An electron moving with velocity of $5.4 × 10^6 m/sec$. What is the de-Broglie wavelength associated with them? $(h=6.63 × 10^{-34} J/s)$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.35 Å
The De-Broglie wavelength (λ) is -
$λ=\frac{h}{P}=\frac{h}{mv}$
$=\frac{6.626×10^{-34}}{9.11×10^{-31}×5.4×10^6}$
$≃1.35Å$