Q. The de-Broglie wavelength of a proton (charge =1.6$\times 10^{- 19}$ C, mass=$1.6 \times 10^{- 27}$ kg) accelerated through a potential difference of 1kV is
A) $600\ \mathring{\mathrm{A}}$
B) 0.9$\times 10^{- 12}$ m
C) $7\ \mathring{\mathrm{A}}$
D) o.9 nm
Explanation
(b) According to de-Broglie hypothesis
$\lambda = \frac{h}{p}$
= $\frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2mqV}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times \left( 1.6 \times 10^{- 27}\ \right)(1.6 \times 10^{- 19}) \times 1000}}$
$= \frac{6.6 \times 10^{- 34}}{7.16 \times 10^{- 22}}$
=0.9$\times 10^{- 12}$ m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.