Question
The de-Broglie wavelength $\lambda $ associated with an electron having kinetic energy $E$ is given by the expression
✓
Answer
(a) $\frac{1}{2}m{v^2} = E $
$\Rightarrow mv = \sqrt {2mE} ;\;\;\therefore \;\;\lambda = \frac{h}{{mv}} = \frac{h}{{\sqrt {2mE} }}$
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