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If the kinetic energy of a free electron doubles, its de-Broglie wavelength $\lambda$ changes by a factor
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The correct answer is:
$\frac{1}{\sqrt{2}}$
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2(K) m}}$
Thus, when kinetic energy $K$ is doubled, the wavelength is changes by a factor of $\frac{1}{\sqrt{2}}$.
Thus, when kinetic energy $K$ is doubled, the wavelength is changes by a factor of $\frac{1}{\sqrt{2}}$.
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