Question

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:

• A. 0.032 nm
• B. 0.40 nm
• C. 2.5 nm
• D. 14.0 nm

Hint:

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.

Answer 1

The Correct Option is B

To determine the wavelength (\( \lambda \)) of a proton moving at a given velocity, we utilize the de Broglie hypothesis, which relates the wave-like and particle-like properties of matter. The de Broglie equation is given by: \[ \lambda = \frac{h}{mv} \] where:

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

Given:

• Mass of proton (\( m \)) = \( 1.673 \times 10^{-27} \, \text{kg} \)
• Velocity of proton (\( v \)) = \( 1.0 \times 10^3 \, \text{m/s} \)

1. Step 1: Substitute the Known Values into the 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 Denominator
   \[ m \times v = 1.673 \times 10^{-27} \, \text{kg} \times 1.0 \times 10^3 \, \text{m/s} = 1.673 \times 10^{-24} \, \text{kg·m/s} \]
3. Step 3: Compute the 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 Meters to Nanometers
   Since \( 1 \, \text{nm} = 1 \times 10^{-9} \, \text{m} \): \[ \lambda = 3.957 \times 10^{-10} \, \text{m} \times \frac{1 \, \text{nm}}{1 \times 10^{-9} \, \text{m}} = 0.3957 \, \text{nm} \]
5. Step 5: Round to Appropriate Significant Figures
   Given the velocity is provided as \( 1.0 \times 10^3 \, \text{m/s} \) (two significant figures), we round the wavelength accordingly: \[ \lambda \approx 0.40 \, \text{nm} \]

Conclusion:
The wavelength associated with a proton moving at \( 1.0 \times 10^3 \, \text{m/s} \) is approximately 0.40 nm. Therefore, the correct option is: \[ \boxed{(2) \, 0.40 \, \text{nm}} \]