MCQ
The de$-$Broglie wavelength $\lambda$ associated with an electron having kinetic energy $E$ is given by the expression
  • $\frac{\mathrm{h}}{\sqrt{2} \mathrm{mE}}$
  • B
    $\frac{2 \mathrm{~h}}{\mathrm{mE}}$
  • C
    $2 \ mhE$
  • D
    $\frac{2 \sqrt{2 \mathrm{mE}}}{\mathrm{h}}$

Answer

Correct option: A.
$\frac{\mathrm{h}}{\sqrt{2} \mathrm{mE}}$
$\frac{\mathrm{h}}{\sqrt{2} \mathrm{mE}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free