Q. A particle moving with a velocity $10^{6}$ m/s will have de-Broglie wavelength nearly [Given, $m = 6.62 \times 10^{- 27}kg,\ h = 6.62 \times 10^{- 34}J - s\rbrack\$
A) $10^{- 9}m$
B) $10^{- 13}m$
C) $10^{- 19}m$
D) 1 Å
Explanation
(b)
de-Broglie wavelength, $\lambda = \frac{h}{p} = \frac{h}{mv}$
$\ \$($\because$momentum $p = mv$)
$\Rightarrow \ \lambda = \frac{6.62 \times 10^{- 34}J - s}{6.62 \times 10^{- 27} \times 10^{6}kg\ m/s}$
$\ \ \ \ \ \ \ \ \ = 10^{- 13}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.