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Q. The de-Broglie wavelength $\lambda$ associated with an electron having kinetic energy $E$ is given by the expression

Dual Nature of Radiation and Matter

Solution:

$\frac{1}{2} m v^{2}=E$
$\Rightarrow m v=\sqrt{2 m E};$
$\therefore E=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}$