Q. The kinetic energy of an electron is $5\ eV.$ Calculate the de-Broglie wavelength associated with it $(h = 6.6 \times 10^{- 34}Js,\ m_{e} = 9.1 \times 10^{- 31}kg)$
A) $5.47\ \mathring{\mathrm{A}}$
B) $10.9\ \mathring{\mathrm{A}}$
C) $2.7\ \mathring{\mathrm{A}}$
D) None of these
Explanation
(a) $\lambda = \frac{h}{\sqrt{2mE}} = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 5 \times 1.6 \times 10^{- 19}}}$
$= 5.469 \times 10^{- 10}m = 5.47\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.