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Recall the de Broglie wavelength formula.

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Answer

0.33 nm\text{0.33 nm}

Step-by-step solution

6 steps

  1. Recall the de Broglie wavelength formula.

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

    λ=hmv\lambda = \frac{h}{mv}

    where:

    • hh is Planck's constant (6.626×10−34 J s6.626 \times 10^{-34} \text{ J s})
    • mm is the mass of the particle
    • vv is the velocity of the particle

  2. Identify the given values and necessary constants.

    • Mass of an electron, me=9.109×10−31 kgm_e = 9.109 \times 10^{-31} \text{ kg}
    • Velocity of the electron, v=2.18×106 m/sv = 2.18 \times 10^6 \text{ m/s}
    • Planck's constant, h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}

  3. Substitute the values into the de Broglie wavelength formula.

    λ=6.626×10−34 J s(9.109×10−31 kg)×(2.18×106 m/s)\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{(9.109 \times 10^{-31} \text{ kg}) \times (2.18 \times 10^6 \text{ m/s})}

  4. Calculate the denominator.

    m×v=(9.109×10−31 kg)×(2.18×106 m/s)m \times v = (9.109 \times 10^{-31} \text{ kg}) \times (2.18 \times 10^6 \text{ m/s}) m×v=(9.109×2.18)×(10−31×106) kg m/sm \times v = (9.109 \times 2.18) \times (10^{-31} \times 10^6) \text{ kg m/s} m×v=19.85762×10−25 kg m/sm \times v = 19.85762 \times 10^{-25} \text{ kg m/s}

  5. Calculate the wavelength.

    λ=6.626×10−34 J s19.85762×10−25 kg m/s\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{19.85762 \times 10^{-25} \text{ kg m/s}} λ=6.62619.85762×10−34−(−25) m\lambda = \frac{6.626}{19.85762} \times 10^{-34 - (-25)} \text{ m} λ=0.33367×10−9 m\lambda = 0.33367 \times 10^{-9} \text{ m}

  6. Convert the wavelength to nanometers (nm).

    Since 1 nm=10−9 m1 \text{ nm} = 10^{-9} \text{ m}:

    λ=0.33367 nm\lambda = 0.33367 \text{ nm}

    Rounding to two significant figures, as suggested by the options:

    λ≈0.33 nm\lambda \approx 0.33 \text{ nm}

    The wavelength will be 0.33 nm\text{0.33 nm}.

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