Q. The de-Broglie wavelength $\lambda$ associated with an electron having kinetic energy $E$ is given by the expression
A) $\frac{h}{\sqrt{2mE}}$
B) $\frac{2h}{mE}$
C) $2mhE$
D) $\frac{2\sqrt{2mE}}{h}$
Explanation
(a) $\frac{1}{2}mv^{2} = E \Rightarrow mv = \sqrt{2mE}\ \therefore\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mE}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.