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).