The wavelength associated with an electron having kinetic energy is given by the expression:

h / \sqrt{2 m E}

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2 h / m E

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2 m h E

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\frac{2 \sqrt{2 m E}}{h}

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Solution

The correct option is A $h / \sqrt{2 m E}$
$\frac{1}{2} m v^{2} = E (g i v e n)$

$m v = \sqrt{2 E m}$

$∴ λ = \frac{h}{m v}$

$= \frac{h}{\sqrt{2 m E}}$
So, the answer is option (A).

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