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: Concept
Use de Broglie's equation: $\lambda = \frac{h}{mv}$.

Step 2: Meaning
Wavelength $\lambda$ is inversely proportional to momentum ($mv$).

Step 3: Analysis
$\lambda = \frac{6.63\times10^{-34}}{6.63\times10^{-31} \times 1\times10^{6}} = 10^{-9}~m$.

Step 4: Conclusion
$10^{-9}~m = 1.0~nm$. Final Answer: (B)