Question

The wavelength of an electron is \( 10^3 \) nm. What is its momentum in kg m s\(^{-1}\)?

• A. \( 6.625 \times 10^{-31} \)
• B. \( 6.625 \times 10^{-37} \)
• C. \( 6.625 \times 10^{-28} \)
• D. \( 6.625 \times 10^{-34} \)

Hint:

The de Broglie wavelength formula \( p = \frac{h}{\lambda} \) is used to calculate the momentum of a particle. Ensure that the wavelength is converted into meters before substitution.

Answer 1

The Correct Option is C

Step 1: Use de Broglie's Equation 
According to de Broglie's hypothesis, the momentum of a particle is given by: \[ p = \frac{h}{\lambda} \] where: - \( h = 6.625 \times 10^{-34} \) Js (Planck's constant), - \( \lambda = 10^3 \) nm \( = 10^{-6} \) m. 

Step 2: Substitute the Given Values 
\[ p = \frac{6.625 \times 10^{-34}}{10^{-6}} \] \[ = 6.625 \times 10^{-28} \text{ kg m s}^{-1} \]

 Step 3: Verify the Correct Answer 
Thus, the momentum of the electron is \( 6.625 \times 10^{-28} \) kg m s\(^{-1}\), which corresponds to Option (3).