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Given: Planck's constant h = 6.626 x 10⁻³⁴ J s, mass of an electron m = 9.11 x 10⁻³¹ kg, and velocity v = 5 x 106 m/s.

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3 parts
iii

1.45×10−10 m\mathbf{1.45 \times 10^{-10} \text{ m}}

iv

5.96×10−19 J\mathbf{5.96 \times 10^{-19} \text{ J}}

v

9.91×10−13 m\mathbf{9.91 \times 10^{-13} \text{ m}}

Step-by-step solution

9 steps

Here are the solutions to the remaining questions from the image.

Part iii· 3 steps

Calculate the wavelength of an electron moving at 5×1065 \times 10^6 m/s.

  1. Use the de Broglie wavelength formula, λ=hmv\lambda = \frac{h}{mv}.

    Given: Planck's constant h=6.626×10−34h = 6.626 \times 10^{-34} J s, mass of an electron m=9.11×10−31m = 9.11 \times 10^{-31} kg, and velocity v=5×106v = 5 \times 10^6 m/s.

  2. Substitute the values into the formula.

    λ=6.626×10−34 J s(9.11×10−31 kg)×(5×106 m/s)\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{(9.11 \times 10^{-31} \text{ kg}) \times (5 \times 10^6 \text{ m/s})}

  3. Calculate the wavelength.

    λ=6.626×10−344.555×10−24 m\lambda = \frac{6.626 \times 10^{-34}}{4.555 \times 10^{-24}} \text{ m} λ=1.45466×10−10 m\lambda = 1.45466 \times 10^{-10} \text{ m} λ=1.45×10−10 m\lambda = \boxed{\mathbf{1.45 \times 10^{-10} \text{ m}}}

Part iv

5.96×10−19 J\mathbf{5.96 \times 10^{-19} \text{ J}}

Find the energy of a photon of frequency 9×10149 \times 10^{14} Hz.

  1. Use the formula for the energy of a photon, E=hνE = h\nu.

    Given: Planck's constant h=6.626×10−34h = 6.626 \times 10^{-34} J s, and frequency ν=9×1014\nu = 9 \times 10^{14} Hz.

  2. Substitute the values into the formula.

    E=(6.626×10−34 J s)×(9×1014 Hz)E = (6.626 \times 10^{-34} \text{ J s}) \times (9 \times 10^{14} \text{ Hz})

  3. Calculate the energy.

    E=59.634×10−20 JE = 59.634 \times 10^{-20} \text{ J} E=5.96×10−19 JE = \boxed{\mathbf{5.96 \times 10^{-19} \text{ J}}}

Part v

9.91×10−13 m\mathbf{9.91 \times 10^{-13} \text{ m}}

Determine the wavelength of a proton moving at 4.0×1054.0 \times 10^5 m/s.

  1. Use the de Broglie wavelength formula, λ=hmv\lambda = \frac{h}{mv}.

    Given: Planck's constant h=6.626×10−34h = 6.626 \times 10^{-34} J s, mass of a proton m=1.672×10−27m = 1.672 \times 10^{-27} kg, and velocity v=4.0×105v = 4.0 \times 10^5 m/s.

  2. Substitute the values into the formula.

    λ=6.626×10−34 J s(1.672×10−27 kg)×(4.0×105 m/s)\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{(1.672 \times 10^{-27} \text{ kg}) \times (4.0 \times 10^5 \text{ m/s})}

  3. Calculate the wavelength.

    λ=6.626×10−346.688×10−22 m\lambda = \frac{6.626 \times 10^{-34}}{6.688 \times 10^{-22}} \text{ m} λ=0.99073×10−12 m\lambda = 0.99073 \times 10^{-12} \text{ m} λ=9.91×10−13 m\lambda = \boxed{\mathbf{9.91 \times 10^{-13} \text{ m}}}
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