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If the kinetic energy of a moving particle is $\mathrm{E}$, then the de-Broglie wavelength is
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Verified Answer
The correct answer is:
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \mathrm{E}}}$
$\quad \mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}$ or $\mathrm{mv}=\sqrt{2 \mathrm{~m} \mathrm{E}}$
so $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \mathrm{E}}}$
so $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \mathrm{E}}}$
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