Question

A sub-atomic particle of mass $6.63\times10^{-31}\ \text{kg}$ is moving with a velocity of $1\times10^{6}\ \text{m s}^{-1}$. What is the de Broglie wavelength (in nm) associated with it ($h=6.63\times10^{-34}\ \text{J s}$)?

• A. 10.0
• B. 1.0
• C. 0.10
• D. 5.0
• E. 0.50

Hint:

$1~nm = 10^{-9}~m$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem applies the de Broglie hypothesis, which states that all matter has wave-like properties. The wavelength of a particle is inversely proportional to its momentum.
Step 2: Key Formula or Approach:
The de Broglie wavelength (\( \lambda \)) is given by the formula: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] where h is Planck's constant, m is the mass of the particle, and v is its velocity. We need to calculate \( \lambda \) and then convert it to nanometers (nm). \[ 1 \text{ nm} = 10^{-9} \text{ m} \] Step 3: Detailed Explanation:
We are given: - Mass, \( m = 6.63 \times 10^{-31} \text{ kg} \) - Velocity, \( v = 1 \times 10^6 \text{ m/s} \) - Planck's constant, \( h = 6.63 \times 10^{-34} \text{ J s} \) Substitute these values into the de Broglie formula: \[ \lambda = \frac{6.63 \times 10^{-34}}{(6.63 \times 10^{-31}) \times (1 \times 10^6)} \] The \( 6.63 \) terms cancel out: \[ \lambda = \frac{10^{-34}}{10^{-31} \times 10^6} \] Simplify the exponents in the denominator: \[ \lambda = \frac{10^{-34}}{10^{-31+6}} = \frac{10^{-34}}{10^{-25}} \] Simplify the fraction: \[ \lambda = 10^{-34 - (-25)} = 10^{-34 + 25} = 10^{-9} \text{ m} \] The question asks for the wavelength in nanometers (nm). Since \( 1 \text{ nm} = 10^{-9} \text{ m} \), our result is: \[ \lambda = 1.0 \text{ nm} \] Step 4: Final Answer:
The de Broglie wavelength is 1.0 nm.