Question:medium

Calculate the wavelength (in nanometer) associated with a proton moving at \(1.0 \times 10^3 \, \text{m/s}\)
Choose the correct answer from the options given below:

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The de Broglie wavelength formula λ = h/mv is essential for understanding the waveparticle duality of matter. It shows that particles with smaller mass or higher velocity have shorter wavelengths.
Updated On: Feb 10, 2026
  • 0.032 nm
  • 0.40 nm
  • 2.5 nm
  • 14.0 nm
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The Correct Option is B

Solution and Explanation

To find the wavelength (\( \lambda \)) of a proton at a specific velocity, we apply the de Broglie hypothesis, linking wave and particle behavior. The de Broglie equation is:

  • \( \lambda \): particle wavelength
  • \( h \): Planck's constant (\( 6.626 \times 10^{-34} \, \text{J·s} \))
  • \( m \): particle mass
  • \( v \): particle velocity

Known Values:

  • Proton mass (\( m \)) = \( 1.673 \times 10^{-27} \, \text{kg} \)
  • Proton velocity (\( v \)) = \( 1.0 \times 10^3 \, \text{m/s} \)
  1. Step 1: Apply de Broglie Equation
    \[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J·s}}{1.673 \times 10^{-27} \, \text{kg} \times 1.0 \times 10^3 \, \text{m/s}} \]
  2. Step 2: Calculate the Product of Mass and Velocity
    \[ m \times v = 1.673 \times 10^{-27} \, \text{kg} \times 1.0 \times 10^3 \, \text{m/s} = 1.673 \times 10^{-24} \, \text{kg·m/s} \]
  3. Step 3: Calculate Wavelength (\( \lambda \))
    \[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J·s}}{1.673 \times 10^{-24} \, \text{kg·m/s}} = 3.957 \times 10^{-10} \, \text{m} \]
  4. Step 4: Convert to Nanometers
    Using \( 1 \, \text{nm} = 1 \times 10^{-9} \, \text{m} \): \[ \lambda = 3.957 \times 10^{-10} \, \text{m} \times \frac{1 \, \text{nm}}{1 \times 10^{-9} \, \text{m}} = 0.3957 \, \text{nm} \]
  5. Step 5: Round for Significant Figures
    With velocity given as \( 1.0 \times 10^3 \, \text{m/s} \) (two significant figures): \[ \lambda \approx 0.40 \, \text{nm} \]

Conclusion:
A proton moving at \( 1.0 \times 10^3 \, \text{m/s} \) has a wavelength of approximately 0.40 nm. Therefore: \[ \boxed{(2) \, 0.40 \, \text{nm}} \]

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