\mathbf{1.7 \times 10^{-5} \text{ m}}

Step-by-step solution

4 steps

  1. Identify the given values.

    The mass of the electron is m=9.1×10−31 kgm = 9.1 \times 10^{-31} \text{ kg}.
    The velocity of the electron is v=4.2×10 ms−1=42 ms−1v = 4.2 \times 10 \text{ ms}^{-1} = 42 \text{ ms}^{-1}.
    Planck's constant is h=6.6×10−34 Jsh = 6.6 \times 10^{-34} \text{ Js}.

  2. Apply the de Broglie wavelength formula.

    The de Broglie wavelength (λ\lambda) is given by the formula:

    λ=hmv\lambda = \frac{h}{mv}

  3. Substitute the values into the formula and calculate the wavelength.

    λ=6.6×10−34 Js(9.1×10−31 kg)×(42 ms−1)\lambda = \frac{6.6 \times 10^{-34} \text{ Js}}{(9.1 \times 10^{-31} \text{ kg}) \times (42 \text{ ms}^{-1})} λ=6.6×10−34 Js382.2×10−31 kg⋅ms−1\lambda = \frac{6.6 \times 10^{-34} \text{ Js}}{382.2 \times 10^{-31} \text{ kg} \cdot \text{ms}^{-1}} λ=6.6×10−343.822×10−29 m\lambda = \frac{6.6 \times 10^{-34}}{3.822 \times 10^{-29}} \text{ m} λ≈1.7268×10−5 m\lambda \approx 1.7268 \times 10^{-5} \text{ m}

  4. Round the answer to an appropriate number of significant figures.

    The given values have two significant figures (9.1, 4.2, 6.6).

    λ≈1.7×10−5 m\lambda \approx 1.7 \times 10^{-5} \text{ m}

    The wavelength is 1.7×10−5 m\mathbf{1.7 \times 10^{-5} \text{ m}}.

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