• # question_answer 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) $\frac{2\sqrt{2mE}}{h}$

[a] $\frac{1}{2}m{{v}^{2}}=E\Rightarrow mv=\sqrt{2mE};\ \ \therefore \ \ \lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$