Q. What is the de-Broglie wavelength (in $\mathring{\mathrm{A}})$ of the $\alpha$-particle accelerated through a potential difference $V$?
A) $\frac{0.287}{\sqrt{V}}$
B) $\frac{12.27}{\sqrt{V}}$
C) $\frac{0.101}{\sqrt{V}}$
D) $\frac{0.22}{\sqrt{V}}$
Explanation
(c) $\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2meV}}$
$= \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times \left( 4 \times 1.66 \times 10^{- 27} \right) \times (2 \times 1.6 \times 10^{- 19}) \times V}}$
$= \frac{0.101}{\sqrt{V}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.