Question

The momentum of an electron is \(6.625 \times 10^{-28} \, kg \cdot ms^{-1}\). What is its wavelength in nm? - (Given: \( h = 6.625 \times 10^{-34} \, J \cdot s\))

• A. \(100\)
• B. \(1000\)
• C. \(500\)
• D. \(6625\)

Hint:

The de Broglie wavelength of a particle is given by: \[ \lambda = \frac{h}{p} \] Always ensure units are consistent, and convert meters to nanometers by multiplying by \(10^9\).

Answer 1

The Correct Option is B

Step 1: Using de Broglie Wavelength Formula The de Broglie wavelength is given by: \[ \lambda = \frac{h}{p} \] where: 
- \( h = 6.625 \times 10^{-34} \, J \cdot s \) (Planck’s constant), 
- \( p = 6.625 \times 10^{-28} \, kg \cdot ms^{-1} \) (momentum of the electron). 

Step 2: Substituting the Given Values \[ \lambda = \frac{6.625 \times 10^{-34}}{6.625 \times 10^{-28}} \] \[ \lambda = 10^{-6} \, m \] Since \( 1 \, m = 10^9 \, nm \), we convert: \[ \lambda = 10^{-6} \times 10^9 = 1000 \, nm \] Thus, the correct answer is \( \mathbf{(2)} \ 1000 \).